55315748-14

MasteringPhysics: Assignment Print View

Projectile Motion Tutorial

Learning Goal: Understand how to apply the equations for 1-dimensional motion to the y and x directions separately in order to derive standard formulae for the range and height of a projectile.

A projectile is fired from ground level at time , at an angle with respect to the horizontal. It has an initial speed . In this problem we are assuming that the ground is level.

Part AFind the time it takes the projectile to reach its maximum height.

Hint A.1 A basic property of projectile motionHint not displayed

Hint A.2 What condition applies at the top?Hint not displayed

Hint A.3 Vertical velocity as a function of timeHint not displayed

Hint A.4 Putting it all togetherHint not displayed

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=1448953 (1 of 97) [12/13/2010 7:01:26 PM]


Hint A.5 A list of possible answersHint not displayed

Express in terms of , , and (the magnitude of the acceleration due to gravity).

ANSWER:

=

Correct

Part BFind , the time at which the projectile hits the ground.

Hint B.1 Two possible approachesHint not displayed

Hint B.2 Some needed kinematicsHint not displayed

Hint B.3 Solving for

Hint not displayed

Express the time in terms of , , and .

ANSWER:

=

Correct

Part C



Find , the maximum height attained by the projectile.

Hint C.1 Equation of motionHint not displayed

Hint C.2 When is the projectile at the top of its trajectory?Hint not displayed

Hint C.3 Finding

Hint not displayed

Express the maximum height in terms of , , and .

ANSWER:

=

Correct

Part D

Find the total distance (often called the range) traveled in the x direction; in other words, find where the projectile lands.

Hint D.1 When does the projectile hit the ground?Hint not displayed

Hint D.2 Where is the projectile as a function of time?Hint not displayed

Hint D.3 Finding the rangeHint not displayed

Hint D.4 A list of possible answersHint not displayed

Express the range in terms of , , and .



ANSWER:

=

Correct

The actual formula for is less important than how it is obtained:1. Consider the x and y motion separately. 2. Find the time of flight from the y-motion 3. Find the x-position at the end of the flight - this is the range.

If you remember these steps, you can deal with many variants of the basic problem, such as: a cannon on a hill that fires horizontally (i.e. the second half of the trajectory), a projectile that lands on a hill, or a projectile that must hit a moving target.

Circular Launch

A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free fall, the ball has a centripetal acceleration of magnitude 2 .

Part A



How far from the bottom of the chute does the ball land?

Hint A.1 Speed of ball upon leaving chuteHow fast is the ball moving at the top of the chute?

Hint A.1.1 Equation of motion

The centripetal acceleration for a particle moving in a circle is , where is its speed and is its instantaneous radius of rotation.

ANSWER:

=

Answer Requested

Hint A.2 Time of free fallHint not displayed

Hint A.3 Finding the horizontal distanceHint not displayed

Your answer for the distance the ball travels from the end of the chute should contain .

ANSWER: =

Correct

Projectile Motion Ranking Task



Six baseball throws are shown below. In each case, the ball is thrown with speed at an angle from the horizontal. In all cases, the baseball is thrown from the same height above the ground. Assume for the basis of these rankings that the effects of air resistance are negligible.

Part ARank these throws based on the maximum height reached by the ball.

Hint A.1 Solving two-dimensional motion problemsA key insight in solving two-dimensional motion problems is the realization that motion in the horizontal direction and motion in the vertical direction are independent. This means that the position, velocity, and acceleration in one direction do not influence the position, velocity, and acceleration in the other direction.

Hint A.2 Finding vector components

Given a vector magnitude and angle , the x and y components of the vector can be determined using the equations and , as shown in the figure.

Rank from largest to smallest. To rank items as equivalent, overlap them.



ANSWER:

View Correct

Part BRank these throws based on the amount of time it takes the ball to hit the ground.

Hint B.1 How to approach the problemHint not displayed

Hint B.2 Compare times to reach the maximum heightHint not displayed

Hint B.3 Compare times to fall from the maximum heightHint not displayed


ANSWER:

View Correct

Shooting over a Hill



A projectile is fired with speed at an angle from the horizontal as shown in the figure .

Part A

Find the highest point in the trajectory, .

Hint A.1 Velocity at the topHint not displayed

Hint A.2 Which equation to useHint not displayed

Express the highest point in terms of the magnitude of the acceleration due to gravity , the initial velocity , and the angle .

ANSWER:

=

Correct

Part B



What is the range of the projectile, ?

Hint B.1 Find the total time spent in airHint not displayed

Hint B.2 Find

Hint not displayed

Express the range in terms of , , and .

ANSWER:

=

Correct

Consider your advice to an artillery officer who has the following problem. From his current postition, he must shoot over a hill of height at a target on the other side, which has the same elevation as his gun. He knows from his accurate map both the bearing and the distance to the target and also that the hill is halfway to the target. To shoot as accurately as possible, he wants the projectile to just barely pass above the hill.

Part C

Find the angle above the horizontal at which the projectile should be fired.

Hint C.1 How to approach the problemHint not displayed

Hint C.2 Set up the ratioHint not displayed

Express your answer in terms of and .

ANSWER: =

Correct

Recall the following trigonometry formulas:



,

,

and

.

In this case, since , you can

draw a right triangle with as one of the angles, an "opposite" side of length , and an "adjacent" side of length . You can then use

this triangle to find and

, after you find the length of the hypotenuse using the Pythagorean Theorem.

Part DWhat is the initial speed?

Hint D.1 How to approach this partHint not displayed

Hint D.2 Find

Hint not displayed

Hint D.3 Find

Hint not displayed

Express in terms of , , and .



ANSWER:

=

Correct

Part E

Find , the flight time of the projectile.

Hint E.1 How to proceedHint not displayed

Express the flight time in terms of and .

ANSWER:

=

Correct

Speed of a Softball

A softball is hit over a third baseman's head with speed and at an angle from the horizontal. Immediately after the ball is hit, the third baseman turns around and runs straight back at a constant velocity , for a time

. He then catches the ball at the same height at which it left the bat. The third baseman was initially from the location where the ball was hit at home plate.

Part A



Find . Use for the magnitude of the acceleration due to gravity.

Hint A.1 Find the initial velocity in the x directionHint not displayed

Hint A.2 Find the initial velocity in the y directionHint not displayed

Hint A.3 Find the total initial velocityHint not displayed

Express the initial speed in units of meters per second to four significant figures.

ANSWER: = 18.77

Correct

Part B

Find the angle in degrees.

Express your answer in degrees to four significant figures.

ANSWER: = 31.51

Correct

Part C



Find a vector expression for the velocity of the softball 0.1 s before the ball is caught.

Hint C.1 vs

Hint not displayed

Hint C.2 Find as a function of timeHint not displayed

Hint C.3 Unit vectorsHint not displayed

Use the notation , , an ordered pair of values separated by commas. Express your answer in units of meters per second to three significant figures.

ANSWER: = 16.0,-8.82

Correct

Part D

Find a vector expression for the position of the softball 0.1 s before the ball is caught.

Hint D.1 Equations of motionHint not displayed

Use the notation , , an ordered pair of values separated by commas, where and are expressed in meters, as measured from the point where the softball initially left the bat. Express your answer to three significant figures.

ANSWER: = 30.4,0.932

Correct

The Archerfish



The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers. When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.

Part AAt what speed should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle above the water surface.

Hint A.1 How to approach the problemHint not displayed

Hint A.2 Find how long it takes the water drop to fall back into the pond

Hint not displayed

Hint A.3 Find how far from the fish the drop fallsHint not displayed

Express your answer in meters per second.

ANSWER: = 3.01

Correct

Some archerfish can "shoot" as far as 3.5 , hitting their targets with reasonable accuracy as far as 1.2 . They have binocular vision, which helps them judge distance.

Part B



Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle above the surface and with the same initial speed as before. At what height above the surface was the insect?


Hint B.2 Find the time it takes the water drop to hit the insectHint not displayed

Express your answer in meters.

ANSWER: = 0.260

Correct

Experiments have shown that the archerfish can predict the point where the disabled prey will fall and hit the water by simply looking at the initial trajectory of the dislodged insect for only 10 . The archerfish then darts to the place where it has "calculated" the insect will hit the water, planning to get there before another fish does.

Horizontal Cannon on a Cliff

A cannonball is fired horizontally from the top of a cliff. The cannon is at height = 90.0 above ground level, and the ball is fired with initial horizontal speed . Assume acceleration due to gravity to be

= 9.80 .



