physics 111 - valparaiso university · physics 111 title page tuesday, november 2, 2002 physics 111...

Physics 111

Title page

Tuesday,November 2, 2002

Physics 111 Lecture 18

• Ch 11: Rotational DynamicsTorqueAngular MomentumRotational Kinetic Energy

• Wednesday, 8 - 9 pm in NSC 118/119• Sunday, 6:30 - 8 pm in CCLIR 468

Help sessions

AnnouncementsTuesNov

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This week’s lab will be on oscillations.There will be a short quiz.

labs

AnnouncementsTuesNov

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Let’s look at a balance scale...

50 g

L

If I hang a 50 g mass atthe end of the left sideof the balance...

250 g

…where should I hangthe 250 g mass to getthe scale to balance?

X?

? Balancing (W1)

Worksheet #1

Ch 11: Rotational DynamicsTuesNov

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50 g

L

250 g

L/5

How did we arrive at such an answer?

W1

W2

=L

2

L1

Perhaps we used one of theseformulae

m1L

1= m

2L

2

how did we get that?


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Let’s restate our balance condition so that weget all the subscript 1’s on the same side andall the subscript 2’s on the same side...

L1W

1= L

2W

2

50 g

L

250 g

L/5

con’t


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W1

W2

=L

2

L1

So in the case of this balance, we have...

L(50 g)(9.8 m/s2 ) = (L / 5)(250 g)(9.8 m/s2 )

490 gm/s2 L = 490 gm/s2 L Good! The balanceis in balance!

L1W

1= L

2W

2

con’t


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50 g

L

250 g

L/5

What happens toour balance if Ichange the masson the left sidefrom 50 g to 25 g?

25 g

L

250 g

L/5

The scale beginsto rotate aroundthe pivot point.

25 g

L

250 g

L/5

Out of balance?


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Let’s look at one more case with our balance...

100 g

L

100 g

X?

50 g

L

X = L/2

How did we arrive atthis answer?

Now where should I hang the 100 gmass to get the scale to balance?

Worksheet #2

? 3 mass balance (W2)


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First, we examined the left side of the balance:

L

1W

1= L(100 g)(9.8 m/s2 ) = 980L gm

s2

soln: left


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100 g

L

100 g

X?

50 g

L

Then, we examined the right side:

L2W

2+ L

3W

3= X (100 g)(9.8 m/s2 ) + L(50 g)(9.8 m/s2 )

980X gm

s2 + 490L gm

s2

soln: right


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100 g

L

100 g

X?

50 g

L

If the scale is balanced, the two sides shouldequal one another:

980L gm

s2 = 980 X gm

s2 + 490L gm

s2

X = L / 2

soln: equal


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100 g

L

100 g

X?

50 g

L

• In our scale experiments, we’ve been playingaround with a physical quantity that we’venot yet encountered formally.

• Our experiments show us that this quantity isrelated to a force applied at a distance.

• The farther away from the pivot point the forceis applied, the greater our new quantity is.

Summary 1


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• We’ve also seen that the quantity is additive.That is, if we have two forces on one side of thepivot at two different distances, the resultingphysical quantity is simply the sum of the twoforce * distance products.

• Before we name our new physical quantity,let’s examine one more thought experiment.

• When present and unbalanced, our new physicalquantity causes the scale to rotate in thedirection of the unbalanced force.

summary 2


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What will eventually happen to this system?(Assume there is SOME friction in the pivot,but that the pivot is free to rotate 360o.)

50 gLet’s put a block on

one side of ourbalance and let go.

L

One sided


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The block eventuallycomes to rest

with the balancealigned vertically.

50 g

L

We still have a mass of 50 gat a distance L from the pivotpoint, so why has the motionstopped? What hashappened to the value of ournew physical quantity?

stops moving


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For this case, we notice thatthe gravitational force on themass is acting along thelength of the balance.

In other words, the force isparallel to the radius of lengthL around the pivot point.

In this case, themagnitude of ournew physicalquantity must be

nothing


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50 g

L

Let’s finally give a name tothis new physical quantity. How about...

Where d is the distance from the pivot point (thepoint of rotation) to the point at which the forceis applied and…

F⊥

is the component of the appliedforce that is perpendicular to theline joining the pivot point to thepoint at which the force is applied.

|τ |= F⊥d

Def: torque


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[|τ |] = [F⊥ ][d]

[|τ |] = N m

Although torque has the same units as energy(J), we generally denote torques as forcesacting at a distance, and therefore leave theunits in the form N m.

Units: torque


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|τ |= F⊥d

50 g

0.5 m

What is themagnitude of thetorque experiencedby this unbalancedbalance?

Fg

Here, the force ofgravity acts

perpendicularlyto the radius joining

the two yellowpoints above.

pivot

|τ |= mgd = (0.05 kg)(9.8 m/s2 )(0.5 m) = 0.245 Nm

|τ |= F⊥d

One-sided balance


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CQ3: torque

Worksheet #3


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Worksheet #3

CQ3 – torque ?

