Dynamics of Rotational Motion - FCPS 10.pdf · Dynamics of Rotational Motion. ... publishing as...

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-WesleyCopyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

PowerPoint® Lectures forUniversity Physics, Twelfth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 10

Dynamics of Rotational Motion

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

Goals for Chapter 10

• To examine examples of torque

• To see how torques cause rotational dynamics (just as linear forces cause linear accelerations)

• To examine the combination of translation and rotation

• To calculate the work done by a torque

• To study angular momentum and its conservation

• To relate rotational dynamics and angular momentum


Homework

• 1, 5, 7, 177, 17, 19, 25, 27, 33, 35, 41, 45, 49

• Read Ch 10

• Read 404 to 414


Introduction

• Skydivers and ice skaters can change their rotational motion without pushing off or holding onto any objects. How can this be so?

• Twisting or turning gives us physical pictures of a new concept, torque.


Torque• A force applied at a right

angle to a lever will generate a torque.

• The distance from the pivot to the point of force application will be linearly proportional to the torque produced.

• As an equation:


Lines of force and calculations of torques• The directions of torques produced

will give us another opportunity to apply the right-hand rule (RHR).

• Torques that rotate counterclockwise are positive.


What would be the sign of torque from each force if the pivot was about point A?


Calculate an applied torque

A weekend plumber, unable to loosen a pipe fitting, slips a piece of scrap pipe over his wrench handle. He then applies his full weight of 900 N to the end of the cheater by standing on it. The distance from the center of the fitting to the pointwhere the weight acts is 0.80 m, and the wrench handle and scrap pipe make an angle of 19o with the horizontal.Find the magnitude and direction of the torque he applies about the center of the pipe fitting.


Torque and Angular Acceleration• Let’s consider a rigid body being made up of many small

particles. One particle with mass m1 has components F1,tan, F1,rad, and F1z of a net force acting on m1.

• From Newton’s 2nd law: F1,tan = m1a1• We also know that a = rα, so we can rewrite Newton’s law

as: F1,tan = m1r1αz.Multiplying both sides by r1:F1,tan r1= m1r1

2αzF1,tan r1 is the torque by the net force about the z axis, τ1z, and m1r1

2 is the moment of inertia of m1 about the z axis, I1.


Τ = Iα and F = ma—Newton’s 2nd Law• Neither F1z nor F1,rad have any effect on the torque about the

z axis. We can say the total torque about the z axis is:

• If we make an equation like this for every particle in the body and then add them:

For a rigid body, the angular acceleration is constant for every particle.

21 1 1 1z z zI m rτ α α= =

( )2iz i i z

iz z

m r

I

τ α

τ α

=

=

∑∑

2 21 2 1 2 1 1 22z z z z z zI I m r m rτ τ α α α α+ = + = +


Torque and a cable about a pulley


Torque and a cable about a pulley

• The figure below shows the same situation that we analyzed in last chapter using the energy method. Assuming that the cable unwinds without slipping, what is its acceleration?


Another look at the unwinding cable

Find the acceleration of mass m.


Homework

1, 5, 7, 13

Read 370 to 377


10.3 | A rigid body in motion about a moving axis• Consider a bike wheel rolling down a hill. Does the wheel have kinetic

energy?• What kind of kinetic energy?

• Both, YES!! The kinetic energy of the rotating wheel is the sum of both its translational and rotational kinetic energies.

• Lets consider a special case of rotational motion, rolling without slipping. If an object rolls without slipping, its translational velocity is the velocity of its center of mass. (see figure 10.14 on 371)

• Consider the motion of the wheel where it contacts the surface. How fast is it moving with respect to the surface?

• It must have a rotational velocity, ω, such that the tangential velocity, vt, corresponding to ω cancels out the velocity of the center of mass, vcm.

• That velocity will be the same for any point tangent to the wheel surface.

2212

21 ωcmIMvK

cm+=

ωRvcm =


Rolling with and without slipping• The following figure graphically explains how to calculate rolling without

slipping.• Rolling with slipping may be calculated. Slipping makes things worse (for

driving and calculations).


A rolling cylindrical shell

• A hollow cylindrical shell with mass M and radius R rolls without slipping with speed vcmon a flat surface. What is its kinetic energy?


Consider the speed of a yo-yo toy• A primitive yo-yo is made by

wrapping a string several times around a solid cylinder with mass M and radius R. You hold the end of the string stationary while releasing the cylinder with no initial motion. The string unwinds but does not slip or stretch as the cylinder drips and rotates.