Part A

Assume that the cannon is fired at time and that the cannonball hits the ground at

time . What is the y position of the cannonball at the time ?


Hint A.2 Identify the knowns and unknownsHint not displayed

Hint A.3 Determine which equation to use to find the height at the requested time

Hint not displayed

Hint A.4 Find

Hint not displayed

Answer numerically in units of meters.

ANSWER: = 67.5

Correct

The same answer can be obtained more easily (perhaps you did it this way) if you notice that . This means that the vertical displacement is given by

and therefore is one-quarter of ; then .

Part B



Given that the projectile lands at a distance = 200 from the cliff, as shown in the figure, find the initial speed of the projectile, .


Hint B.2 Knowns and unknownsHint not displayed

Hint B.3 The equation to useHint not displayed

Express the initial speed numerically in meters per second.

ANSWER: = 46.7

Correct

Part C

What is the y position of the cannonball when it is at distance from the hill? If you need to, you can use the trajectory equation for this projectile, which gives in terms of directly:

.

You should already know from the previous part.

Express the position of the cannonball numerically in meters.

ANSWER: = 67.5

Correct

Not surprisingly, the answer to this part is the same as that in Part A because a projectile travels equal horizontal distances in equal amounts of time.

Newton's 1st and 2nd Laws



Learning Goal: To understand the meaning and the basic applications of Newton's 1st and 2nd laws.

Newton's laws are fundamental in mechanics. Their mathematical expressions are very simple but conceptual understanding of Newton's laws, which is necessary for solving nontrivial problems, is not simple at all.

Newton's 1st law The common textbook statement of Newton's 1st law may seem rather straightforward. Here it is:An object has a constant velocity (possibly zero) if and only if the net force acting on the object is zero. In other words, if the vector sum of the forces applied to the object is zero, the object would be either at rest or at constant velocity (that is, the object would have zero acceleration). If such a sum is not zero, the object cannot possibly be moving at a constant velocity.

Frames of reference The statement of Newton's 1st law becomes a bit more complicated in actual applications. Imagine yourself in a car. To understand Newton's 1st law fully, we need the concept of a frame of reference. A frame of reference is a set of coordinates used to measure distances and times. In your frame of reference, any distance would be measured relative to you. For example, the radio in the car is 0.75 m to the right of you. The radio is at rest in your frame of reference, because the radio doesn't change its distance or direction from you.In your frame of reference, the car is always at rest. It is entirely possible that the net force acting on the car is not zero: The car may (in the frame of reference of an observer standing on the ground) be accelerating, turning, or braking. Yet in your frame of reference, the car would remain at rest because, relative to you, it is not moving at all. So, the car is at rest or accelerating, depending upon who you ask.

Inertial frames of reference It's tempting to ignore this difficulty by saying that the frame of reference attached to the car is somehow wrong. The observer on the ground, in contrast, is right: The observer sees the motion of the car as it really is. However, such a line of reasoning seems flawed, because it raises the question of how to determine which frames of reference are "right" and which ones are "wrong."This is what Newton's 1st law settles. Newton established the concept of an inertial frame of reference. An inertial frame of reference, by definiton, is one in which the statement of Newton's 1st law is, in fact, true.

Newton's 2nd law It is important to know that the frame of reference being used is, in fact, inertial. Only then does Newton's 2nd law work in a simple and elegant form. Newton's 2nd law establishes the relationship between the net force acting on an object, the mass of the object, and its acceleration:

,or



.

Note that Newon's 2nd law allows one to find the magnitude of the object's acceleration. It also establishes the fact that the acceleration of an object has the same direction as the net force acting on the object.

Applying Newton's laws in inertial and noninertial frames If the frame of reference is not inertial, using Newton's 2nd law to calculate acceleration is still possible but may be far more complicated. Objects that experience zero net force may accelerate, and objects that move at constant velocity may experience a net force not equal to zero. The important question is: Which frames of reference are inertial and which ones are not? This also raises the following question: Are there any inertial frames of reference in this universe?Newton postulated that inertial frames of reference do exist. This statement, coupled with the definition of inertial frames of reference, may be considered a more proper way to state Newton's 1st law.Only an experiment can establish whether a particular frame of reference is inertial (or, to be precise, "inertial enough" for the purposes needed). Let us go back to the car example. The frame of reference attached to the ground, we would usually say, is inertial. That is, if we get an object and make sure that all external forces acting on it add up to zero, we can then observe that the object is, in fact, moving at constant velocity (or, possibly, remaining at rest). In most problems that we will be solving, the frame of reference of the earth will be considered an inertial frame of reference. For all practical purposes, this means that Newton's 2nd law will work in it.However, it is instructive to understand that the earth provides a reference frame that is less than "perfectly inertial." An observer on the sun, for instance, would notice that the object in question does, in fact, have an acceleration: the centripetal acceleration associated with the orbital motion of the earth around the sun! The best inertial frame of reference is the one assoicated with distant stars and any other frame of reference that is moving at a constant velocity relative to distant stars.The conceptual questions that follow should help you learn to apply Newton's 1st and 2nd laws properly. Note that, throughout this problem, we will assume that the frame of reference associated with the earth is perfectly inertial.

Part AWhich object provides an inertial frame of reference?

ANSWER: the tip of the moving second hand of a clock a rock thrown vertically upward a pendulum swinging with no air resistance a skydiver falling at terminal velocity

Correct

Assuming that the earth provides an inertial frame of reference, an object moving at a constant velocity relative to the earth would also provide an inertial frame of reference.



Part BYou are conducting an experiment inside an elevator that can move in a vertical shaft. A load is hung vertically from the ceiling on a string, and is stationary with respect to you. The tension in the string is measured to be 10% less than the weight of the load. No other forces are acting on the load. Which of the following statements about the elevator are correct?

Check all that apply.

ANSWER: The elevator is an inertial frame of reference.The elevator is not an inertial frame of reference.The elevator may be at rest for the duration of the entire experiment.The elevator may be moving at a constant velocity upward.The elevator may be moving at a constant velocity downward.The elevator must be accelerating.

Correct

Part CYou are conducting an experiment inside an elevator that can move in a vertical shaft. A load is hung vertically from the ceiling on a string. The tension in the string is measured to be exactly equal to the weight of the load. No other forces are acting on the load. Which of the following statements about the elevator are correct?


ANSWER: The elevator is an inertial frame of reference.The elevator is not an inertial frame of reference.The elevator may be at rest.The elevator may be moving at a constant velocity upward.The elevator may be moving at a constant velocity downward.The elevator may be accelerating.The elevator must be accelerating.

Correct

Part D



You are conducting an experiment inside a train car that may move horizontally along rail tracks. A load is hung from the ceiling on a string. The load is not swinging, and the string is observed to make a constant angle of with the horizontal. No other forces are acting on the load. Which of the following statements are correct?


ANSWER: The train is an inertial frame of reference.The train is not an inertial frame of reference.The train may be at rest.The train may be moving at a constant speed in a straight line.The train may be moving at a constant speed in a circle.The train must be speeding up.The train must be slowing down.The train must be accelerating.

Correct

Since the tension and the weight are not directed opposite to each other, the net force cannot possibly be zero--and yet the load is at rest relative to the train car. Therefore, the car is not an inertial frame of reference. It must be accelerating relative to the earth, although it is not clear exactly how.

Part EConsider the train car described in the previous part. Another experiment is conducted in it: A net force of is applied to an object of mass . Can you determine the acceleration of the object with respect to the train, and, if so, what is its value?

ANSWER:Yes; .

Yes; .

Yes; .

No; there is not enough information. Correct

The train car is not an inertial frame of reference, so would not work here.



Part F

A 1000-kg car is moving along a straight road down a slope at a constant speed of

. What is the net force acting on the car?

ANSWER:

Correct

The car has zero acceleration; therefore, it experiences zero net force. According to Newton's 1st law, no net force is required to maintain a constant velocity (in an inertial frame of reference, of course). The car has a constant veclocity relative to the earth; therefore, the car is also an inertial frame of reference.

Part GConsider two cars moving along the same straight road in opposite directions. Car A has a mass of and has a constant speed of ; car B has a mass of and a

constant speed of . Whar can you say about the net forces on the cars?

ANSWER: Car A experiences greater net force than car B. Car B experiences greater net force than car A. Both cars experience equal net forces.

Correct

Each car has zero acceleration; therefore, the net force on each car, according to Newton's 1st law, is zero.

