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Physics 203 – College Physics I Department of Physics – The Citadel

Physics 203College Physics I

Fall 2012

S. A. Yost

Chapter 8 Part 2

Rotational Motion

Physics 203 – College Physics I Department of Physics – The Citadel

AnnouncementsProblem set HW08 is due Thursday.

It covers Chapter 8 except angular momentum, rolling, and some torque problems.

Today: Rotational Motion, Ch. 10 , mostly sec7 – 8.

Thursday: chapter 9, sec. 1, 2, and 4 .

Please read these sections before class. A problem set HW09 on them and the remainder of Ch. 8 will be posted today and due next Tuesday.

Deadline for making up Exam 2: Wed. 5 PM.

Physics 203 – College Physics I Department of Physics – The Citadel

Rotational Analogs of Linear Motion

Linear Motion Rotational Motion

Kinematics: x , v, a , , q w a

Dynamics: m, F, Kt I, t, Kr

F = ma = t Ia

Kt = ½ mv 2 Kr = ½ I w 2

W = Fx W = t q

P = F v P = tw

Physics 203 – College Physics I Department of Physics – The Citadel

F

Torque

Rq

t = R F sin q .

The torque is defined to be the perpendicular component of the force times the distance from the pivot to where it acts:

t = R F^

F^

where F^ = F sin q. Then

Counter-clockwise torque is considered to be positive, as for angles.

Physics 203 – College Physics I Department of Physics – The Citadel

F

Torque

Rq

t = R F sin q

The torque can also be expressed in terms of the magnitude of the force and the distance from the axis to the line of the force.

The distance R ^ is called the lever arm of the torque.

t = R F^

R ^

t = R ^ F

Physics 203 – College Physics I Department of Physics – The Citadel

Tightening a Nut with a Wrench

Which use of the wrench is most effective for tightening the nut?

Which is least effective?

Which of A and D is more effective?

Choose E if they are the same.

The lever arms are the same.

A B

C D

Physics 203 – College Physics I Department of Physics – The Citadel

QuestionA force is applied to the rim of two wheels.

Assuming the only significant mass is in the rim, what force F2 will give the wheels identical angular accelerations?

(A) 0.25 N (B) 0.5 N (C) 1.0 N (D) 2.0 N (E) 4.0 N

I = mR2

Physics 203 – College Physics I Department of Physics – The Citadel

Question t = RF = I a = mR2 a.

= = → =

F2 = 2 F1 = 2 N.

I = mR2

t2 R2F2 mR22 a F2 R2

t1 R1F1 mR12 a F1 R1

Physics 203 – College Physics I Department of Physics – The Citadel

Example

Mass falling on rope wrapped around a massive pulley.

Assume the pulley is a uniform disk as shown.

What is the acceleration of the hanging mass?

m

M

R

a

Physics 203 – College Physics I Department of Physics – The Citadel

Example

Isolate the hanging mass:

Newton’s Law:

M a = Fnet = Mg – FT

where FT is the tension in the rope. M

a

mg

FT

Physics 203 – College Physics I Department of Physics – The Citadel

Example

Isolate the pulley: = t I awith I = ½ m R2, t = RFT , a = a/R.

Then RFT = (½ mR2)(a/R).

Therefore, FT = ½ ma.

mR

FT

a

Physics 203 – College Physics I Department of Physics – The Citadel

Example

Combine results:

M a = Fnet = Mg – FT

= Mg – ½ ma.

Then (M + m/2) a = Mg.

Result: a =

mR

FT

a1 + m/2M

g M

Physics 203 – College Physics I Department of Physics – The Citadel

Example

Note that the tension does not need to be the same on two sides of a massive pulley.

Net torque =

R(F1 – F2) = Ia.

m

R

F1

F2

R

a

Physics 203 – College Physics I Department of Physics – The Citadel

Rigid Body Motion

The relation t = I a holds for rigid body rotation in any inertial frame.

This always holds in the CM frame of the rigid body, even if it is accelerating.

The energy of a rigid body can be expressed as a sum K = K cm + K rot with

K cm = ½ mvcm2, K rot = ½ Iw 2.

“Newton’s Law” F = m acm (ch. 7), t = I . a

Physics 203 – College Physics I Department of Physics – The Citadel

Dumbbell

A force F is applied for time t to a dumbbell in one of two ways shown.

Which gives the greater speed to the center of mass?

(a) A (b) B

(c) the same

m

m

m

m

F

F

A

B Dp = Ft → →

→

→

→

Physics 203 – College Physics I Department of Physics – The Citadel

Dumbbell

A force F is applied for time t to a dumbbell in one of two ways shown.

Which gives the greater energy to the dumbbell?

