physics 2211: lecture 37

Lecture 37, Physics 2211 Spring 2005© 2005 Dr. Bill Holm

Physics 2211: Lecture 37Physics 2211: Lecture 37

Work and Kinetic Energy Rotational Dynamics Examples

Atwood’s machine with massive pulleyFalling weight and pulley

Translational and Rotational Motion CombinedRotation around a moving axisImportant kinetic energy theorem


WorkWork

daxis

r

F

ˆ ds rd

Consider the work done by a force acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement d:

F

cos F F rdd dsW

2co s rdF in sF rd

dW d

W will be negative if and have opposite signs!

W F r

Analog of

For constant : W =

sin rF d


Work & Kinetic EnergyWork & Kinetic Energy

Recall the Work/Kinetic Energy Theorem: WNET = K

This is true in general, and hence applies to rotational motion as well as linear motion.

So for an object that rotates about a fixed axis:

2 212 NET f iW K I


Example: Disk & StringExample: Disk & String

A massless string is wrapped 10 times around a disk of mass M = 40 g and radius R = 10 cm. The disk is constrained to rotate without friction about a fixed axis though its center. The string is pulled with a force F = 10 N until it has unwound. (Assume the string does not slip, and that the disk is initially not spinning).

How fast is the disk spinning after the string has unwound?

F

RM


Disk & StringDisk & String

The work done is W = The torque is = RF (since = 90o)The angular displacement is

2 rad/rev x 10 rev.

F

RM

So W = (.1 m)(10 N)(20rad) = 62.8 J


Disk & StringDisk & String

RM

Recall that I for a disk about

its central axis is given by:

212I MR

So 2 21 12 2 K MR

NET22

4 62.8 J4W

.04 kg .1 m

MR792.5 rad/s

7568 rpm

21262.8 NETW J K I


ExampleExampleWork & EnergyWork & Energy

Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same distance. Disk 1 has a bigger radius, but both have the same moment of inertia. Both disks rotate freely around axes though their centers, and start at rest.Which disk has the biggest angular velocity after the pull ?

(1)(1) disk 1

(2)(2) disk 2

(3)(3) same FF

1 2


ExampleExample SolutionSolution

FF

1 2

d

The work done on both disks is the same!W = Fd

The change in kinetic energy of each will therefore also be the same since W = K.

So since I1 = I2

1 = 2

But we know 21

2K I


Review: Torque and Angular Review: Torque and Angular AccelerationAcceleration

This is the rotational analog of FNET = ma

Torque is the rotational analog of force:Torque is the rotational analog of force:The amount of “twist” provided by a force.

Moment of inertiaMoment of inertia I I is the rotational analog of massis the rotational analog of massIf I is big, more torque is required to achieve a given

angular acceleration.

NET I


ExampleExample RotationsRotations

Two wheels can rotate freely about fixed axles through their centers. The wheels have the same mass, but one has twice the radius of the other. Forces F1 and F2 are applied as shown. What is F2 / F1 if the angular acceleration of the wheels is the same?

(a) 1

(b) 2

(c) 4

F1

F2


ExampleExample SolutionSolution

We know I

so

mRF

mRFR 2

1

2

1

2

1

2

RR

mRmR

FF

F1

F2

Since R2 = 2 R12

FF

1

2

I mR2but andFR


Work & PowerWork & Power

The work done by a torque acting through a displacement is given by:

The power provided by a constant torque is therefore given by:

dW dP

dt dt

W


Atwood’s Machine with Massive PulleyAtwood’s Machine with Massive Pulley

A pair of masses are hung over a massive disk-shaped pulley as shown.Find the acceleration of the blocks.

For the hanging masses use F = ma T1 m1g = m1(-a)

T2 m2g = m2a

I 1

22MR(Since for a disk)

Ia

RMRa

1

2

I Ia

R For the pulley use

T1R - T2Rm2m1

R

M

y

x

m2m1

m2g

aT1

m1g

a

T2

T1 T2


Atwood’s Machine with Massive PulleyAtwood’s Machine with Massive Pulley

m2m1

R

M

y

x

m2m1

m2g

aT1

m1g

a

T2

T1 T2

am m

m m Mg

1 2

1 2 2

We have three equations and three unknowns (T1, T2, a). Solve for a.