You are using a wrench andtrying to loosen a rusty nut.Which of the arrangementsshown is most effective inloosening the nut? List in orderof descending efficiency thefollowing arrangements:

PI, Mazur (1997)

≥≥

1) 1, 2, 3, 4 5) 1, 3, 4, 22) 4, 3, 2, 1 6) 4, 2, 3, 13) 2, 1, 4, 3 7) 2, 3, 4, 14) 2, 4, 1, 3 8) 2, 3, 1, 4


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The tangential force resultsin a tangential acceleration. Ft

= mat

Torque & alpha

Ft

pinned totable

r

m


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It also creates a torque about the pinned point.

|τ |=|

F

t||r |= m |

a

t||r |

con’t


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Ft

pinned totable

r

m

Recalling that tangential acceleration is relatedto angular acceleration, we get:

|τ |= m |

a

t||r |= m(|

r ||α |) |

r |= mr 2 |

α |

con’t


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Ft

pinned totable

r

m

This expression is valid solong as our connectingrod/string is massless.

τ α= mr2

con’t


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Ft

pinned totable

r

m

If we examine a more general system and plotthe relationship, we find

α

τThe slope of thisline is known asthe moment ofinertia, I.

For our previousexample,

I mr= 2Note: r is the distance tothe axis of rotation.

Moment of inertia


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2Imr= [I ] = [m][r 2]

[I ] = kg m2

If we have an extended object, we can computeits moment of inertia about some axis from thesum of the moments of each of the individualpieces of that object:

I = m

1r12 + m

2r

22 + m

3r

32 + ...+ m

Nr

N2 = m

ir

i2

i=1

N

∑

Units: I


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So we can rewrite ourexpression for torquein terms of our newquantity, the moment ofinertia...

τ α= I

Notice the similarityto our linear motionexpression of

F ma=

Back to torque


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Worksheet #4

CQ4: two wheels


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Worksheet #4

CQ4 – two wheels ?

Two wheels with fixed hubs, each having a mass of 1 kg,start from rest, and forces are applied as shown. Assume thehubs and spokes are massless, so that the rotational inertiais I = mR2. In order to impart identical angularaccelerations, how large must F2 be?

PI, Mazur (1997)

≥≥

1. 0.25 N2. 0.5 N3. 1 N4. 2 N5. 4 N


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For the ball to fall into thebasket, the linear accelerationof the end of the stick mustexceed that of gravity.

Is that possible?

Soln to Demo


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The moment of inertia of a stickrotated about its end is given by

I = 1

3ML2

1) Yes 2) No

Soln to Demo


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τ

net= F⊥R = I

α

The force creating the torque in this case isthat of gravity, acting at the center of the stick.

τ

net= Mg(cosθ )

L2= 1

3ML2 α

Soln to Demo


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To get the linearacceleration of the end ofthe stick, we simply multiplyby the length of the stick.

α = 3g2L

(cosθ )

a

tan= αL = 1.5g(cosθ )

So, if the cos θ > 2/3, the end of the stickwill move with a linear acceleration > g.

θ < 480

Recall in ourdiscussion of force,we discovered

F = ma = m Δv

Δt=ΔpΔt

p mv=

where p was themomentumin the system

Angular Momentum


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We can construct a completely analogousargument to define angular momentum

τ = I α = I Δ

ωΔt

=ΔLΔt L I= ωwhere L is the

angular momentum

con’t


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L = I

ω

[L] = [I ][

ω ]

[L] = (kg m2 )(rad/s) = kg m2

s

Units: Angular Momentum


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When there is no NET external torqueon a system, the angular momentumof the system will be conserved.

So when examining isolated systems,total energy, linear momentum, and angularmomentum are all conserved!

L I I Li i f f= = =ω ω

Conservation of AngularMomentum


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We’ve studied this quantityalready in looking at massesat some fixed distance, r,from an axis of rotation.

K mv= 12

2

When looking at an object rotating at a constantangular velocity ω at a distance r from the axis,

K m r mr I= = =12

2 12

2 2 12

2( )ω ω ω

Rotational Kinetic Energy


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K = 1

2Iω 2

[K ] = [ 1

2][I][ω 2 ]

[K ] = (kg m2 )(rad/s)2

[K ] = kg m2

s2= Nm = J

Units: rotational kinetic energy


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So we’ve defined another form of kinetic energy.

Before, we studied translational kinetic energy.Now we have rotational kinetic energy, too.

So the total kinetic energy of a system nowincludes two terms (if the system is rotatingand translating):

K K Ktot r t= +

Total kinetic energy


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Worksheet #5

CQ5: L and Krot


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CQ5 – L and Krot ?

A figure skater stands on one spot on the ice (assumedfrictionless) and spins around with her arms extended.When she pulls in her arms, she reduces her rotationalinertia and her angular speed increases so that herangular momentum is conserved. Compared to her initialrotational kinetic energy, her rotational kinetic energyafter she has pulled in her arms must be

PI, Mazur (1997)

≥≥

1. the same.2. larger because she’s rotating faster.3. smaller because her rotational inertia is smaller.


