lecture24 - physics.udel.edunowak/phys207/05s/lecture24.pdf(a) case 1 (b) case 2 (c) same l l f f...

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Physics 207: Lecture 24

Recap: Rotational dynamics and torque

Work and energy with exampleMany body dynamics examples

Weight and massive pulley Rolling and sliding examples Rotation around a moving axis: Puck on ice Rolling down an incline Bowling ball: sliding to rollingAtwood’s Machine with a massive pulley

Today’s Agenda

Announcements

• No labs next week, May 2 – 5• Exam 3 review session: Wed, May 4 from 8:00 – 9:30 pm; here

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Review: Torque and Angular Acceleration

τNET = Iα

This is the rotational analogue of FNET = maTorque is the rotational analogue of force:

The amount of “twist” provided by a force.Moment of inertia I is the rotational analogue of mass

If I is big, more torque is required to achieve a given angular acceleration.

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Torque

= r F sin φ= r sin φ F

τ = rpF

Equivalent definitions!

rp = “distance of closest approach”

τ = rFθ

Recall the definition of torque:

rφ

rp

F

φFθ

Fr φ

4

Torque

τ = r Fsin φ

So if φ = 0o, then τ = 0

And if φ = 90o, then τ = maximum

r

F

rF

5

Lecture 23, Act 3Torque

In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same.

(a) case 1

(b) case 2

(c) sameL

L

F F

axis

case 1 case 2

6

Lecture 23, Act 3Solution

Torque = F x (distance of closest approach)

LF F

case 1 case 2

L

Torque is the same!

The applied force is the same.The distance of closest approach is the same.

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Torque and the Right Hand Rule:

The right hand rule can tell you the direction of torque:Point your hand along the direction from the axis to the point where the force is applied.Curl your fingers in the direction of the force.Your thumb will point in the directionof the torque.

r

F

x

y

z

τ

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The Cross Product

We can describe the vectorial nature of torque in a compact form by introducing the “cross product”.

The cross product of two vectors is a third vector:

A X B = C

The length of C is given by:C = AB sin φ

The direction of C is perpendicular to the plane defined by A and B, and inthe direction defined by the right handrule.

A

B

C

φ

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The Cross Product

Cartesian components of the cross product:

C = A X B

CX = AY BZ - BY AZ

CY = AZ BX - BZ AX

CZ = AX BY - BX AY

A

B

C

Note: B X A = - A X B

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Torque & the Cross Product:

r

F

x

y

z

τ

So we can define torque as:

τ = r X F= rF sin φ

τX = rY FZ - FY rZ = y FZ - FY z

τY = rZ FX - FZ rX = z FX - FZ x

τZ = rX FY - FX rY = x FY - FX y

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Comment on τ = Iα

When we write τ = Iα we are really talking about the zcomponent of a more general vector equation. (Recall that we normally choose the z-axis to be the the rotation axis.)

τz = Izαz

We usually omit the z subscript for simplicity.

z

αz

τz

Iz

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Example

To loosen a stuck nut, a (stupid) man pulls at an angle of 45o on the end of a 50 cm wrench with a force of 200 N.

What is the magnitude of the torque on the nut?If the nut suddenly turns freely, what is the angular acceleration of the wrench? (The wrenchhas a mass of 3 kg, and its shape is that of a thin rod).

L = 0.5 m

F = 200 N

45o

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Example

L = 0.5m

F = 200 N45o

α

Torque τ = LFsin φ = (0.5 m)(200 N)(sin 45) = 70.7 Nm

If the nut turns freely, τ = IαWe know τ and we want α, so we need to figure out I.

( )( ) 222 kgm250m50kg331ML

31 ..I ===

α = 283 rad/s2

So α = τ / I = (70.7 Nm) / (0.25 kgm2)

14

Work

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

dW = F.dr = FR dθ cos(β)= FR dθ cos(90-φ)= FR dθ sin(φ)= FR sin(φ) dθ

dW = τ dθ

We can integrate this to find: W = τθAnalogue of W = F •∆rW will be negative if τ and θ have opposite signs!

