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

PowerPoint® Lectures for University Physics, Twelfth Edition – Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 10

Dynamics of Rotational Motion

Goals for Chapter 10 •  Torque: “angular force” •  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

Review from Ch 9

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l = Rθ

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atan =Δ vtanΔt

= Rα

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arad =v2tanR

= Rω 2

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vtan =ΔlΔt

= Rω

Relation between linear quantities and angular quantities (by geometry)

Kinematic equation just like before: θ like displacement Δx ω like velocity α like acceleration

Kinetic energy of a rotating object about a fixed (stationary) axis:

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KErot =12Iω 2

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I = miri2

i∑ Moment of inertia

or “angular mass” (DEPENS ON THE AXIS CHOSEN)

Torque: Angular force •  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 is linearly proportional to the torque produced.

Lines of force and calculations of torques

The direction of the torque is given by the RHR

Just like moment of inertia, torque is always measured with respect to a point/axis around which the object rotates (Pivot Point)

Magnitude of torque is given by:

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τ = rF sinθ = rFtan

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τ = r × F

Torque of force F wrt. applied a distance r away:

Calculating an applied torque

THREE WAYS TO CALCULATE THE MAGNITUDE (EQUIVALENT) 1st: find the “leverage arm” (shortest distance between PP and line determined by the force, i.e. perpendicular to the force and going through PP) and then multiply by the magnitude of F:

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l = (0.80)cos(19) m

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l = (0.80)cos(19) m

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⇒τ = Fl = (900 N)(0.80)cos(19 ) m = 680 Nm2nd: find the tangential component of the force (perpendicular to the line connecting the PP and where the force is applied) and multiply by this distance (note that this distance is not the “line of action”)

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Ftan = (900 N) cos (19 ) = 851 N

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⇒τ = Ftanr = (851 N)(0.80) m = 680 Nm

Ftan

3rd: Just find the magnitude of the cross product by multiplying F, r, and sine of the angle between them

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⇒τ = Frsin(θF −r ) = (900 N)(0.80 m) sin(71 ) = 680 Nm

Newton’s 2nd Law in angular sense: Σ τ = Iα is just like Σ F = ma

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τ∑z

= Iα z

Sum of torques on a rigid object AROUND a PP (O in the figure)

Moment of inertia AROUND the same PP (O in the figure)

The label z is to remind you that the actual torque and angular velocity vector point perpendicular to the x-y plane defined by the force and the leverage arm

This angular equation is IN ADDITION to any linear force which moves the center of mass but as long as the axis of rotation is fixed we do not need to worry about the linear motion. This changes when the rotating object is also moving

Flywheel problem from Ch 9 (using work energy theorem)

The cable is wrapped around a cylinder. If it unwinds 2.0 m by pulling it with a force of 9.0 N and it starts at rest, what is its final angular velocity and velocity of the cable? (use work energy theorem)

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Wtotal = KE f −KEi

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FΔx =12Iω 2 − 0

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ω =2FΔxI

=4FΔxmR2 = 20 rad/s

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v = Rω = (20 rad/s)(0.060 m) =1.2 m/s

Flywheel problem using torque (using work energy theorem)

The cable is wrapped around a cylinder. If it unwinds 2.0 m by pulling it with a force of 9.0 N and it starts at rest, what is its final angular velocity and velocity of the cable? (use work energy theorem)

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τ = FR = (9.0 N)(0.06 m)

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I =12MR2

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τ = Iα ⇒ α =τI

=2FRMR2 =

2FMR

= 6.0 rad/s2

Use α to get acceleration of the cable:

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atan = Rα = (0.06m)(6.0 rad/s2) = 0.36 m/s2

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v2 = v02+ 2atan (x − x0)Then use kinematics

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v = 0 + 2(0.36 m/s2)(2 m) =1.2 m/s

What is the velocity of the block when it hits the ground?

The work done by the cable is zero since the two tension forces cancel each other out so energy is conserved

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KEi + PEi = KE f + PE f

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0 +mgh =12mv2 +

12Iω 2 + 0

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0 +mgh =12mv2 +

1212MR2

⎛

⎝ ⎜

⎞

⎠ ⎟ vR⎛

⎝ ⎜

⎞

⎠ ⎟ 2

+ 0

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v = 2mgh(m +M/2)

Another look at the unwinding cable What is the linear acceleration of the block?

These are two coupled objects; one rotates and the other moves linearly

For the rotating wheel we have:

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τ = Iα

TR =12MR2 a

R⎛

⎝ ⎜

⎞

⎠ ⎟ ⇒ T =

12Ma

For the block we have:

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mg −T = ma

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mg − 12Ma = ma

⇒ a =mg

m + M /2

Combine the two equations to get

A rigid body in motion about a moving axis

When an object is moving and rotating at the same time we can show that the total kinetic energy is:

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KE =12mvcm

2 +12Icmω

2

That is, there are two parts, the motion of the center of mass and the rotation about the center of mass. Keep in mind that it is ONLY ABOUT THE CENTER OF MASS when this applies. Any other point of rotation it will not work.

Rolling with and without slipping

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KE =12mvcm

2 +12Icmω

2 =12mvcm

2 +12IcmR2v2

When rolling without slipping then

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vcm = Rω This is the condition to roll without slipping.

Then, if you are rolling without slipping the kinetic energy is

Also note that when one is rolling without slipping (i.e. rolling down an incline) the friction force is static so no work is done by it and energy is conserved in this case.

