Uniform Circular Motion

ObjectivesRecognize that an object moving in a circular path experiences an acceleration even if that object is moving with a constant speed.Be able to relate the acceleration of the object to its constant speed and radius of circular path. Recognize that the acceleration of an object moving in such a manner is always directed toward the center of the circular path.By the end of this lesson, you should

Acceleration RevisitedRecall that acceleration is defined to be the time rate of change in velocity of an object. This means that any time an object is changing its velocity, it is accelerating. So, what are the distinct ways that an object can change its velocity? We have seen that an object that changes speed experiences acceleration, but is that the only way?

Distinct Ways That A Vector Quantity Can Be ChangedRecall that velocity is a vector quantity. A vector can be changed by (i) Changing the magnitude of the vector (ii) Changing the direction of the vector (iii) Changing both the magnitude and direction of the vector

Distinct Ways That The Velocity of An Object Can Be ChangedAn object will therefore experience an acceleration if (i) The speed of the object can change while it travels in a straight line (ii) The direction of travel of the object can change while it travels at constant speed (iii) The object can change speed and change direction at the same time

Uniform Circular MotionWe have examined situations (i) and (iii) in some detail and will return to situation (iii) in much more detail later in the term, but the focus of this lesson is that of uniform circular motion. An object moving with this type of motion travels at constant speed along a circular path.

Warning!An object moving along a circular path is not traveling in a straight line. This means that you cannot use the straight-line kinematics formulas developed in previous lessons to describe the motion of such an object. Always remember the limitations of the models we develop in order to avoid the pitfalls that many a student before you has fallen into. You must use the equations for circular motion which we are about to demonstrate/derive.

Uniform Circular Motion AnalyzedConsider an object moving with a constant speed in a circular path. At time t1 the object has a velocity v1. At time t2 the object has a velocity v2. These velocity vectors have the same magnitude since the speed is constant.v1

v2

Uniform Circular Motion AnalyzedThe acceleration of the object is given by:

Thus, the direction of the acceleration is the same as the direction of the velocity change (v2 -v1). v1

v2

Uniform Circular Motion AnalyzedTo get the direction of the velocity change, we note that v2 - v1 = v2 + (-v1), where -v1 is a vector that is identical to v1 but points in the opposite direction. v1

v2

-v1

Dv

Uniform Circular Motion AnalyzedThe acceleration of the object points in the direction of the velocity change. Now comes the placement of the acceleration vector. Where does it go? Well, remember that the expression we are using gives the average acceleration of the object. v1

v2

-v1

Dv

Uniform Circular Motion AnalyzedWe want the direction of the instantaneous acceleration. We note that if an object moves with an acceleration that is constant in magnitude, the average and instantaneous value are the same only at the midpoint of the time interval. v1

v2

-v1

Dv

Uniform Circular Motion AnalyzedBecause this object is moving at a constant speed, the middle of the time interval happens to coincide to the middle of the distance interval. We move the velocity change vector to the middle of the distance interval and re-labeling it as the acceleration vector. v1

v2

-v1

a

Uniform Circular Motion AnalyzedWe note that the diagram can be rotated about any angle without changing the relative orientation of the vectors. As we rotate the diagram, we note one important feature: the acceleration vector for this object always points toward the center of the circular path!

Centripetal AccelerationSuch an acceleration that always points toward the center of a circular path is called a center-seeking, or centripetal, acceleration. It can be shown that the magnitude of the centripetal acceleration of an object traveling in a circular path of radius, r, at a constant speed, v, is given by:

Centripetal AccelerationRemember that this centripetal acceleration always points toward the center of the circular path, or radially inward. This is a unit vector that points in the radial (center seeking) direction, then the centripetal acceleration can be written as:

PeriodThere is a special time of travel that is of interest when the motion of an object repeats itself. This special time is called the period of the motion and is the time required for the motion of the object to complete itself one time. You can determine the constant speed of an object traveling in a circular path in terms of the time required for the object to complete one circular path. Hence, circumference divided by time is the speed of the object.

SpeedIf an object travels at constant speed, that constant speed can be found by taking the distance traveled and dividing by the associated time of travel. If the time that is used to calculate the constant speed is the period, or time to complete one circular path, then the correct distance to use would be the circumference of the circular path.

Speedrv

ExampleWhat is the acceleration of a person standing at the earths equator?

Try It On Your OwnAssume that the earth travels in a perfectly circular orbit around the sun. Determine a solution to the question of What is the acceleration that the earth experiences as it travels around the sun.

Try It On Your OwnTest tubes are put in a centrifuge and spin in a circle of radius 20.0 cm. How long should it take a test tube to complete one revolution if you want the contents in the test tube to experience a 25 g acceleration?

A Quick SummaryAn object traveling in a circular path at constant velocity does indeed experience an acceleration. This acceleration is toward the center of the circular path at all times. The velocity vector for the object is always tangent to the circle. NOTE: The expressions from straight line kinematics cannot be applied to the motion of such an object.