T1 m1g = m1a (1)

T2 m2g = m2a (2)

T1 - T2 = 1/2 Ma (3)


Falling weight & pulleyFalling weight & pulley

A mass m is hung by a string that is wrapped around a pulley of radius R attached to a heavy flywheel. The moment of inertia of the pulley + flywheel is I. The string does not slip on the pulley.

Starting at rest, what is the speed of the mass after it has fallen a distance L.

I

m

R

T

mg

a

L

T



For the hanging mass use F = mamg - T = ma

For the pulley + flywheel use = I = TR = I

Realize that a = R

amR

mRg

2

2 I

TRa

RI

Now solve for a using the above equations.

I

m

R

T

mg

a

L

T



Using 1-D kinematics we can solve for the speed of the mass after it has fallen a distance L:

2 20 2fv v a y

2fv aL

where2

2

mgRa

mR

I

2

2

2f

mgLRv

mR

I

I

m

R

T

mg

a

L

T



Conservation of Energy Solution:

2 21 12 2E mv mgy I

0initialE mgL

where v R

2

212 2

vE mR mgy

R I

2

212 2

ffinal

vE mR

R I

initial finalE E y = 0

2

2

2f

mgR Lv

mR

I

I

m

R

T

mg

a

L

T


Rotation around a moving axisRotation around a moving axis

A string is wound around a puck (disk) of mass M and radius R. The puck is initially lying at rest on a frictionless horizontal surface. The string is pulled with a force F and does not slip as it unwinds.

What length of string L has unwound after the puck has moved a distance D?

F

RM

Top view



The CM moves according to F = MA AF

M

D AtF

Mt

1

2 22 2 The distance moved by the CM is thus

F

M A

RI 1

22MR

The disk will rotate about its CM according to = I 21

2

2RF F

MR MR

I

So the angular displacement is 2 21

2

Ft t

MR



So we know both the distance moved by the CM and the angle of rotation about the CM as a function of time:

D

F F

L

The length of string

pulled out is L = R:

L D2

(b)2

Ft

MR

Divide (b) by (a):2

D R

2 R D

(a)2

2

FD t

M


Comments on CM accelerationComments on CM acceleration

We just used = I for rotation about an axis through the CM even though the CM was accelerating! The CM is not an inertial reference frame! Is this OK??

(After all, we can only use F = ma in an inertial reference frame).

YES!YES! We can always write = I for an axis through the CM.This is true even if the CM is accelerating.We will prove this when we discuss angular momentum!

F

R

M A


Important kinetic energy theoremImportant kinetic energy theorem

Consider the total kinetic energy of a system of two masses:

2 21 11 1 2 22 2K m v m v

Now write the velocities as a sum of

the velocity of the center of mass and

a velocity relative to the center of mass

1 1CMv v u

2 2CMv v u

so

2 2 21 1 11 2 1 1 2 2 1 1 2 22 2 2CM CMK m m v m u m u v m u m u

2 2 21 1 1 1 12CM CMv v v v u v u

2 2 22 2 2 2 22CM CMv v v v u v u

= KCM = 0 *= KREL

1 1 2 21 1 2 2 1 1 2 2 1 1 2 2

1 2

0CM CM CM CM

m v m vm u m u m v v m v v m v m v Mv M Mv

m m

*


Thus

KREL is the kinetic energy due to motion relative to the center of mass.

So 212 CM RELK Mv K

is the kinetic energy of the center of mass (M is total mass).212 CMMv

2 2 21 1 11 2 1 1 2 22 2 2CMK m m v m u m u

= KCM = KREL

Important kinetic energy theoremImportant kinetic energy theorem


Connection with rotational motionConnection with rotational motion

For a solid object rotating about its center of mass:

212REL i i

i

K m u

2 21 12 2CM i i

i

K Mv m u

KCM KREL

where i iu r

2 2 2 21 12 2REL i i i i

i i

K m r m r

but2CM i i

i

I m r

21CM2 I

RELK


For a solid object which rotates about its center or mass and whose CM is moving:

VCM

Translational & rotational motion Translational & rotational motion combinedcombined

2 21 12 2NET CM CMK MV I

physics 2211: lecture 37

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