Consider the speed of a yo-yo toy What is the speed of the Yo-yo at the bottom (use conservation of energy)

Why conservation of energy: the hand is not moving so it does no work on the system. You may be confused about the tension but keep in mind that it is an internal force so the sum of the upper and lower tension is zero.

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Ei = E f

0 + Mgh =12Mvcm

2 +12Icmω

2

0 + Mgh =12Mvcm

2 +1212MR2

⎛

⎝ ⎜

⎞

⎠ ⎟ vcmR

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

Mgh =34Mvcm

2

⇒ vcm =43gh

The race of objects with different moments

Let’s figure out which circular object has the largest velocity when it reaches the bottom of the ramp. They roll without slipping so energy is conserved.

h

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Ei = E f

0 + Mgh =12Mvcm

2 +12Icmω

2

0 + Mgh =12Mvcm

2 +12Icm

vcmR

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

=12Mvcm

2 +12IcmMR2

vcm2

Mgh =12(1+

IcmMR2

)Mvcm2

⇒ vcm =2gh

1+IcmMR2

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vcmsphere =

2gh

1+25

=10gh7

vcmring =

2gh1+1

= gh

vcmcylinder =

2gh

1+12

=4gh3

Combined translation and rotation: dynamics

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F ∑ = m a cm

τz∑ = Icmαz

When an object is moving and rotating then the dynamics are determined by considering the linear acceleration of the center of mass and the rotation about the center of mass (i.e. the torques considered are around the center of mass)

The yo-yo (again) This time let’s calculate the acceleration

The linear Newton’s equation read:

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y : Fy∑ = Macm−yMg −T = Macm

+

+

-

-

The rotational Newton’s equation read:

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θ : τ z∑ = Iαz

τMg +τT = Iα z

0 + RT =12MR2 acm

R

T =12Macm

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⇒ Mg − 12Macm = Macm

acm =23g

Consider the acceleration of a rolling sphere

The rotational Newton’s equation read:

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θ : τ z∑ = Iαz

τFN +τMg +τ fric = Iα z

0 + 0 + RFf =25MR2 acm

R

Ff =25Macm

The linear Newton’s equations read:

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x : Fx∑ = macm−x

y : Fy∑ = 0

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Mgsinβ − Ff −stat = macmFN −Mgcosβ = 0

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⇒ Mgsinβ − 25Macm = Macm

acm =57gsinβ

Work and power in rotational motion Work done by a torque

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Wτ = τzdθθ1

θ 2

∫

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Wτ = τz (θ2 −θ1)Work done by a any torque Work done by a constant torque

Power delivered by a torque

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Pτ =dWdt

= τzω z

Angular Momentum

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τ = r × F Recall that torque was defined as

Similarly the angular momentum of a particle is

Also, just as before for linear momentum we can show that the rate of change of the angular momentum on a particle is equal to the torque on that particle

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d L

dt=

ddt

( r × m v ) =d r dt× m v + r × m d v

dt= v × m v + r × m a = 0 +

r × F

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⇒ τ =

d L

dtFOR A RIGID BODY ONE USES THE MOMENT OF INERTIA

TO DEFINE THE ANGULAR MOMENTUM

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L = I ω

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L = r × p = r × m v

Conservation of angular momentum

Before we saw that if the external forces on a system are zero then linear momentum is conserved. Similar for angular momentum.

If the external torques on a system are zero then the TOTAL angular momentum of the systems is conserved.

Note that in this case the TORQUE has to be zero. There can be still forces acting on the system but they do not generate any torque (e.g. force due to gravity on the cat which acts at the center of mass)

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I1ω1 = I2ω 2

How a car’s clutch work

The clutch disk and the gear disk is pushed into each other by two forces that do not impart any torque, what is the final angular velocity when they come together?

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Lz−before = Lz−afterIAωA + IBωB = (IA + IB )ω final

⇒ ω final =IAωA + IBωB

(IA + IB )

Angular momentum conservation in collisions

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 intelf in the door.

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Lz−before = Lz−aftermBvB l = (Idoor +mBl

2)ω f

mBvB l = (13Md2 +mBl

2)ω f

⇒ ω f =mBvB l

(13Md2 +mBl

2)= 0.4 rad/s

Exercise 10.16 A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.00-kg weight by a thin, light wire that passes without slippage over a frictionless pulley (the figure ). The pulley has the shape of a uniform solid disk of mass 2.10 kg and diameter 0.600 m. Find the tensions in the cables and the acceleration.

Problem worked out in class

Exercise 10.20 A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the string is held in place and the hoop is released from rest. After the hoop has descended 90.0 cm, calculate the speed of its center of mass.

Problem worked out in class

Exercise 10.22 A hollow, spherical shell with mass 2.25 kg rolls without slipping down a slope angled at 31.0 degrees. Find the acceleration, the frictional force and the minimal coefficient of static friction to prevent it from slipping.

Problem worked out in class

Exercise 10.39 Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly 1014 times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 6.0×105 km (comparable to our sun); its final radius is 17 km. If the original star rotated once in 32 days, find the angular speed of the neutron star.

Problem worked out in class

Exercise 10.70 A thin-walled, hollow spherical shell of mass m and radius r starts from rest and rolls without slipping down the track shown in the figure . Points A and B are on a circular part of the track having radius R. The diameter of the shell is very small compared to h0 and R, and rolling friction is negligible. What is the minimum height for which this shell will make a complete loop-the-loop on the circular part of the track?

Problem worked out in class

Gyroscopic precession •  The precession of a

gyroscope shows up in many “common” situations.