Uniform Circular Motion And Newtons Laws of MotionSince Newton's 2nd law is about unbalanced forces then you must have an unbalanced force to cause a centripetal acceleration. We will call this unbalance force a CENTRIPETAL FORCE!Another Objective: To understand what a Centripetal Force is and why it is pointing toward the center of the circle.

ObjectivesSolve problems involving Newtons Laws of motion for objects moving with uniform circular motion in a systematic way.By the end of this part of the lesson, you should be able to

Centripetal ForceThe key to solving a problem involving an object moving with uniform circular motion is to realize that the object is traveling in a circular path and therefore experiences a centripetal acceleration. The net force acting on the object must point toward the center of the circular path.

WARNING!Remember that in your Free Body Diagram you include all of the forces acting on the object. For every force on your FBD, you can point to an object that exerts that force on the object. For example, a normal force is exerted by a surface; a gravitational force is exerted by the earth; tension is exerted by a string; etc. You might be tempted to include a centripetal force in the FBD. Dont do this! The centripetal force is a combination of the forces already included in your FBD.

An EggcitingExampleA crate of eggs sits in the middle of the flatbed of a truck as the truck negotiates a curve in the road. The curve may be regarded as an arc of a circle of radius 35 m. If the coefficient of static friction between the crate and the flatbed is 0.6, what is the maximum speed the truck can have if the crate is not to slide during the maneuver?

SolutionStart by drawing a FBD of the crate of eggs:+y+xNfwIn the y-direction, the crate doesnt move, so vy = 0 m/s = constant, and ay = 0 m/s2.

Along the x-axis, there is only one force, so:Now, since the object is moving in a circular path at constant speed:Substituting what is known about the frictional force and the acceleration:

The truck can travel as fast as 14.3 m/s around the curve without the crate of eggs moving.

Try It On Your OwnA merry go round turns completely every 12 s. If a 45 kg child sits on the horizontal floor of the merry go round 3 m from the center, what minimum coefficient of static friction is necessary to keep the child from slipping?

Frames of Reference Video: by Professors, Humes and Ivy.http://archive.org/details/frames_of_reference

Engineering a CurveAn engineer needs to design a curved exit ramp for a highway in such a way that the car will not have to rely on friction to round the curve without skidding. Suppose a typical car rounds the curve with a speed of 30 mi/h (13.4 m/s) and that the radius of the curve is 50 m. At what angle should the curve be banked.

A drawing of the situation and the FBD looks like:qqwNrNyNxw+x+yThe forces are broken into components along the standard axes so that components of force will point toward the center of the circular path.

Now, write the components of the Normal force in terms of the angle:qNyNxNSince there is no motion of the car along the y-axis, vy = 0 m/s = constant. This means there is no component of acceleration along this axis and:

Along the x-axis, the x-component of the Normal force causes the centripetal acceleration:Substituting for the components of Normal force in terms of the banking angle:(1)(2)

Dividing equation (1) by (2)

Substituting numbers:

Try It On Your OwnAn amusement park ride consists of a rotating circular platform 8 m in diameter, from which seats are suspended at the end of a 2.5 m long chain. When the ride rotates, each chain holding a seat makes an angle of 28 with the vertical. If a 45 kg child is sitting in a seat while the ride is going, what is the tension in the chain for that seat?

SummaryProblems involving objects moving with uniform circular motion are solved in the same manner as any problem involving Newtons Laws. The forces, or combinations of all or parts of forces, cause a centripetal acceleration. You should not include a centripetal force in your FBD along with all of the other forces; the centripetal force is already there.

Here is a flow chart showing topics from uniform circular motion which hopefully you will read about and understand during your study.

http://hyperphysics.phy-astr.gsu.edu/hbase/images/rotcon.gifcle*SCHWINN; Insert the graphic above minus the the green letters and the mathematical symbols for integration put an X in the blue area and a V for the velocity area. http://www.opamp-electronics.com/tutorials/images/semiconductor/03356.png ASK THE QUESTION: but is that the only way? And then say what if you are not going in a strait line direction?*During a race a car can change speeds and directions. The first part of this course we worried about only changing speeds when we had an acceleration. We will now have to consider when the vector quantity of velocity has a magnitude (speed) and or a directional change. If it has a directional change it will have an acceleration which we will consider to be a Centripetal acceleration.*An object will therefore experience an acceleration if (i) The speed of the object can change while it travels in a straight line (ii) The direction of travel of the object can change while it travels at constant speed (iii) The object can change speed and change direction at the same time

*http://igphysics.com/physics/Camb/paper%203/p3%20jun%202007/img%20p3%20jun%202007/image002.jpg Schwinn: you could animate a picture of a car moving around a circular track here. You could show a picture of the speedometer and a compass. The speedometer could have a constant value but the compass would have to change as the direction of the car changes*An object moving along a circular path is not traveling in a straight line. This means that you cannot use the straight-line kinematics formulas developed in previous lessons to describe the motion of such an object. Always remember the limitations of the models we develop in order to avoid the pitfalls that many a student before you has fallen into. You must use the equations for circular motion which we are about to demonstrate/derive.*You can combine the next couple of slides by animating the motion of the velocity vectors around the circle. Read the description and it explains how to move the vectors.*The vector V1 is subtracted from the vector v2 and shown as a change in velocity. Invert Vector V1 and then slide it over to the tip of vector V2. Then add the vector called the change in velocity vector.*The acceleration of the object points in the direction of the velocity change. Now comes the placement of the acceleration vector. Where does it go? Well, remember that the expression we are using gives the average acceleration of the object.

