APC PHYSICS CHAPTER 11 Mr. Holl Rotation

Student Notes

11-1 Translation and Rotation

All of the motion we have studied to this point was linear or translational. Rotational motion isthe study of spinning bodies such as wheels, gears, jet engines, helicopter blades, planets and spinningathletes such as figure skaters and gymnasts.

11-2 Rotational Variable

In this chapter we deal with rigidbodies rotating around a fixed axis, calledthe axis of rotation. The diagram to theright shows two masses moving in rotational motion around a fixed point oraxis of rotation. All of the formulasyou have learned for translation havecorresponding formulas that deal withrotation, and every variable has acorresponding rotational variable. Ingeneral, Greek letters are used to represent rotational quantities.

1) Angular Position - where an objectis located angularly as measuredfrom a reference line. The variableis è and is measured in radians.Remember, è = s/r where s is thetranslational distance (arc length)from the reference point and r is theradius of the path followed. One complete revolution is 2ð rad. è isnot reset at zero each revolution, soif you have gone around the circle 3times your location is 6ð rad.

2) Angular Displacement - As you movebetween two points your angular locationchanges just as your linear position changedin translational problems.

2 1Äè = è - è

3) Average Angular Velocity - The average rate of change in angular displacement.The symbol used is lowercase omega - ù.

4) Instantaneous Angular Velocity - rate of change of angular displacement at any instance, oraverge ù as Ät approaches zero.

The unit for both average and instantaneous angular velocity is radians/second,abbreviated rad/s.

5) Average Angular Acceleration - Average rate of change of angular velocity.

6) Instantaneous Angular Acceleration - Rate of change of angular velocity at any given instant.

The unit for angular acceleration is rad/s .2

Example 11-1) The diagram to the right shows the overheadview of a child on a merry-go-round. If theride is rotating at a constant 12 rev/min:

a) What is its angular velocity in rad/s?

b) What is the child’s angular displacement in 4 sec?

c) How far has the child actually traveled?

Example 11-2) The angular position of a reference line on a spinning wheel is given by the equation: è =t - 27t+ 4, where t is in seconds.3

a) Find ù(t) and á(t)

d) At what times is the angular velocity zero?

e) Describe the wheels motion when t $ 0.

Example 11-3) A child’s top is spun with an angular acceleration of á = 5t - 4t where3

the coefficients are compatible with seconds and radians. At t= 0 the top has an angular velocityof 5 rad/s, and a position of è = 2 rad with respect to some reference line.

a) Obtain an expression for the angular velocity ù(t) of the top.

b) Obtain an expression for the angular position of the top, è(t).

c) After 2 seconds find the angular acceleration, angular velocity and final angulardisplacement of the top.

11-3 Angular Quantities as Vectors

When dealing with linear vectors we expressed directions as positive and negative, with angularquantities we use clockwise and counterclockwise about the axis.

When describing rotational vectors we establish the direction of ù using a right-hand rule. When your fingers are aligned in the direction of the rotation, your thumb will point in the direction ofthe angular velocity vector. The vector defines the axis of rotation.

ex) When looking down at a playing CD, what is the direction of the velocity vector?

ex) In which direction is the object to the rightrotating if the velocity vector is in the +x dir.?

11-4 Rotation with Constant Angular Acceleration

Just as with translational motion we had a group of formulas which were useful as long as theacceleration was constant, the same group can be derived for angular motion. These are:

Example 11-4) A grindstone has a constant angular acceleration of á = 0.35 rad/s It starts from rest with2

oan arbitrary horizontal line at è = 0.

a) What is the angular displacement è of the reference line at t= 18 s?

b) What is the wheel’s angular velocity at t= 18 s?

oExample 11-5) Lets assume that the grindstone in 11-4 does not start at rest but has an initial speed of ù= -4.6 rad/s.

a) What will the grindstone be doing over the first few seconds?

b) At what time will the grindstone be at rest?

c) How many revolutions will the grindstone make before coming to rest?

Example 11-6) During the analysis of a helicopter engine you determine that the rotor’s velocity changesfrom 320 rev/min to 225 rev/min during the 1.5 minutes of observation.

a) What is the average angular acceleration of the rotor?

b) Assuming that the rotational acceleration remains constant, how long will it takethe rotor to stop?

c) How many revolutions will it make before coming to rest. from the original 320rev/s?

11-5 Relating Angular and Linear Variables

In the diagram to the right supposethat the two masses are traveling with a constantand equal angular velocity. This means that theirrelative positions will remain constant becauseeach will pass through the same angulardisplacement è in the same time interval.