(a) A (b) B

(c) the same

m

m

m

m

F

A

B

→

F→

→

Physics 203 – College Physics I Department of Physics – The Citadel

Dumbbell

The total kinetic energy is

Case A:

K = Ktrans + Krot = ½ mvcm2 + ½ Iw2

Case B: no rotation:

K = ½ mvcm2

There is more energy in case A.

m

m

m

m

A

B

F→

F→

Physics 203 – College Physics I Department of Physics – The Citadel

Rolling

When an object rolls, its circumference moves a distance 2pr every period, so w and v are related:

v = 2pr/T = rw

2pr 2pr 2pr 2pr

Physics 203 – College Physics I Department of Physics – The Citadel

Rolling

A solid wheel and a hollow wheel roll down a ramp, starting from rest at the same point.

Which gets to the bottom faster?

Physics 203 – College Physics I Department of Physics – The Citadel

Rolling

If an object with mass m and moment of inertia I rolls down an inclined plane of height h and length L, how fast is it rolling when it gets to the bottom?

m,I

Lh

Physics 203 – College Physics I Department of Physics – The Citadel

Rolling

Energy conservation:

Ui = Ktrans + Krot

mgh = ½ mv2 + ½ Iw2.

Rolling: w = v/R.

mgh = ½ (m + I/R2) v2.

√ v = 2gh

1 + I/(mR2)

m,I

Lh

Physics 203 – College Physics I Department of Physics – The Citadel

Rolling

The solid wheel gets to the bottom first, because the object with the smaller moment of inertia relative to its mass and size moves faster.

Physics 203 – College Physics I Department of Physics – The Citadel

Rotational Analog of Momentum

Linear Motion: (one dimension)

Momentum: p = mv

Impulse: Dp = Ft

Rotational Motion: (fixed axis)

Angular momentum: L = Iw

DL = t t

Physics 203 – College Physics I Department of Physics – The Citadel

Angular Momentum

Units of angular momentum:

L = I w = [kg m2][s-1] = kg . m2/s

DL = t t = [mN][s] = Nms = J.s

When there is no external torque on a system, angular momentum is conserved.

In particular, this applies to collisions between rigid bodies.

Physics 203 – College Physics I Department of Physics – The Citadel

Figure Skater

A figure skater increases her rotational rate from 1.0 rev/s to 2.5 rev/s in 1.5 s. Her initial moment of inertia was 4.6 kg ∙ m2.

(a) What was her final moment of inertia?

Physics 203 – College Physics I Department of Physics – The Citadel

Figure Skater

A figure skater increases her rotational rate from 1.0 rev/s to 2.5 rev/s in 1.5 s. Her initial moment of inertia was 4.6 kg ∙ m2.

I1w1 = I2 w2

w2 = 2.5 w1

I2 = I1 / 2.5 = 1.84 kg ∙ m2

≈ 1.8 kg ∙ m2

Physics 203 – College Physics I Department of Physics – The Citadel

Figure Skater

A figure skater increases her rotational rate from 1.0 rev/s to 2.5 rev/s in 1.5 s. Her initial moment of inertia was 4.6 kg ∙ m2.

(b) What average power did she apply to pull in her arms?

Physics 203 – College Physics I Department of Physics – The Citadel

Figure Skater

P = W/t, W = DK = ½ Iw22 – ½ Iw1

2.

w1 = 1.0 rev/s (2p rad/rev) = 2.0 p rad/s

w2 = 2.5 rev/s (2p rad/rev) = 5.0 p rad/s

I1 = 4.6 kg ∙ m2, I 2 = 1.84 kg ∙ m2

W = 277 J – 90.8 J ≈ 186 J, t = 1.5 s.

P = 124 W.

Physics 203 – College Physics I Department of Physics – The Citadel

Analog of Inelastic Collision

A bar of length 2R is dropped onto a rotating disk of radius R.

Suppose M = 2m. If the disk initially rotates at 120 rpm, how fast does it rotate if the stick drops onto it and rotates together with the disk?

R w0

2R

M

m

Physics 203 – College Physics I Department of Physics – The Citadel

Analog of Inelastic Collision

Angular momentum is conserved: I1w1 = I2 w2 .

I1 =½ MR2 = mR2 and I2 = Ibar + Idisk with

Ibar = mL2/12 = mR2/3, Idisk = ½ MR2 = mR2

I2 = 4mR2/3

w2= ( I1 / I2) w1 = ¾ w1

= ¾ (120 rpm) = 90 rpm.

RwfM

m

Physics 203 – College Physics I Department of Physics – The Citadel

Angular Quantities as Vectors

If the axis is not fixed, we have to specify a direction for angular displacements and velocities. The convention use a vector pointing along the axis.

For fixed axis rotations, the vectors q and w are parallel, but they won’t be if the axis direction changes.

wq

w = d q /dtq

w→

→

→ →

→ →

Right-Hand Rule