Part H



In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly remain at rest?The forces applied are as follows:

Hint H.1 Using the net forceHint not displayed


ANSWER: 2 N; 2 N200 N; 200 N200 N; 201 N2 N; 2 N; 4 N2 N; 2 N; 2 N2 N; 2 N; 3 N2 N; 2 N; 5 N200 N; 200 N; 5 N

Correct

Part IIn an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly move at a constant velocity of ?The forces applied are as follows:

Hint I.1 Using the net forceHint not displayed




ANSWER: 2 N; 2 N200 N; 200 N200 N; 201 N2 N; 2 N; 4 N2 N; 2 N; 2 N2 N; 2 N; 3 N2 N; 2 N; 5 N200 N; 200 N; 5 N

Correct

You should have noticed that the sets of forces applied to the object are the same as the ones in the prevous question. Newton's 1st law (and the 2nd law, too) makes no distinction between the state of rest and the state of moving at a constant velocity (even a high velocity). In both cases, the net force applied to the object must equal zero.Although some of the questions in this problem may have seemed tricky and unfair, the subtleties here are important in improving conceptual understanding. That understanding, in turn, will enable you to correctly solve complex computational problems using Newton's laws.

Free-Body Diagrams and Newton's Laws

When solving problems involving forces and Newton’s laws, the following summary of things to do will start your mind thinking about getting involved in the problem at hand.

Problem Solving: Free-Body Diagrams and Newton's Laws

1. Draw a sketch of the situation. 2. Consider only one object (at a time), and draw a free-body diagram for that body, showing all the forces acting on that body. Do not show any forces that the body exerts on other bodies. If several bodies are involved, draw a free-body diagram for each body separately, showing all the forces acting on that body.

3. Newton's second law involves vectors, and it is usually important to resolve vectors into components. Choose an x and y axis in a way that simplifies the calculation.

4. For each body, Newton's second law can be applied to the x and y components separately. That is the x component of the net force on that body will be related to the x component of that body's acceleration: , and similarly for the y direction.

5. Solve the equation or equations for the unknown(s).

Apply these steps Use the steps outlined above to find the magnitude of the acceleration of a chair and the magnitude of the normal force acting on the chair: Yusef pushes a chair of mass



= 55.0 across a carpeted floor with a force of magnitude = 152 directed

at = 35.0 below the horizontal . The magnitude of the frictional force between the carpet and the floor is

= 106 .

Part AIdentify and sketch all the external forces acting on the chair. Because the chair can be represented as a point particle of mass , draw the forces with their tails centered on the black dot in the middle of the chair. Be certain to draw your forces so that they have the correct orientation.

Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The length of the vectors will not be graded.

ANSWER:

View All attempts used; correct answer displayed

Part B



Which set of coordinate axes is the most convenient to use in this problem?

Hint B.1 Determine the direction of the accelerationHint not displayed

ANSWER:

Correct

Now that you have selected a coordinate system, you should resolve the forces into x and y components so that you can apply Newton's second law to each coordinate direction independently.

Part C



Use the component form of Newton's second law to write an expression for the x component of the net force, .

Hint C.1 Find the x component of the pushing forceHint not displayed

Express your answer in terms of some or all of the variables: , , , ,

and .

ANSWER: =

Correct

Part DUse the component form of Newton's second law to write an expression for the y component of the net force, .

Hint D.1 Find the y component of the pushing forceHint not displayed

Express your answer in terms of some or all of the variables: , , , ,

and .

ANSWER: =

Correct

You have created two equations that describe the motion of the chair:

and

Solve these equations to find and .

Part E



What is the magnitude of the acceleration of the chair? What is the magnitude of the normal force acting on the chair?

Hint E.1 Find the component of the acceleration in the y directionHint not displayed

Hint E.2 Find the weight of the chairHint not displayed

Express your answers, separated by a comma, in meters per second squared and newtons to three significant figures.

ANSWER: , = 0.328,626

Correct ,

A free-body diagram is a useful way to begin all problems involving forces. This drawing will help you to easily identify the most appropriate coordinate axes and to resolve any 2 dimensional vectors into components. Then you can apply Newton's second law to each coordinate direction to set up equations which will allow you to solve for any unknown quantities.

Binary Star System

A binary star system consists of two stars of masses and . The stars, which gravitationally attract each other, revolve around the center of mass of the system. The star with mass has a centripetal acceleration of magnitude . Note that you do not need to understand universal gravitation to solve this problem.

Part AFind , the magnitude of the centripetal acceleration of the star with mass .

Hint A.1 What causes acceleration?Hint not displayed

Hint A.2 Relationship between forces on each starHint not displayed

Hint A.3 How to relate acceleration and forceHint not displayed



Hint A.4 Putting it togetherHint not displayed

Express the acceleration in terms of quantities given in the problem introduction.

ANSWER:

=

Correct

To make sure you understand this result, consider the actual gravitational force acting on each star. The magnitude of the gravitational force on either star due to the other one is given by

,

where is the separation between the stars. Now, consider Newton's 2nd law for the star of mass . The net external force is

, so .Now consider Newton's 2nd law applied to the star of mass . Once again, the net external force acting on this star will be , so Newton's 2nd law for the star of mass

is .You can see that the same force, , appears in both the equation for the star of mass and that for the star of mass . (Think about how this relates to Newton's 3rd law.) You can therefore write . Solving for the acceleration

you find the equation . Note that you did not need to know the exact

form of the gravitational force, nor did you need to know or . Newton's 3rd law allows you to realize that is the same for the two stars, and Newton's 2nd law allows you to solve for in terms of , , and .

Blocks in an Elevator Ranking Task



Three blocks are stacked on top of each other inside an elevator as shown in the figure. Answer the following questions with reference to the eight forces defined as follows. ● the force of the 3 block on the

2 block, ,

● the force of the 2 block on the

3 block, ,


1 block, ,


3 block, ,


1 block, ,

● the force of the 1 block on the 2 block, ,

● the force of the 1 block on the floor, , and

● the force of the floor on the 1 block, .

Part AAssume the elevator is at rest. Rank the magnitude of the forces.

Hint A.1 Newton's 3rd lawHint not displayed

Hint A.2 Contact forcesHint not displayed


ANSWER:

View Correct



Part BNow, assume the elevator is moving upward at increasing speed. Rank the magnitude of the forces.

Hint B.1 Effects of accelerationHint not displayed


ANSWER:

View Correct

Velocity from Force Diagram Ranking Task

Below are birds-eye views of six identical toy cars moving to the right at 2 . Various forces act on the cars with magnitudes and directions indicated below. All forces act in the horizontal plane and are either parallel or at 45 or 90 degrees to the car's motion.

Part ARank these cars on the basis of their speed a short time after the forces are applied.


Hint A.2 Summing force vectorsHint not displayed




ANSWER:

View Correct

Pushing a Chair along the Floor

A chair of weight 150 lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of = 35.0 directed at an angle of 42.0 below the horizontal and the chair slides along the floor.

Part AUsing Newton's laws, calculate , the magnitude of the normal force that the floor exerts on the chair.


Hint A.2 Choosing the correct free-body diagramHint not displayed

Hint A.3 Find the vertical net forceHint not displayed

Hint A.4 Find the vertical component of the force that you exert on the chair

Hint not displayed

Express your answer in newtons.

ANSWER: = 173

Correct



Centripetal Acceleration Explained

Learning Goal: To understand that centripetal acceleration is the acceleration that causes motion in a circle.

Acceleration is the time derivative of velocity. Because velocity is a vector, it can change in two ways: the length (magnitude) can change and/or the direction can change. The latter type of change has a special name, the centripetal acceleration. In this problem we consider a mass moving in a circle of radius with angular velocity ,

.The main point of the problem is to compute the acceleration using geometric arguments.

Part A

What is the velocity of the mass at a time ? You can work this out geometrically with the help of the hints, or by differentiating the expression for given in the introduction.



Hint A.1 Direction of the velocityHint not displayed

Hint A.2 SpeedHint not displayed

Express this velocity in terms of , , , and the unit vectors and .

ANSWER: =

Correct

Assume that the mass has been moving along its circular path for some time. You start timing its motion with a stopwatch when it crosses the positive x axis, an instant that corresponds to . [Notice that when , .] For the remainder of this problem, assume that the time is measured from the moment you start timing the motion. Then the time refers to the moment a time before you start your stopwatch.

Part B



What is the velocity of the mass at a time ?

Express this velocity in terms of , , , and the unit vectors and .

ANSWER: =

Correct

Part C

What is the average acceleration of the mass during the time interval from to ?

Hint C.1 Average accelerationHint not displayed

Express this acceleration in terms of , , , and the unit vectors and .