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Worksheet #5

E K K U Utot r t g spring= + + +( )

So, when we examine the total energy ofa system, we must take care to include therotational kinetic energy term as well.

The total energy (including rotational kineticenergy) will be conserved so long as NO non-conservative forces are acting upon the system.

Energy Conservation


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We have studied objects in “equilibrium” before.

What is the defining characteristicof an object in equilibrium?

And from our earlier studies, the fact that itdid not accelerate indicated that:

Fnet = 0

Equilibrium


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But is this a sufficient condition to ensurethat in fact an object does not acceleratein some way?

This “zero net force” conditiononly guarantees that the center of massof the object on which the forces arebeing exerted does not accelerate.

Let’s look at the meter stick again...

Necessary but not sufficient


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Exert a constant forcetangential to the circleof motion.

Holding pencil through thepivot point.

Draw a free-bodydiagram for thissystem.

Assume the system isOriented in a horizontal plane

Top view

Rotating ruler (W6)

Worksheet #6


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Is the net force in this system zero?

So why does the ruler rotate withincreasing speed?

Because the net torque is NOT zero!

We have a new equilibrium condition:τ net = 0

New equilibrium condition


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For an object to be in STATIC equilibrium(i.e., at rest, not moving, not rotating), thefollowing TWO conditions must be met.

Fnet = 0

τ net = 0

(We’re free to pick ANY origin about which tocomputer our torques in this case.)

New equilibrium


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A student sits on a rotating stool holding twoweights, each of mass 3.00 kg. When his armsare extended horizontally, the weights are each1.00 m from the axis of rotation, and he rotatesat an angular speed of 0.750 rad/s. Themoment of inertia of the student + stool is 3.00kg m2 and is assumed to be constant. Thestudent now pulls the weights horizontally in toa distance of 0.300 m from the rotation axis. (a)What’s the new angular speed? (b) What are theinitial and final kinetic energies of this system?

? Angular momentum (P1)

Problem #1


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(a) What’s the new angular speed? (b) What are theinitial and final kinetic energies of this system?

soln Angular momentum (P1)


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Knowns: Istudent = 3.00 kg m2

m = 3.00 kg ri = 1.00 m rf = 0.30 m ωi = 0.75 s-1

Unknowns: ωf

Pictorial Representation:

Initially: ωi

rim

Finally: ωf

rf m



Itotal = Iweights + Istudent = 2 (mr2) + (3 kg m2)

Itotal = 2 (3 kg)(1 m)2 + (3 kg m2) = 9.00 kg m2Initially:

Itotal = 2 (3 kg)(0.3 m)2 + (3 kg m2) = 3.54 kg m2Finally:

Ii ωi = If ωf

ωf=

Ii

If

ωi= (9kg m2 )

(3.54kg m2 )(0.75rad/s) = 1.91rad/s


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Ki = 0.5 Ii ωi2 = 0.5 (9 kg m2)(0.75 rad/s)2 = 2.53 J

Initially:

Finally:

Kf = 0.5 If ωf2 = 0.5 (3.54 kg m2)(1.91 rad/s)2 = 6.45 J


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A uniform horizontal beam weighs 300 N, is 5.00 mlong, and is attached to a wall by a pin connectionthat allows the beam to rotate. Its far end issupported by a cable that makes an angle of 53o

with the horizontal. If a 600-N person stands1.50 m from the wall, find the tension in the cableand the force exerted by the wall on the beam.

? Reaction forces (P2)

1.5 m

5.0 m

53o

Problem #2


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A uniform horizontal beam weighs 300 N, is 5.00 mlong, and is attached to a wall by a pin connectionthat allows the beam to rotate. Its far end issupported by a cable that makes an angle of 53o

with the horizontal. If a 600-N person stands1.50 m from the wall, find the tension in the cableand the force exerted by the wall on the beam.

Soln Reaction forces (P2)

Free-bodydiagramfor theplank

5.0 m 53o

T

1.5 mFc

N

f

WW

2.5 m


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From the free-bodydiagram we can writedown the torque equationand Newton’s 2nd Law inthe x and y directions.

5.0 m 53o

T

1.5 mFc

N

f

WW

2.5 m

x

y

Compute the torques about the left end of theplank, where the plank meets the wall:

τnet

= f (0m) − Fc(1.5m) −W (2.5m) + (T sin530 )(5.0m) = 0

0 = 0 − (600N)(1.5m) − (300N)(2.5m) + (4.0m)T

T = 412.5 N


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5.0 m 53o

T

1.5 mFc

N

f

WW

2.5 m

x

y

Newton’s 2nd Law in the x-direction:

Fnet ,x

= N − T cos530 = 0

0 = N − (412.5N)cos530

N = 248 N


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5.0 m 53o

T

1.5 mFc

N

f

WW

2.5 m

x

y

Newton’s 2nd Law in the y-direction:

Fnet , y

= f + T sin530 − Fc−W = 0

0 = f + (412.5N)sin530 − 600N − 300N

f = 571 N


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physics 111 - valparaiso university · physics 111 title page tuesday, november 2, 2002 physics 111...

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