• Use energy considerations to find the speed vcm after it has dropped a distance h.


The race of objects with different moments

• Which will reach the bottom of the incline first: a disk or ring with the same mass?


Combined Translational and Rotational Motion

• We can analyze the motion of rotating objects by combining Newton 2nd Law and its rotational analog:

• These are valid only if the axis through the center of mass is an axis of symmetry and the axis does not change direction.

cmext aMF =∑ zcmz I ατ =∑


The yo-yo (again)• Find the downward acceleration of the cylinder and the

tension in the string.


Consider the acceleration of a rolling sphere • A solid bowling ball rolls without slipping down the return

ramp at the side of the alley. The ramp is inclined at an angle β to the horizontal.

• What is the ball’s acceleration?


Consider the acceleration of a rolling sphere II

• When friction is present, it gives the force required to produce a torque and hence rotation of an object.

• If the surface deforms, the normal force is no longer through the axis of rotation and produces a force and torque that opposes the torque from friction.


10.3 Homework

• 10.19 and 10.25

• Read 377 to 387


Work and power in rotational motion • Consider a tangential force, like that in Figure 22a. The disk rotates

through an infinitesimal angle dθ during an infinitesimal time dt.• The work done (W=Fs) is dW=Ftands where ds is the infinitesimal

arclength. See Figure 22b• If dθ is measured in radians then ds = Rdθ and

• Since FtanR is torque τz then

• Integrating both sides yields an expression for work:

• For a constant torque we can say:

θτθ

ddWRdFdW

z== tan

∫=2

1

θ

θ

θτ dW z

( ) θτθθτ Δ=−= zzW 12


Work-Kinetic Energy with Torque• We need to change the integrand in our work equation from

angular displacement into angular velocity. Let’s start by saying that τz is the net torque on the object.

• Since τz is the net torque, the total work done on the rigid body is:

• The power associated with work done by torque:

( ) zz z z z z

d dd I d I d I d I ddt dtω θτ θ α θ θ ω ω ω= = = =

2

1

2 21 12 12 2tot z zW I d I I

ω

ω

ω ω ω ω= = −∫

z zP τ ω=zdW ddt dt

θτ=


Example 10.10• The power output of an automobile engine is advertised to be

200 hp at 6000 rpm. What is the corresponding torque?

• An electric motor exerts a constant torque of 10 Nm on a grindstone mounted on its shaft. The moment of inertia of the grindstone about the shaft is 2.0 kgm2.

• If the system starts from rest, find the work done by the motor in 8.0 seconds and the kinetic energy at the end of this time.

• What was the average power delivered by the motor?

• For Homework:

• 10.27 and 10.33 Read 387 to 391


Angular Momentum• Every rotational quantity we have encountered has a linear

analog. The analog of linear momentum p is angular momentum L. The relation is the same as that of torque to force :

z r F

L r p r mv

τ = ×

= × = ×


Torque and Angular Momentum• When a net force acts on a particle, its velocity and

momentum change, so its angular momentum may also change. We can show that the rate of change of angular momentum is equal to the torque of the net force. We take the time derivative of the angular momentum equation using the rule for the derivative of a product:

• The first term is zero and F = ma so:

• The rate of change of angular momentum of a particle equals the torque of the net force acting on it.

( ) ( )dL dr dvmv r m v mv r madt dt dt

⎛ ⎞ ⎛ ⎞= × + × = × + ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

dL r Fdt

τ= × =


Torque and Angular Momentum• Consider a thin slice of a rotating body lying in the xy-plane.Each particle in the slice moves in a circle centered at the origin, and at each instant its velocity vi is perpendicular to its position vector ri, as shown. Remember that the magnitude of L = mvr sin φ, for our situation, φ, is 90o. We can express the angular momentum as Li = mi(riω)ri = miri

2ω. The total angular momentum is the sum of this equation:

( )2i i iL L m r Iω ω= = =∑ ∑


Torque and Angular Momentum

• Equations on 381. Memorize them.


Angular Momentum and Torque• A turbine fan in a jet engine has a momentum of

inertia of 2.5 kgm2 about its axis of rotation. AS the turbine is starting up, its angular velocity as a function of time is

a) Find the fan’s angular momentum as a function of time, and find its value at time t = 3.0 s.

b) Find the net torque acting on the fan as a function of time, and find the torque at time t = 3.0 s.