R

F

dr = R dθdθaxis

φ

β

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Work & Power

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

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

W = τθ

P dWdt

ddt

= = =τθ

τω

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Work & Kinetic Energy:

Recall the Work/Kinetic Energy Theorem: ∆K = WNET

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:

( ) NET2i

2f W

21K =−=∆ ωωI

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Example: 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

R M

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Disk & String...

The work done is W = τ θThe torque is τ = RF (since φ = 90o)The angular displacement θ is2π rad/rev x 10 rev.

F

R M

θτ

So W = (.1 m)(10 N)(20π rad) = 62.8 J4342144 344 21

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Disk & String...

WNET = W = 62.8 J = ∆K =12

2I ω

Recall that I for a disk aboutits central axis is given by:

I =12

2MR

∆K MR W= ⎛⎝⎜

⎞⎠⎟

=12

12

2 2ωSo

( )( )( )

ω = =4 4 62 8

04 12 2W

MRJ

kg.

. .ω = 792.5 rad/s

ω

R M

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Lecture 23, Act 4Work & 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 ?

(a) disk 1

(b) disk 2

(c) sameFF

ω1ω2

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FF

ω1ω2

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.

But we know ∆K = 12

2I ω

So since I1 = I2

ω1 = ω2

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Lecture 24, Act 1Rotations

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) 4F1

F2

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We know

I = mR2but andτ = FR

τ α= I

soαα

mRFmRFR 2

==

1

2

1

2

1

2

RR

mRmR

FF

==αα

F1

F2

Since R2 = 2 R12

FF

1

2 =

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Falling weight & pulley

A mass m is hung by a string that is wrapped around a pulley of radius Rattached 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, how long does it take for the mass to fall a distance L.

I

m

R

T

mg

α

aL

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Falling weight & pulley...

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

For the pulley + flywheel use τ = Iα

τ = TR = IαRealize that a = αR

Now solve for a using the above equations.

I

m

R

T

mg

α

aL

a mRmR

g=+

⎛

⎝⎜

⎞

⎠⎟

2

2 I

TR aR

= I

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Falling weight & pulley...

Using 1-D kinematics (Lecture 2) we can solve for the time required for the weight to fall a distance L: I

m

R

T

mg

α

aL

a mRmR

g=+

⎛

⎝⎜

⎞

⎠⎟

2

2 I

L at=12

2 t La

=2

where

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Rotation 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 horizontalsurface. 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?

FR

M

Top view

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Rotation around a moving axis...

The CM moves according to F = MA

F

M A

A FM

=

D At FM

t= =12 2

2 2The distance moved by the CM is thus

α RI =12

2MR

MRF2

MR21

RFI 2

===τ

αThe disk will rotate about its CM according to τ = Iα

θ α= =12

2 2t FMR

tSo the angular displacement is

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Rotation around a moving axis...

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

D FM

t=2

2θ =

FMR

t 2

F

θ

Divide (b) by (a):

(a) (b)

θD R

=2

R Dθ = 2

L

The length of stringpulled out is L = Rθ:

L D= 2

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Comments on CM acceleration:

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

FR

M Aα

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Rolling

An object with mass M, radius R, and moment of inertia Irolls without slipping down a plane inclined at an angle θwith respect to horizontal. What is its acceleration?

Consider CM motion and rotation about the CM separately when solving this problem (like we did with the lastproblem)...

θ

R

I

M

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Rolling...

Static friction f causes rolling. It is an unknown, so we must solve for it.First consider the free body diagram of the object and use FNET = MACM :

In the x direction Mg sin θ - f = MA

Now consider rotation about the CMand use τ = Iα realizing that

τ = Rf and A = αR R

M

θ

f

Mg

y

x

Rf AR

= I f AR

= I 2

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Rolling...