ANSWER: =

Correct

Part D



What is the magnitude of this acceleration in the limit of small ? In this limit, the average acceleration becomes the instantaneous acceleration.

Hint D.1 Expansion of

For small times (or more precisely when ), what is the first term in the Taylor

series expansion for ?

Hint D.1.1 Taylor series expansionHint not displayed


ANSWER: = Answer not displayed


ANSWER: =

Correct

Part EConsider the following statements: a. The centripetal acceleration might better be expressed as because it is a vector. b. The magnitude of the centripetal acceleration is .

c. The magnitude of the centripetal acceleration is .

d. A particle that is going along a path with local radius of curvature at speed

experiences a centripetal acceleration . e. If you are in a car turning left, the force you feel pushing you to the right is the force that causes the centripetal acceleration. In these statements refers to the component of the velocity of an object in the direction toward or away from the origin of the coordinate system or the rotation axis. Conversely, refers to the component of the velocity perpendicular to .

Identify the statement or statements that are false.



ANSWER: a onlyb onlyc onlyd onlye onlyb and ec and ed and e

Correct

That's right; the true statements are therefore: ● a. The centripetal acceleration might better be expressed as because it is a

vector. ● c. The magnitude of the centripetal acceleration is .

● d. A particle that is going along a path with local radius of curvature at speed

experiences an acceleration . There is so much confusion about centripetal force that you should probably ban this term from your vocabulary and thought processes. If you are in a car turning left, your centripetal acceleration is to the left (i.e., inward) and some real force must be applied to you to give you this acceleration--typically this would be provided by friction with the seat. The force you "feel" pushing you to the right is not a real force but rather a "fictitious force" that is present if you are in an accelerating coordinate system (in this case the car). It is best to stick to inertial (i.e., nonaccelerating) coordinate systems when doing kinematics and dynamics (i.e., calculations).

Pushing a Block

Learning Goal: To understand kinetic and static friction.A block of mass lies on a horizontal table. The coefficient of static friction between the block and the table is . The coefficient of kinetic friction is , with .

Part A



If the block is at rest (and the only forces acting on the block are the force due to gravity and the normal force from the table), what is the magnitude of the force due to friction?

Hint A.1 Consider the type of friction at restHint not displayed

ANSWER: = 0 Correct

Part BSuppose you want to move the block, but you want to push it with the least force possible to get it moving. With what force must you be pushing the block just before the block begins to move?

Hint B.1 Consider the type of friction to start movementHint not displayed

Express the magnitude of in terms of some or all the variables , , and , as well as the acceleration due to gravity .

ANSWER: =

Correct

Part CSuppose you push horizontally with half the force needed to just make the block move. What is the magnitude of the friction force?

Hint C.1 What level of force is required?Hint not displayed

Express your answer in terms of some or all of the variables , , and , as well as the acceleration due to gravity .

ANSWER:

=

Correct



Part DSuppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration of the block after it begins to move.

Hint D.1 Calculate applied forceHint not displayed

Hint D.2 Consider applied force and kinetic frictionHint not displayed

Hint D.3 Calculate net horizontal forceHint not displayed


ANSWER: =

Correct

At the Test Track

You want to test the grip of the tires on your new race car. You decide to take the race car to a small test track to experimentally determine the coefficient of friction. The racetrack consists of a flat, circular road with a radius of 45 . The applet shows the result of driving the car around the track at various speeds.

Part AWhat is , the coefficient of static friction between the tires and the track?


Hint A.2 Use the applet to find the speedHint not displayed

Hint A.3 Find an expression for Hint not displayed



Express your answer to two significant figures.

ANSWER: = 0.91

Correct

A Ride on the Ferris Wheel

A woman rides on a Ferris wheel of radius 16 that maintains the same speed throughout its motion. To better understand physics, she takes along a digital bathroom scale (with memory) and sits on it. When she gets off the ride, she uploads the scale readings to a computer and creates a graph of scale reading versus time. Note that the graph has a minimum value of 510 and a maximum value of 666 .

Part AWhat is the woman's mass?


Hint A.2 Find the extreme points on the circular pathHint not displayed

Hint A.3 Find the acceleration of the womanHint not displayed

Hint A.4 Mass and weight



Hint not displayed

Express your answer in kilograms.

ANSWER: = 60

Correct

Block on an Incline Adjacent to a Wall

A wedge with an inclination of angle rests next to a wall. A block of mass is sliding down the plane, as shown. There is no friction between the wedge and the block or between the wedge and the horizontal surface.

Part A

Find the magnitude, , of the sum of all forces acting on the block.

Hint A.1 Direction of the net force on the blockHint not displayed

Hint A.2 Determine the forces acting on the blockHint not displayed

Hint A.3 Find the magnitude of the force acting along the direction of motion



Hint not displayed

Express in terms of and , along with any necessary constants.

ANSWER: =

Correct

Part B

Find the magnitude, , of the force that the wall exerts on the wedge.

Hint B.1 The force between the wall and the wedgeHint not displayed

Hint B.2 Find the normal force between the block and the wedgeHint not displayed

Hint B.3 Find the horizontal component of the normal forceHint not displayed

Express in terms of and , along with any necessary constants.

ANSWER: =

Correct

Your answer to Part B could be expressed as either or . In either form, we see that as gets very

small or as approaches 90 degrees ( radians), the contact force between the

wall and the wedge goes to zero. This is what we should expect; in the first limit ( small), the block is accelerating very slowly, and all horizontal forces are small. In the second limit ( about 90 degrees), the block simply falls vertically and exerts no horizontal force on the wedge.

Board Pulled Out from under a Box



A small box of mass is sitting on a board of mass and length . The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box is . The coefficient of kinetic friction between the board and the box is, as usual, less than .Throughout the problem, use for the magnitude of the acceleration due to gravity. In the hints, use for the magnitude of the friction force between the board and the box.

Part A

Find , the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board).

Hint A.1 Condition for the board sliding out from under the boxHint not displayed

Hint A.2 Find the acceleration of the box in terms of

Hint not displayed

Hint A.3 Find the largest acceleration of the boxHint not displayed

Hint A.4 Find the sum of horizontal forces on the boardHint not displayed

Hint A.5 Find the acceleration of the board for large

Hint not displayed




Express your answer in terms of some or all of the variables , , , , and . Do not include in your answer.

ANSWER: =

Correct

Conical Pendulum I

A bob of mass is suspended from a fixed point with a massless string of length (i.e., it is a pendulum). You are to investigate the motion in which the string moves in a cone with half-angle .

Part AWhat tangential speed, , must the bob have so that it moves in a horizontal circle with the string always making an angle from the vertical?

Hint A.1 What's happening here?



In this situation, which of the following statements is true?

ANSWER: The bob has no acceleration since its velocity is constant. The tension in the string is less than . A component of the tension causes acceleration of the bob. If the tension in the string would be greater than

. Correct

Hint A.2 Find the vertical acceleration of the bobWhat is , the vertical component of the acceleration of the bob?

ANSWER: = 0

Correct

Hint A.3 Find the tension in the string

Find the magnitude, , of the tension force in the string.

Hint A.3.1 What approach to useHint not displayed


ANSWER: =

Correct

Hint A.4 Find the horizontal acceleration of the bob



Find a general expression for , the magnitude of the bob's centripetal acceleration, as a function of the tangential speed of the bob.

Hint A.4.1 Find the radius of the bob's motionHint not displayed

Express your answer in terms of and some or all of the variables , , and .

ANSWER:

=

Correct

Hint A.5 Find the horizontal force

Find the magnitude, , of the inward radial force on the bob in the horizontal plane.


ANSWER: =

Correct


ANSWER:

=

Correct

Part B



How long does it take the bob to make one full revolution (one complete trip around the circle)?

Hint B.1 How to approach the problemSince the speed of the bob is constant, this is a relatively simple kinematics problem. You know the speed, which you found in the previous part, and you can calculate the distance traveled in one revolution (i.e., the circumference of the circle). From these two you can calculate the time required to travel that distance.


ANSWER:

Correct

Hanging Chandelier

A chandelier with mass is attached to the ceiling of a large concert hall by two cables. Because the ceiling is covered with intricate architectural decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung the chandelier couldn't attach the cables to the ceiling directly above the chandelier. Instead, they attached the cables to the ceiling near the walls. Cable 1 has tension

and makes an angle of with

the ceiling. Cable 2 has tension

and makes an angle of with the ceiling.

Part A



Find an expression for , the tension in cable 1, that does not depend on .