( )3240 rad

z s tω =


Conservation of Angular Momentum

• When the net external torque acting on a system is zero, the total angular momentum of the system is constant.

• Consider Willie spinning on the stool with his legs and arms extended and rotating counterclockwise about an axis through his center of mass. When he pulls his legs and arms in his moment of inertia changes from a large I1 to a much smaller I2. The only external forces acting on him are his weight and the normal force from the stool, both of which are acting through the axis of rotation, and give no torque.

• His angular momentum, Lz = Iωz, does not change. When Idecreases, ω must increase such that: I1ω1=I2ω2

0 0dLif thendt

τ = =∑


Conservation of Angular Momentum

• With Newton’s 3rd law, we can find that the total angular momentum of a system of several interacting bodies is constant. The torques of the internal forces can transfer angular momentum from one body to the other, but they can’t change the total angular momentum of the system.


Anyone can be a ballerina? An acrobatic physics teacher stands at the center of a turntableholding his arms extended horizontally with a 5.0 kg dumbbell in each hand. He is set rotating about a vertical axis , makingone revolution in 2.0 s. Find the teacher’s new angular velocity if he pulls the dumbbells in to his stomach, and discuss how this affects the kinetic energy. His moment of inertia is 3.0 kgm2

(without the dumbbells) when his arms are outstretched, dropping to 2.2 kgm2 when his hands are at his stomach. The dumbbells are 1.0 m from the axis initially and 0.20 m from it at the end.


A rotational “collision” 1The figure shows two disks: one an engine flywheel, and the other a clutch plate attached to a transmission shaft. Their moments of inertial are IA and IB respectively. We then push the disks together with forces acting along the axis, so as not to apply any torque on either disk. The disks rub against each other and eventually reach a common final angular speed ω. Derive an expression for ω.


This is how a car’s clutch works!Suppose flywheel A has a mass of 2.0 kg, a radius of 0.20 m, and an initial angular speed of 50 rad/sand that clutch plate B has a mass of 4.0 kg, a radius of 0.10 m, and an initial angular speed of 200 rad/s. Find the common final angular speed ω after the disks are pushed into contact.What happens to the kinetic energy during this process?


Angular momentum in a crime bust• A door 1.00 m wide, of mass 15 kg, is hinged at one

side so that it can rotate without friction about a vertical axis. It is unlatched. A police officer fires a bullet with a mass of 10 g and a speed of 400 m/s into the exact center of the door, in a direction perpendicular to the plane of the door.

• Find the angular speed of the door just after the bullet embeds itself in the door.

• Is kinetic energy conserved?


Homework

• 35, 37, 41, 45


Gyroscopic precession • Consider the figure below (also on page 387). The

normal force adds no torque (its through the pivot) only the weight adds to the torque. The torque is about the y-axis. Why?

• Because torque is along y.

• The angular momentum is:

• Direction?• Lets see this.

dLdt

dL dt

τ

τ

=

=


Gyroscopic precession • What happens if the gyro is spinning initially? Li is

along x-axis, the torque supplied by the weight makes changes the angular momentum direction to make is spin towards the y-axis (counterclockwise from above).

The magnitude of the angular momentum vector does not change, just direction. Why?This is just like centripetal acceleration causing circular motion, just a change in direction, not magnitude.


A rotating flywheel• At the instant shown below the gyroscope has angular

momentum L. A short time later the angular momentum is L + dL. The small change in angular momentum is given by dL= τdt, which is always perpendicular to dL. The flywheel axis turned through a small angle dφ = |dL|/|L|. The rate the axis moves is called the precession angular speed.


Precession Angular Speed

• The precession speed is inversely proportional to the angular velocity of the spinning gyro. The faster the spin, the slower the precession. As the gyroscope spin slows down, the precession speed increases.

• Think of a top: As it spin starts, it processes (wobbles) slowly. After some time, the wobble increases. This is due to friction slowing the spin of the top.

z

z

dL Ld wrdt dt L I

τφω

Ω = = = =


Another gyroscopic processionThe figure below shows a top view of a cylindrical gyroscope wheel that has been set spinning by an electric motor. The pivot is at O, and the mass of the axle is negligible.

a) As seen from above, is the precession clockwise or counter clockwise?

b) If the gyro takes 4.0 s for one revolution of precession, at what angular speed does the wheel spin?


Homework

• 1, 5, 7, 13, 19, 25, 27, 33, 35, 37, 41, 45, 49

• Read 401 - 414

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