*Note you can now slide the entire triangle back toward the midpoint between where the first and second dots appear. Then you will be able to draw the ACCELERATION vector as is seen on the next slide.*Because this object is moving at a constant speed, the middle of the time interval happens to coincide to the middle of the distance interval. We move the velocity change vector to the middle of the distance interval and re-labeling it as the acceleration vector.

*Rotate the diagram and you can see that as long as the object is moving in a constant speed that the acceleration vector will point to the center of the circle along a radius toward the origin Hence we call this a center seeking acceleration or a centripetal acceleration.*Such an acceleration that always points toward the center of a circular path is called a center-seeking, or centripetal, acceleration. It can be shown that the magnitude of the centripetal acceleration of an object traveling in a circular path of radius, r, at a constant speed, v, is given by:

*Recall from our experience with linear motion that speed was the distance traveled divided by the time it took to travel the distance. This was defined as the velocity. Here we have set up the same scenario with the distance traveled actually being the circumference of the circular path and the time being called the Time Period, which is the amount of time required to make a complete cycle around the circle. We call this value of velocity the velocity tangent to the circular path or a linear velocity for the object in circular motion.*You will need to look up the radius of the earth as well as think about how long it takes the earth to rotate on its axis daily. Using this information and the equation from the previous slide what is the correct answer to this question?

http://www.hort.purdue.edu/newcrop/tropical/lecture_02/02m.jpg*Once you have calculated the velocity of the earth as it travels around the sun. (A process similar to the process required for calculating the rotational speed of the earth at the equator) you can use the centripetal acceleration equation to solve this problem. *Recall that an acceleration of 25 g means that it is 25 time the acceleration rate due to gravity. If we assume that value to be approximately 10 m/s2 (9.8 m/s2 actually) then the acceleration required is 250 m/s2. A direct plug into the equation for centripetal acceleration will give you a value of an approximate time.**Another Objective: To understand what a Centripetal Force is and why it is pointing toward the center of the circle.

*Solve problems involving Newtons Laws of motion for objects moving with uniform circular motion in a systematic way. Here is an illustration of a classic physics course experiment where you determine the force needed to keep the swinging rubber stopper in a circle of known radius. The experiment involves spinning the swinging rubber stopper in a horizontal circle above your head. If you swing the stopper as a fast enough speed the 100 gram weight will counter act the force needed to hold the stopper in place. The 100 gram weight is the centripetal force. Like wise you can see that the earth is supplying a centripetal force to keep the moon from flying off of its orbital path. This is a gravitational force which is also called a centripetal force. Remember that the Centripetal Force is directed inward to ward the center of the circle while the velocity of the object is always tangent to the radius of the circle.

*http://stockcarscience.com/images/Centripetal_Force.gif SCHWINN: If you could make an animation of an object that would actually circle with the axis and velocity and centripetal force created it would be fabulous. *Remember that in your Free Body Diagram you include all of the forces acting on the object. For every force on your FBD, you can point to an object that exerts that force on the object. For example, a normal force is exerted by a surface; a gravitational force is exerted by the earth; tension is exerted by a string; etc. You might be tempted to include a centripetal force in the FBD. Dont do this! The centripetal force is a combination of the forces already included in your FBD.

*The solution to the problem is on the upcoming page.*Follow the path of the equations used and you can see how fast the truck can travel and safely haul its eggs!

*If you live near and part that has a merry go round it is sometimes fun to just go and set on the merry go round and get pushed around but it. Feeling the kinesthetic properties of rotation can be a great learning experience too. While you are setting on the moving merry go round try playing catch with somebody on the same merry go round, or somebody standing on the ground beside the merry go round. Think about the perspective of what the ball would look like from each of these positions. *These are classic 1960 films that have been archived. Professors Humes and Ivy do an excellent job getting you to think about the merry-go-round experience plus other ideas related to frames of reference. Millions of Physics teachers and students have enjoyed these films in the past and you might consider doing the same.*Solution In this previous problems, a single force was responsible for causing the object to travel in a circular path. In this case, a component of a force causes the circular motion. In other cases, combinations of forces might cause the centripetal acceleration.

*The necessary banking angle is 20.1 degrees as was just derived from the mathematical interpretation of this banked curve problem.*