Hint A.1 Find the sum of forces in the x directionHint not displayed

Hint A.2 Find the sum of forces in the y directionHint not displayed


Express your answer in terms of some or all of the variables , , and , as well as the magnitude of the acceleration due to gravity .

ANSWER:

=

Correct

Kinetic Friction in a Block-and-Pulley System

Consider the system shown in the figure . Block A has weight and block B has weight . Once block B is set into downward

motion, it descends at a constant speed. Assume that the mass and friction of the pulley are negligible.

Part A



Calculate the coefficient of kinetic friction between block A and the table top.


Hint A.2 Find the net force on block A Hint not displayed

Hint A.3 Find the net force on block B Hint not displayed

Express your answer in terms of some or all of the variables , , and (the acceleration due to gravity).

ANSWER: =

Correct

Part BA cat, also of weight , falls asleep on top of block A. If block B is now set into downward motion, what is the magnitude of its acceleration?


Hint B.2 How to find the massHint not displayed

Hint B.3 Find the net force on block A and the catHint not displayed

Hint B.4 Find the net force on block B Hint not displayed

Express your answer in terms of some or all of the variables , , and (the acceleration due to gravity).



ANSWER: =

Correct

Kinetic Friction Ranking Task

Below are eight crates of different mass. The crates are attached to massless ropes, as indicated in the picture, where the ropes are marked by letters. Each crate is being pulled to the right at the same constant speed. The coefficient of kinetic friction between each crate and the surface on which it slides is the same for all eight crates.

Part ARank the ropes on the basis of the force each exerts on the crate immediately to its left.

Hint A.1 General problem-solving strategyHint not displayed

Hint A.2 Evaluate the effect of frictionHint not displayed

Hint A.3 Examine the top chain of cratesHint not displayed




ANSWER:

View Correct

Pushing a Lawnmower

Consider a lawnmower of weight which can slide across a horizontal surface with a coefficient of friction . In this problem the lawnmower is pushed using a massless handle, which makes an angle with the horizontal. Assume that , the force exerted by the handle, is parallel to the handle.Take the positive x direction to be to the right and the postive y direction to be upward.

Part A



Find the magnitude, , of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

Hint A.1 How to approach this problemHint not displayed

Hint A.2 Compute the sum of vertical forcesHint not displayed

Hint A.3 Compute the normal forceHint not displayed

Hint A.4 Compute the sum of horizontal forcesHint not displayed

Express the required force in terms of given quantities.

ANSWER: =

Correct

Part B

The solution for has a singularity (that is, becomes infinitely large) at a certain angle

. For any angle , the expression for will be negative. However, a

negative applied force would reverse the direction of friction acting on the lawnmower, and thus this is not a physically acceptable solution. In fact, the increased normal force at these large angles makes the force of friction too large to move the lawnmower at all.Find an expression for .


ANSWER: =

Correct



You should have found that , the force required to push the lawnmower at constant speed, was

.

Note that this expression becomes infinite when the denominator equals zero:,

or

.

(The phrase " has a singularity at angle " means that " goes to infinity at

a certain angle .") It's not too hard to understand what this means. Suppose you were pushing straight down on the lawnmower ( degrees). It obviously wouldn't move. But, according to the equation for , when you plug in degrees, you get a negative force (which doesn't make sense).The more vertical you push, the harder it gets to move the lawnmower. At

, it gets impossible to move it. The force required to move it goes to infinity; you have to push infinitely hard.

Static Friction and Frictional Force Ranking Task

Below are six crates at rest on level surfaces. The crates have different masses and the frictional coefficients [given as ] between the crates and the surfaces differ. The same external force is applied to each crate, but none of the crates move.

Part ARank the crates on the basis of the frictional force acting on them.

Hint A.1 The static friction relationshipHint not displayed




ANSWER:

View Correct

The Window Washer

A window washer of mass is sitting on a platform suspended by a system of cables and pulleys as shown . He is pulling on the cable with a force of magnitude . The cables and pulleys are ideal (massless and frictionless), and the platform has negligible mass.

Part A



Find the magnitude of the minimum force that allows the window washer to move upward.

Hint A.1 Find a simple expression for the tensionHint not displayed

Hint A.2 Upward forces on window washerHint not displayed

Hint A.3 All forces on window washerHint not displayed

Hint A.4 What about the platform?Hint not displayed

Express your answer in terms of the mass and the magnitude of the acceleration due to gravity .

ANSWER: =

Correct

Two Blocks and Two Pulleys

A block of mass is attached to a massless, ideal string. This string wraps around a massless pulley and then wraps around a second pulley that is attached to a block of mass

that is free to slide on a frictionless table. The string is firmly anchored to a wall and the whole system is frictionless. Use the coordinate system indicated in the figure when solving this problem.



Part AAssuming that is the magnitude of the horizontal acceleration of the block of mass , what is , the tension in the string?

Hint A.1 Which physical principle to useHint not displayed

Hint A.2 Force diagram for the block of mass Hint not displayed

Express the tension in terms of and .

ANSWER: =

Correct

Part B



Given , the tension in the string, calculate , the magnitude of the vertical acceleration of the block of mass .

Hint B.1 Which physical principle to useHint not displayed

Hint B.2 Force diagram for the block of mass Hint not displayed

Express the acceleration magnitude in terms of , , and .

ANSWER:

=

Correct

Part CGiven the magnitude of the acceleration of the block of mass , find , the magnitude of the horizontal acceleration of the block of mass .

Hint C.1 Method 1: String constraint (uses calculus)Hint not displayed

Hint C.2 Method 2: Intuition (does not involve calculus)Hint not displayed

Express in terms of .

ANSWER: =

Correct

Part D



Using the result of Part C in the formula for that you previously obtained in Part A, express as a function of .


ANSWER: =

Correct

Part EHaving solved the previous parts, you have all the pieces needed to calculate , the magnitude of the acceleration of the block of mass . Write an expression for .

Hint E.1 How to approach this problemHint not displayed

Express the acceleration magnitude in terms of , , and .

ANSWER: =

Correct

Two Masses, a Pulley, and an Inclined Plane

Block 1, of mass , is connected over an ideal (massless and frictionless) pulley to block 2, of mass , as shown. Assume that the blocks accelerate as shown with an acceleration of magnitude and that the coefficient of kinetic friction between block 2 and the plane is .



Part A

Find the ratio of the masses .

Hint A.1 Draw a free-body diagramHint not displayed

Hint A.2 Apply Newton's 2nd law to block 2 in the direction parallel to the incline

Hint not displayed

Hint A.3 Find an expression for the friction forceHint not displayed

Hint A.4 Find the normal forceHint not displayed

Hint A.5 Apply Newton's 2nd law to block 1 in the vertical directionHint not displayed

Hint A.6 Solve for the unknown tension

Hint not displayed


Express your answer in terms of some or all of the variables , , and , as well as the magnitude of the acceleration due to gravity .

ANSWER:

=

Correct

Velocity of a Roller Coaster Ranking Task



Six roller-coaster carts pass over the same semicircular "bump." The mass of each cart (including passenger) and the normal force of the track on the cart at the top of each bump are given in the figures.

Part ARank the speeds of the different carts as each passes over the top of the bump.

Hint A.1 Newton's 2nd lawHint not displayed

Hint A.2 Determine the net force on the cartHint not displayed


ANSWER:

View Correct

Taking the expressions for the net force on the cart and the centripetal acceleration of the cart and substituting into Newton's 2nd law,

,

results in



.

Since the radius is the same for every cart, we can ignore and write

.

But is simply so

,

or

.

Therefore, larger implies smaller .

Friction in a Bone Joint Vector Drawing

The lubrication of bone joints is a subject of ongoing medical research. Two bones connected at a joint do not touch. The bones are covered in articular cartilage, and are surrounded by lubricating synovial fluid. Rheumatoid arthritis results in overproduction of synovial fluid, swollen joints, and difficult and painful movement. Other joint disorders degrade the synovial fluid, directly increasing the friction between the bones, resulting in painful motion.

Part ATo measure the effective coefficient of friction in a bone joint, a healthy joint (and its immediate surroundings) can be removed from a fresh cadaver. The joint is inverted, and a weight is used to apply a downward force on the head of the femur into the hip

socket. Then, a horizontal force is applied and increased in magnitude until the femur

head rotates clockwise in the socket. The joint is mounted in such a way that will cause clockwise rotation, not straight-line motion to the right. The friction force will point in a direction to oppose this rotation. Draw vectors indicating the normal force (magnitude and direction) and the frictional

force (direction only) acting on the femur head at point A. Assume that the weight of the femur is negligible compared to the applied downward force.