We have two equations:

We can combine these to eliminate f:

f I AR2=

+=

IMR sinMR

gA 2

2 θ

θ

A R

I

M

θθ

sing75

MR52

MR

sinMRgA

22

2

=+

=

For a sphere:

sin -Mg f mAθ =

34


Two uniform cylinders are machined out of solid aluminum. One has twice the radius of the other.

If both are placed at the top of the same ramp and released, which is moving faster at the bottom?

(a) bigger one

(b) smaller one

(c) same

35


Consider one of them. Say it has radius R, mass M and falls a height H.

H

Energy conservation: - ∆U = ∆K MgH MV= +12

12

2 2I ω

I = 12

2MR ω = VR

but and

MgH MR VR

MV= ⎛⎝⎜

⎞⎠⎟

+12

12

12

22

22

MgH MV MV MV= + =14

12

34

2 2 2

36


H

MgH MV= 34

2So: gH V= 34

2

V gH=43

So, (c) does not depend on size,

as long as the shape is the same!!

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Sliding to Rolling

A bowling ball of mass M and radius R is thrown with initial velocity v0. It is initially not rotating. After sliding with kinetic friction along the lane for a distance D it finally rolls without slipping and has a new velocity vf. The coefficient of kinetic friction between the ball and the lane is µ.

What is the final velocity, vf, of the ball?

vf= ωRω

f = µMgv0

D

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Sliding to Rolling...

While sliding, the force of friction will accelerate the ball inthe -x direction: F = -µMg = Ma so a = -µgThe speed of the ball is therefore v = v0 - µgt (a)Friction also provides a torque about the CM of the ball.Using τ = Iα and remembering that I = 2/5MR2 for a solid sphere about an axis through its CM:

D

x

αµτ 2MR52

MgR == R2g5µ

α =

f = µMg

tR2g5

t0µ

αωω =+= (b)

v f= ωRω

v0

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Sliding to Rolling...

We have two equations:

Using (b) we can solve for t as a function of ω:

Plugging this into (a) and using vf = ωR (the condition for rolling without slipping):

D

x

tR2g5µ

ω =v v gt= −0 µ (a) (b)

t Rg

=25

ωµ

v vf =57 0

f = µMg

Doesn’t depend on µ, M, g!!

vf= ωRω

v0

40


A bowling ball (uniform solid sphere) rolls along the floor without slipping.

What is the ratio of its rotational kinetic energy to its translational kinetic energy?

I = 25

2MRRecall that for a solid sphere about

an axis through its CM:

(a) (b) (c) 255

1 12

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The total kinetic energy is partly due to rotation and partly due to translation (CM motion).

rotational

K

translational

K

12

2I ω +12

2MVK =

42


ω = VR

Since it rolls without slipping:

rotational

K

Translational

K

12

2I ω +12

2MVK =

2

2

TRANS

ROT

MV21

I21

KK ω

=52

MVRVMR

52

2

2

22

=⎟⎠⎞

⎜⎝⎛

=

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Atwoods Machine with Massive Pulley:

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

Find the acceleration of the blocks.

m2m1

R

M

y

x

m2g

aT1

m1g

a

T2

For the hanging masses use F = ma-m1g + T1 = -m1a-m2g + T2 = m2a

α

= =I aR

MRa12

= I aR

I =12

2MR(Since for a disk)

For the pulley use τ = Iα

T1R - T2R

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Atwoods Machine with Massive Pulley...

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

-m1g + T1 = -m1a (1)

-m2g + T2 = m2a (2)

T1 - T2 (3)

am m

m m Mg=

−+ +

⎛⎝⎜

⎞⎠⎟1 2

1 2 2

=12

Ma

m2m1

R

M

y

x

α

m2m1

m2g

aT1

m1g

a

T2

lecture24 - physics.udel.edunowak/phys207/05s/lecture24.pdf(a) case 1 (b) case 2 (c) same l l f f...

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