But the linear distance traveled by the outer masswill be significantly larger than that of the innermass due to the fact that it is traveling a long acircular path of larger radius!

We can use the relationship between èand arc length to convert angularquantities into linear quantities.

1) Position or Distance

è = s/r therefore s = èr where s is the actual distance traveled

2) Speed

Differentiating the equation for distance we getbut ds/dt is the linear speed, and dè/dt is the angular velocityso:

v = ùr

3) Period

To this point we have always found the period, the time it takes to make one revolution, using theformula T = 2ðr/v, but if v = ùr than r = v/ù. If we plug this back into the equation we get:

4) Acceleration

a) Tangential (linear) - This is the rate at which the object is speeding up or slowing down. If we differentiate v = ùr with respect to time we get

Since dv/dt = a and dù/dt =áwe get:

a = ár

b) Radial or Centripetal Acceleration - When an object is traveling along a circular path italso has a centripetal or radial (along the radius) component of acceleration. We know

rthat we can find this using a = v /r. But we also now know that v = ùr so:2

This is the acceleration responsible for the change in direction rather than the change in speed.

Example 11-7) The diagram to the right showsa centrifuge used to train astronautsto get use to high accelerations and g forces. The radius of the deviceis r =15 m.

a) What constant angular velocity must the centrifuge have if therider is to experience 11g?

b) What is the tangential acceleration of the astronaut if the centrifuge goes from rest to thisangular velocity in 2 minutes?

11-6 Moment of Inertia

As we’ve seen so far, there seem to be angular quantities that correspond with the linear quantities we dealt with in the first 10 chapters. This is also true when it comes to mass. Wedefined mass (inertial mass) as a measure of how difficult it was to accelerate the object. The greater themass, the harder (more force needed) it is to accelerate the object.

When dealing with rotation there is also a similar quantity. It is called rotational Inertia, ormore often, moment of inertia. The greater the moment of inertia of an object, the harder it is to makeit accelerate angularly (á).

The moment of inertia (I), is determined by not only the objects mass, but the location of themass (r). The farther the mass is from the axis of rotation, the greater its moment of inertia. For a singlepoint mass (like a ball at the end of a string, or a coin on the edge of a turntable) the moment of inertia isfound using:

I = mr 2

If several masses are located in the system then the moment of inertia is:

Example 11-8) The diagram to the right shows a 5 kg mass attached to a massless rodwith a length of 60 cm. The axis ofrotation is at the other end of the rod.What is the moment of inertia of the system?

Example 11-9) 3 children of masses shown are sittingon a massless merry-go-round. Determinethe moment of inertia of the system.

Calculating Moment of Inertia

For any rigid body the moment of inertia is found using the formula on the previous page. If wechoose an infinitely small unit of mass (again our differential mass), dm, we can replace the Gwith an integral to get:

This equation allows us to find the moment ofinertia for any standard shape object .

Example 11-10) The small, thin rod shown to theright has a length of L and a mass of M.

What is the moment of inertia of the rodis the axis of rotation is placed in thecenter of the rod.

What is the rod were to be rotated around one end?

Example 11-11) Find the moment of inertia for a diskwith a mass of M, a thickness of Land a radius of r.

Fortunately for us we do not have to run this integration everytime. The work has already been done for most rigid solids andthe results listed below. I will provide you with this list for allquizzes and tests and the AP gives you one too.

Rotational Inertia for Some Objects

The Parallel Axis Theorem

If you know the rotational inertia of a body about any axis that passes through its center of mass,you can find its rotational inertia about any other axis parallel to that axis with the parallel axis theorem.

cmWhere M is the mass, I is the moment of inertia through thecenter of mass and h is the perpendicular distance to the newaxis.

Example 11-12) Using your information from example 11-10, find the moment of inertia of the rod if it isrotated ½ way between the center and either end.

Example 11-13) In an exciting physics contest a race down a hill is set between a ring (hoop), a disk(cylinder), and a sphere all of equal mass and radius. Place your bets! Who will win the race????

11-7 Rotational Kinetic Energy

A rotating object, such as a lawn mower blade certainly has kinetic energy even though the bladeas a whole has no translational velocity. If we look at any little piece of the blade it has a mass, and aspeed. Add these all up and we should have the total kinetic energy of the blade due to its rotation.