Hint A.1 Determining the direction of the frictional forceHint not displayed

Draw the vectors starting at the black dot. The location, orientation and relative length of the vectors will be graded.



ANSWER:


Part B

The horizontal force would rotate the femur head clockwise in the hip socket, but the

frictional force acts to prevent this clockwise rotation. The apparatus is designed such

that when , the femur head rotates clockwise. A sample of data collected at impending slippage of the femur is shown in the figure. Based on these data, what is the approximate coefficient of static friction between the femur head and the hip socket?

Hint B.1 Identifying the downward and horizontal forcesHint not displayed

Hint B.2 Determine the coefficient of frictionHint not displayed

Enter your answer numerically to one significant figure.



ANSWER: = 0.009

Correct

Banked Frictionless Curve, and Flat Curve with Friction

A car of mass = 1500 traveling at 40.0 enters a banked turn covered with

ice. The road is banked at an angle , and there is no friction between the road and the car's tires.

Part A

What is the radius of the turn if = 20.0 (assuming the car continues in uniform circular motion around the turn)?

Hint A.1 How to approach the problemYou need to apply Newton's 2nd law to the car. Because you do not want the car to slip as it goes around the curve, the car needs to have a net acceleration of magnitude pointing radially inward (toward the center of the curve).

Hint A.2 Identify the free-body diagram and coordinate system



Which of the following diagrams represents the forces acting on the car and the most appropriate choice of coordinate axes?

ANSWER: Figure AFigure BFigure C

Correct

The choice of coordinate system shown in this free-body diagram is the most appropriate for this problem. The car must have a net acceleration toward the center of the curve to maintain its motion and not slip. This implies that the net force must be along the x axis.

Hint A.3 Calculate the normal force



Find , the magnitude of the normal force between the car and the road. Take the positive x axis to point horizontally toward the center of the curve and the positive y axis to point vertically upward.

Hint A.3.1 Consider the net forceThe only forces acting on the car are the normal force and gravity. There must be a net acceleration in the horizontal direction, but because the car does not slip, the net acceleration in the vertical direction must be zero. Use this fact to find .

Hint A.3.2 Apply Newton's 2nd law to the car in the y directionWhich equation accurately describes the equation for the net force acting on the car in the y direction?

ANSWER:

Correct

Express your answer in newtons.


Hint A.4 Determine the acceleration in the horizontal planeHint not displayed

Express your answer in meters.

ANSWER: = 34.6

Correct

Part B



Now, suppose that the curve is level ( ) and that the ice has melted, so that there is a coefficient of static friction between the road and the car's tires. What is , the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from slipping? Assume that the car's speed is still 40.0 and that the radius of the curve is given by the value you found for in Part A.

Hint B.1 How to approach the problemYou need to apply Newton's 2nd law to the car. Because you do not want the car to slip as it goes around the curve, the car needs to have a net acceleration of magnitude pointing radially inward (toward the center of the curve).

Hint B.2 Identify the correct free-body diagramHint not displayed

Hint B.3 Calculate the net forceHint not displayed

Hint B.4 Calculate the friction forceHint not displayed

Express your answer numerically.

ANSWER: = 0.364

Correct

Rolling Friction and Bicycle Tires



Two bicycle tires are set rolling with the same initial speed of 4.00 along a long, straight road, and the distance each travels before its speed is reduced by half is measured. One tire is inflated to a pressure of 40 and goes a distance of 17.3 ; the other is at 105 and goes a distance of 92.6 . Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be = 9.80 .

Part AWhat is the coefficient of rolling friction for the tire under low pressure?

Hint A.1 How to approach the problemThere are two main parts to this problem: 1. Use what you know about rolling friction and the normal force to find an expression for the acceleration that involves . 2. Find an equation that allows you to relate the bike's acceleration to its initial velocity, its final velocity, and the distance it travels.

Hint A.2 How to eliminate the massTo solve for the acceleration , use the following equations:

, , and ,where is the mass of the object and is the normal force.You should now be able to find an expression for that does not depend on the bike's mass.

Hint A.3 Kinematic equationThe best kinematic equation to use in this problem is

,since it relates acceleration to quantities given in the problem. Solve this equation for in terms of the distance traveled and the initial and final velocities of the bike.

ANSWER: =

3.54×10−2 Answer Requested

Part B



What is the coefficient of rolling friction for the second tire (the one inflated to 105 )?

Hint B.1 How to approach the problemYou should solve this part using the same steps you used in Part A.

ANSWER: =

6.61×10−3 Answer Requested

Suspending a Speaker

A loudspeaker of mass 15.0 is suspended a distance of = 1.40 below the ceiling by two cables that make equal angles with the ceiling. Each cable has a length of = 3.50 .

Part A



What is the tension in each of the cables?


Hint A.2 Identify the forcesHint not displayed

Hint A.3 Find the net vertical forceHint not displayed

Hint A.4 Determine

Hint not displayed

Use 9.80 for the magnitude of the acceleration due to gravity.

ANSWER: = 184

Correct

Linear and Rotational Kinematics Ranking Task

The pulley in the figure represents different pulleys with outer radius and inner radius indicated in the table. The horizontal rope is pulled to the right at a constant speed that is the same in each case, and none of the ropes slips in its contact with the pulley.



Part ARank these scenarios on the basis of the speed of the block.

Hint A.1 Relating the two rope speedsBy pulling the horizontal rope at constant speed, the pulley is given a constant angular velocity. This angular velocity in turn causes the rope attached to the block to wind up at a constant speed. Since both ropes are attached to the same pulley, each of their speeds must satisfy the relationship

,where is the angular velocity of the pulley and is the radius for the rope.


ANSWER:

View Correct

Several points on the pulley are indicated in the figure. Each letter designates a point on either the pulley or one of the two ropes. The horizontal rope is pulled to the right at a constant speed, and neither rope slips in its contact with the pulley.

Part B



Rank the designated points on the basis of their speed.

Hint B.1 Determining speedSince each point is located on the same pulley (or is located on a rope attached to the same pulley), each point’s speed is determined by its radial distance from the rotation axis via

.


ANSWER:


Part CRank the designated points on the basis of the magnitude of their acceleration.

Hint C.1 Constant speedHint not displayed


ANSWER:




Playing in the FieldFour children are playing in a field. The children form a line, holding hands. The player at the front of the line starts to spin around faster and faster, causing the others to run in circle, as shown in the figure.

Part AWhile the line of children is rotating, which of the following statements are correct?


Hint A.2 Angular accelerationHint not displayed

Hint A.3 Relation between linear and angular velocitiesHint not displayed


ANSWER: The player at the front of the line has the smallest angular acceleration.The player at the front of the line has the smallest linear velocity.All the children have the same angular acceleration. All the children have the same linear velocity.

Correct



Part BNow consider the children's linear accelerations. Which of the following statements are correct?


Hint B.2 Tangential accelerationHint not displayed

Hint B.3 Radial accelerationHint not displayed


ANSWER: The last child in the line has the greatest tangential acceleration.The last child in the line has the greatest radial acceleration.All the children have the same tangential acceleration. All the children have the same radial acceleration.

All attempts used; correct answer displayed

The last child in the line, being the farthest away from the axis of rotation, has the greatest radial acceleration. The force needed to produce this acceleration is provided by the pull of the rest of the children in the line. It won't take long before this acceleration becomes too high, especially for the final two children in the line. At this point, those children will not be able to apply enough force to hold on and will have to let go.

Rotational Kinematics Ranking Task



The figure shows angular position versus time graphs for six different objects.

Part ARank these graphs on the basis of the angular velocity of each object. Rank positive angular velocities as larger than negative angular velocities.

Hint A.1 Determining angular velocity from an angular position versus time graph

Hint not displayed


ANSWER:


Part B



Rank these graphs on the basis of the angular acceleration of the object. Rank positive angular accelerations as larger than negative angular accelerations.

Hint B.1 Determining angular acceleration Hint not displayed


ANSWER:

View Correct

The End of the Song

As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins down with a constant angular acceleration.

Part AIf the CD rotates at 500 (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 with constant angular acceleration, what is , the magnitude of the angular acceleration of the CD, as it spins to a stop?

Hint A.1 Angular accelerationSince the CD spins down with a constant angular acceleration, the instantaneous angular acceleration is equal to the average angular acceleration. Thus, the change in angular speed of the CD measured in a time interval divided by the length of the time interval yields the acceleration of the CD as it spins to a stop.

Express your answer in radians per second squared.