1 1 2 2 3 3K = ½m v + ½m v + ½m v + ...2 2 2

i ior K = G ½m v i = 1 to n2

iThe problem is that the v is not the same for all particle. We can get around this by expressingthe velocity in angular terms.

i iv = ùr therfore K = G ½m (ùr )2

If we rearrange and move the r over with the mass and pull the ½ out of the sum we get2

i iK = ½ G m r ù2 2

i iBut we have just seen that G m r = I so K = ½ Iù2 2

Example 11-14) The giant Ferris Wheel at Great Adventure has a mass of 12,000 kg and a diameter of 40m. It rotates at a rate of 1.2 RPM. Assuming that the mass is concentrated on the outeredge (like a hoop) find:

a) The moment of inertia of the wheel.

b) The angular velocity of the wheel.

c) The rotational kinetic energy of the wheel.

11-8 Torque

When trying to make an object rotatewe must apply a torque. A torque is theproduct of a force, and the distance the forceis applied from the axis of rotation. Thegreater the force or the greater the distance(sometimes called the lever arm) the greaterthe torque produced.

The diagram tot he right shows an overheadview of a door. If we grab the outer edge pull perpendicular the torque createdis ô = rF

Suppose I now pull not perpendicularly, butat an angle ö from the direction of r. Now we must resolve F into two components,

r tF (radial component) and F (tangentialcomponent). The radial component createsno torque. the torque is created by thetangential component.

This means that the torque we create is

tô = r F . Looking at the diagram we can

tsee that F = F sinö, so torque can be written as:

ô = r F sinö

where ö is the angle between the directionof the lever arm and the direction ofthe force.

Another way to visualize this is to extendthe direction of the force and until you canfind the perpendicular distance to the axis.This rz is known as the moment arm and

torque can be written as ô = rzF. As shown in the diagram to the right,rz = r sinö so no matter which way youlook at it the torque formula comes outthe same.

The definition of torque ô = r F sinö should look familiar! It is our definition of a cross product,therefore we can also write torque as ô = r x F, and its direction is given by the right hand rule!

Example 11-15) The diagram to the right showstwo forces applied to a wheelwith a radius of 20 cm.

a) Determine the magnitudeand direction of the torquecreated by each force.

b) What is the net torque acting on the wheel?

11-9 Newton’s Second law For Rotation (Rotational Dynamics)

t tNewton’s Second Law states F =ma where the (t) stands for translational.

tIf ô = rF then we can rewrite the

tformula for torque as ô = ma r

t = but a ár giving ô = m(ár)r or ô = mr á2

Since mr is the moment of inertia, I2

we can again rewrite the formula as ô = Iá

or more correctly G ô = Iá

The sum of the torques acting on an object (net torque) is always equal to the moment of inertia timesthe angular acceleration.

Example 11-16) If the wheel in example 11-15 is taken to be a solid disc with a mass of 6 kg and a radiusof 40 cm, what will be its angular acceleration?

Example 11-17) The diagram to the right showsa uniform disk with a mass M = 2.5 kgand a radius of R = 20 cm. The disk is mounted on a fixed horizontal axis.A block with a mass of 1.2 kg hangsby a massless rope wrapped around thedisk. Find:

a) The downwards acceleration ofthe mass.

b) The angular acceleration of the disk.

c) The tension in the rope.

Example 11-18) The Atwood’s machine to the rightshows two masses attached witha string that passes over a pulleythat has a mass of 2 kg and a radiusof 20 cm and a frictional toque of8.0 mN (WOW - We’ve beenwaiting all of our physics lives for a pulleywith mass and friction!!!!) Determine:

a) The acceleration of the system.

1b) The tension in the rope attached to m .

2c) The tension in the rope attached to m .

11-10 Work and Rotational Kinetic Energy

When we did work accelerating an objectlinearly we found the work by using; W = FCd = Fd cosèwhen the force was constant, and

(s) when the force was changing; W = IF ds (s = distance)

Taking this concept to rotation we know that

(è)s = rè therefore ds = drè = r dè giving W =IF r dè

But Fr = ô so the work done accelerating

an object angularly becomes

For Power, average power was rate atwhich work was done

Instantaneous power was force timesinstantaneous velocity.

When dealing with rotation, poweris torque times instantaneous angularvelocity.

WORK - ENERGY THEOREM

(ô = Iá)

Example 11-19) A bicycle wheel (hoop) has a mass of 8 kg and a radius of 50 cm. It startsat rest and the pedaler exerts a torque of 3.2 Nm to the wheel for 10 s.

a) What is the moment of inertia of the wheel?

b) What is the angular acceleration of the wheel?

c) What is the angular velocity of the wheel after the 10 s?

d) What is the angular displacement of the wheel after the 10 s?

e) How much work was done accelerating the wheel?

f) What is the average power of the pedaler?

G) If the accelerating torque is now removed, and the wheel comes to rest after 30seconds, what is the average frictional torque acting on the wheel?