ANSWER: = 20.1

Correct

Part B



How many complete revolutions does the CD make as it spins to a stop?

Hint B.1 Find the angular displacementConsider a point P on the CD and take its angular coordinate to be zero when the CD reaches the end of the last song. What is the angular displacement undergone by point P as the CD spins to a stop?

Hint B.1.1 Angular displacement under constant accelerationHint not displayed

Express your answer in radians.

ANSWER:Answer not displayed

Your answer should be an integer.

ANSWER: 10.0 Correct revolutions

A Spinning Grinding Wheel

At time a grinding wheel has an angular velocity of 22.0 . It has a constant

angular acceleration of 34.0 until a circuit breaker trips at time = 1.60 . From then

on, the wheel turns through an angle of 437 as it coasts to a stop at constant angular deceleration.

Part A

Through what total angle did the wheel turn between and the time it stopped?


Hint A.2 Choose the most appropriate kinematic equation



Choose the most appropriate kinematic equation to use to determine the angle through which the wheel turns during the period of constant acceleration. Note that all of these equations are correct for the case of constant angular acceleration.

Hint A.2.1 Factors to finding the correct equationHint not displayed

The variables are (final angle), (initial angle), (final angular velocity), (initial angular velocity), (constant angular acceleration), and (time).

ANSWER:

Correct

Express your answer in radians.

ANSWER: 516 Correct

Part BAt what time does the wheel stop?

Hint B.1 What is the angular velocity when the wheel begins to slow down?

Calculate the angular velocity of the wheel when the circuit breaker trips.

Hint B.1.1 The correct kinematic equationHint not displayed

Express your answer in radians per second.

ANSWER:Answer not displayed



Hint B.2 Solving for timeAfter the initial time 1.60 , the wheel begins to decelerate at a constant rate, say . One can solve for the additional amount of time that it takes for the wheel to slow down to zero by looking at both the equation for angular velocity and the

equation as a function of time. We know that during deceleration

the wheel passes through the angle 437 and that the final angular velocity is . With these two equations, we can solve for the two unknowns (time and deceleration). Remember to set the initial velocity to the velocity when the circuit breaker trips.

Express your answer in seconds.

ANSWER: 13.0 Answer Requested

Part CWhat was the wheel's angular acceleration as it slowed down?

Hint C.1 Calculating the decelerationIn solving Part B you had to use the angular velocity of the wheel when the circuit breaker tripped. To find the deceleration (which is a constant), just divide by the total time it took for the wheel to stop spinning. Remember that deceleration here is a negative acceleration.

Express your answer in radians per second per second.

ANSWER: -6.68 Answer Requested

Acceleration in an Ultracentrifuge



Part AFind the required angular speed, , of an ultracentrifuge for the radial acceleration of a point 1.80 from the axis to equal 5.00×105 g (where is the acceleration due to gravity).

Hint A.1 Find the tangential speedFind the tangential speed of a point at radius 1.80 from the axis.

Hint A.1.1 Centripetal accelerationHint not displayed

Hint A.1.2 Acceleration of gravityHint not displayed

Express your answer numerically in meters per second.


Hint A.2 Converting tangential speed into revolutionsTo find the number of revolutions per second from the tangential speed, one just has to divide by the distance traveled in a single revolution ( ).

Express your answer numerically in revolutions per minute.

ANSWER: =

1.58×105 Answer Requested

An Electric Ceiling Fan



An electric ceiling fan is rotating about a fixed axis with an initial angular velocity of 0.280 . The angular acceleration is 0.917 . Its blades form a circle of diameter 0.700 .

Part ACompute the angular velocity of the fan after time 0.201 has passed.

Hint A.1 Angular velocity and accelerationHint not displayed

Express your answer numerically in revolutions per second.

ANSWER: 0.464 Correct

Part BThrough how many revolutions has the blade turned in the time interval 0.201 from Part A?

Hint B.1 Angle and angular velocityHint not displayed

Express the number of revolutions numerically.

ANSWER: 7.48×10−2 Correct

Part C



What is the tangential speed of a point on the tip of the blade at time = 0.201 ?

Hint C.1 Relating angular and linear speedHint not displayed

Hint C.2 Converting revolutions to radiansHint not displayed

Express your answer numerically in meters per second.

ANSWER: =

1.02 All attempts used; correct answer displayed

Part DWhat is the magnitude of the resultant acceleration of a point on the tip of the blade at time = 0.201 ?

Hint D.1 How to approach the problemSince the fan blade is both moving in a circle and speeding up, the tip of the blade must have both tangential and radial acceleration. Add them to find the total acceleration.Keep in mind that acceleration is a vector, and in order to find the total acceleration, one must use vector addtion (that is, one may not simply add the magnitudes).

Hint D.2 Find the centripetal accelerationCalculate the magnitude of the instantaneous centripetal acceleration of the point at the end of the fan blade. This is the acceleration perpendicular to the direction of motion.

Hint D.2.1 Definition of centripetal accelerationHint not displayed

Express your answer numerically in meters per second squared.


Hint D.3 Find the tangential acceleration



Calculate the magnitude of the instantaneous tangential acceleration (along the direction of motion) of a point on the tip of the blade at time 0.201 .

Hint D.3.1 Definition of tangential accelerationHint not displayed

Hint D.3.2 Converting revolutions to radiansHint not displayed

Express your answer numerically in meters per second squared.


Hint D.4 Calculating the vector sumNotice that the centripetal and tangential accelerations are perpendicular. Thus, you can think of them as the two components of the total acceleration . This makes the

magnitude of the total acceleration , where is the magnitude of

the tangential acceleration and is the magnitude of the centripetal acceleration.

Express the acceleration numerically in meters per second squared.

ANSWER: =

3.60 Answer Requested

Acceleration of a Pulley



A string is wrapped around a uniform solid cylinder of radius , as shown in the figure . The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass . Note that the positive y direction is downward and counterclockwise torques are positive.

Part AFind the magnitude of the angular acceleration of the cylinder as the block descends.

Hint A.1 How to approach the problem1. The block does not rotate. To analyze its motion, you should use Newton's second law in its linear form: . 2. The pulley rotates. To analyze its motion, you should use Newton's second law in its angular form: . 3. Using the geometry of the situation, you need to find the relationship between and

. 4. Finally, solve the system of three equations to obtain an expression for .

Hint A.2 Find the net force on the blockThe block has two forces acting on it: the tension of the string and its own weight. What is the net force acting on the block? Use the coordinate system shown in the figure.

Express your answer in terms of , (the magnitude of the acceleration due to gravity), and (the tension in the string).

ANSWER:

Correct

Hint A.3 Find the net torque on the pulley



The tension in the string produces a torque that acts on the pulley. What is the torque?

Hint A.3.1 Formula for torqueHint not displayed

Express your answer in terms of the cylinder's radius and the tension in the string.

ANSWER:

Correct

The moment of inertia of a uniform cylinder about its axis is equal to .

Substituting this into the above equation gives

.

Hint A.4 Relate linear and angular accelerationThe string does not stretch. Therefore, there is a geometric constraint between the linear acceleration and the angular acceleration . What is the cylinder's angular acceleration in terms of the linear acceleration of the block?

Express your answer in terms of and . Be careful with your signs.

ANSWER: =

Correct

From this equation, . Substitute for in the force equation for the block.

Hint A.5 Putting it togetherHint not displayed

Express your answer in terms of the cylinder's radius and the magnitude of the acceleration due to gravity .

ANSWER:

=

Correct



Note that the magnitude of the linear acceleration of the block is , which does not

depend on the value of .

Balancing Torques Ranking Task

A sign is to be hung from the end of a thin pole, and the pole supported by a single cable. Your design firm brainstorms the six scenarios shown below. In scenarios A, B, and D, the cable is attached halfway between the midpoint and end of the pole. In C, the cable is attached to the mid-point of the pole. In E and F, the cable is attached to the end of the pole.

Part ARank the design scenarios (A through F) on the basis of the tension in the supporting cable.


Hint A.2 The mathematical relationshipHint not displayed


ANSWER:

View Correct

Pivoted Rod with Unequal Masses



The figure shows a simple model of a seesaw. These consist of a plank/rod of mass and length allowed to pivot freely about its center (or central axis), as shown in the diagram. A small sphere of mass

is attached to the left end of the rod, and a small sphere of mass is attached to the right end. The spheres are small enough that they can be considered point particles. The gravitational force acts downward. The magnitude of the acceleration due to gravity is equal to .

Part A

What is the moment of inertia of this assembly about the axis through which it is pivoted?

Hint A.1 How to approach the problemThe moment of inertia of the assembly about the pivot is equal to the sum of the moments of inertia of each of the components of the assembly about the pivot point. That is, the total moment of inertia is equal to the moment of inertia of the rod plus the moment of inertia of the particle of mass plus the moment of inertia of the particle of mass , all measured with respect to the pivot point.

Hint A.2 Find the moment of inertia due to the sphere of mass What is the moment of inertia of the particle of mass measured about the pivot point?

Hint A.2.1 Formula for moment of inertiaHint not displayed

Express your answer in terms of given quantities.

ANSWER: =

Correct

Hint A.3 Find the moment of inertia due to the sphere of mass



What is the moment of inertia of the particle of mass measured about the pivot point?

Express your answer in terms of given quantities.

ANSWER: =

Correct

Hint A.4 Find the moment of inertia of the rodWhat is the moment of inertia of the rod about the pivot point?

Hint A.4.1 General formula for the moment of inertia of a rodHint not displayed

Express in terms of and .

ANSWER: =

Correct

Express the moment of inertia in terms of , , , and . Keep in mind that the length of the rod is , not .

ANSWER: =

Correct

Part B



Suppose that the rod is held at rest horizontally and then released. (Throughout the remainder of this problem, your answer may include the symbol , the moment of inertia of the assembly, whether or not you have answered the first part correctly.) What is the angular acceleration of the rod immediately after it is released?

Hint B.1 How to approach the problemThe forces acting on the system (spheres and rod) are the weights of the spheres and the rod, and the reaction force from the pivot. Find the torque due to each of these forces about the pivot point and add them with the correct signs. Finally, use Newton's second law for rotational motion: .

Hint B.2 Find the torque due to the sphere of mass Hint not displayed

Hint B.3 Find the torque due to the sphere of mass Find the torque about the pivot due to the particle of mass .

Express your answer in terms of given quantities. Keep in mind that the positive direction is counterclockwise.

ANSWER: =

Correct

Hint B.4 Torque due to forces acting on the rodHint not displayed

Hint B.5 Relating the angular acceleration to the net torqueLet the net torque acting on the system about the pivot point be denoted by . Find an expression for .

Express your answer in terms of the system's moment of inertia and its resulting angular acceleration . (Use in your answer, not the expression for you found in Part A.)

ANSWER: =

Correct



Take the counterclockwise direction to be positive. Express in terms of some or all of the variables , , , , , and .

ANSWER:

=

Correct

Substituting for , the value obtained in Part A yields

.

A large angular acceleration is often desirable. This can be accomplished by making the connecting rod light and short (since both and appear in the denominator of the expression for ). For a seesaw, on the other hand, and are usually chosen to be as large as possible, while making sure that the "rod" does not get too heavy and unwieldy. This ensures that the angular acceleration is quite low.

Pebble in a Rolling Tire--Finding Velocity and Acceleration

You are to find the coordinates of a pebble stuck in the tread of a rolling tire that is rotating counterclockwise (i.e., in the positive sense) with angular velocity . The tire rolls without slipping on the ground (which is at ). The outer radius of the tire is . At time , the pebble is at the top of the tire, as shown.

Part A



Find the velocity of the axle of the tire relative to a fixed point on the ground, . Note the order of the subscripts: velocity of axle measured relative to the ground.

Hint A.1 Speed of center of mass

For a wheel of radius that is rolling without slipping, there is a relationship between the speed of its center of mass ( ) and the angular speed . Find in terms of and .


Express your answer in terms of , , and and/or .

ANSWER: . =

Answer Requested

The pebble and tire have now rolled as shown in the figure. Answer the following questions for .

Part B



Find the position vector of the pebble relative to the initial point of contact between the wheel and ground at a time , .

Hint B.1 How to startThis is a relative motion problem. You can find the position of the pebble relative to the ground by adding vectorially the following two vectors: (i) the position vector of the pebble relative to the axle and (ii) the position vector of the axle relative to the ground. In other words,

.

Hint B.2 Position of pebble relative to axle ( )

Hint not displayed

Hint B.3 Position of axle relative to ground ( )

Hint not displayed

Hint B.4 Some possible answersHint not displayed

Express the position vector of the pebble in terms of , , , and the unit vectors and/or of the xy coordinate system shown.

ANSWER: =

Answer Requested

Part C



Find , the velocity vector of the pebble with respect to a fixed point on the ground, in

terms of the unit vectors and of the xy coordinate system shown.

Hint C.1 Velocity as a time derivative of a position

The velocity vector is the time derivative of the position vector, i.e. . To take

the derivative of a vector means to differentiate its x and y components, so for example

.

Express the velocity vector in terms of , , , and and/or .

ANSWER: =

Answer Requested

Part D

Now find , the acceleration vector of the pebble with respect to a fixed point on the ground.

Hint D.1 Finding accelerationThe acceleration vector is the derivative of the velocity vector.

Express your answer in terms of , , and and/or of the xy coordinate system shown.

ANSWER: =

Answer Requested

Part E



Now find the magnitude of the acceleration vector.

Hint E.1 Definition of magnitudeThe magnitude of a vector is the square root of the sum of the squares of its x and y components. For example, if the vector is , then the magnitude of the

vector is . You will also find the trig identity helpful

in simplifying your answer.

Your answer should be independent of time.

ANSWER:

Answer Requested

This is the centripetal acceleration of the pebble. Any object moving in uniform circular motion will always experience centripetal acceleration, as given by your answer here.

Hoop on a Ramp

A circular hoop of mass , radius , and infinitesimal thickness rolls without slipping down a ramp inclined at an angle with the horizontal.

Part A



What is the acceleration of the center of the hoop?

Hint A.1 How to approach the problemDraw a diagram showing the forces and torques on the hoop. Write the corresponding force and torque equations. Consider the condition for no slipping. Finally, solve this system of equations for .

Hint A.2 Find the torque about the center of massWrite an expression for the total torque on the hoop about its center of mass. (By convention, a positive torque produces a counterclockwise rotation, and a negative torque produces a clockwise rotation.)

Hint A.2.1 A formula for the magnitude of torqueHint not displayed

Hint A.2.2 Existence and direction of the frictional forceHint not displayed

Express the torque in terms of given quantities and the force of friction .


Hint A.3 Find an expression for the torqueComplete the general equation of rotational dynamics relating an object's moment of inertia and angular acceleration to the total torque acting on the object.

Express the torque in terms of the moment of inertia and angular acceleration .


Hint A.4 What is the moment of inertia of the hoop?



What is the moment of inertia of the circular hoop?



Hint A.5 Find the frictional forceUsing Newton's second law to relate the forces acting on the hoop to the hoop's acceleration, find an expression for the force of friction .

Hint A.5.1 Existence and direction of the frictional forceHint not displayed

Hint A.5.2 Find the total force on the hoopHint not displayed

Express your answer in terms of , , , and .


Hint A.6 Find the linear accelerationIf you've answered the previous parts, the only missing link is the relationship between linear acceleration and angular acceleration. Find the linear acceleration in terms of the angular acceleration . Use a coordinate system in which the positive x axis points down the ramp, and keep in mind that counterclockwise angular acceleration is positive.




Express the acceleration in terms of physical constants and all or some of the quantities , , and .



ANSWER: =

Correct

So the acceleration is independent of the hoop characteristics, that is, the mass and size (radius) of the hoop. This is quite generally true for objects freely rolling down a ramp; the acceleration depends only on the distribution of mass, for example, whether the object is a disk or a sphere, but within each class the acceleration is the same. For example, all spheres will accelerate at the same rate, though this rate is different from the rate for (all) disks.

Part BWhat is the minimum coefficient of (static) friction needed for the hoop to roll without slipping? Note that it is static and not kinetic friction that is relevant here, since the bottom point on the wheel is not moving relative to the ground (this is the meaning of no slipping).


Hint B.2 Find the maximum value of the frictional forceHint not displayed

Hint B.3 What is the normal force?Hint not displayed

Hint B.4 Putting it all togetherHint not displayed

Express the minimum coefficient of friction in terms of all or some of the given quantities , , and .

ANSWER: =

Correct

Part C



Imagine that the above hoop is a tire. The coefficient of static friction between rubber and concrete is typically at least 0.9. What is the maximum angle you could ride down without worrying about skidding?

Express your answer numerically, in degrees, to two significant figures.

ANSWER: = 61

Correct

When roads are wet or icy though, the coefficient of friction between rubber and concrete drops to about 0.3 (or less), making skidding likely at much smaller angles.


55315748-14

Documents