UB, Phy101: Chapter 9, Pg 1 Physics 101: Chapter 9 l Today’s lecture will cover Textbook Sections 9.1 - 9.6.

UB, Phy101: Chapter 9, Pg 1

Physics 101: Physics 101: Chapter 9 Chapter 9

Today’s lecture will cover Textbook Sections 9.1 - 9.6


Rotation Summary Rotation Summary (with comparison to 1-D kinematics)(with comparison to 1-D kinematics)

Angular Linear

constant

tαωω 0

0 021

2t t

constanta

v v at 0

x x v t at 0 021

2

And for a point at a distance R from the rotation axis:

x = Rv = Ra = R

See text: chapter 8

See Table 8.1

Δθ2 22

0 x2a vv 22

0


New concept: TorqueNew concept: Torque

See text: chapter 9

Rotational analog of force

Torque = (magnitude of force) x (lever arm)

= F l


Comment on axes and signComment on axes and sign(i.e. what is positive and negative)(i.e. what is positive and negative)

Whenever we talk about rotation, it is implied that there is a rotation “axis”.

This is usually called the “z” axis (we usually omit the z subscript for simplicity).

Counter-clockwise (increasing ) is usuallycalled positive.

Clockwise (decreasing ) is usuallycalled negative. z


Chapter 9, PreflightChapter 9, Preflight

The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case.

In which of the cases is the torque on the nut the biggest?

1. Case 1 2. Case 2 3. Case 3

CORRECT



The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case.

In which of the cases is the torque on the nut the smallest?

1. Case 1 2. Case 2 3. Case 3

CORRECT




Static EquilibriumStatic Equilibrium

A system is in static equilibrium if and only if:

acm = 0 Fext = 0 = 0 ext = 0 (about any axis)

torque about pivot due to gravity:

g = mgd

(gravity acts at center of mass)

Center of mass

pivotd

W=mg

This object is NOT in static equilibrium


Center of mass

pivotd

W=mg

Torque about pivot 0

Center of mass

pivot

Torque about pivot = 0

Not in equilibrium Equilibrium


Homework HintsHomework Hints

Painter is standing to the right of the support B.

FA FB

Mg mg

What is the maximum distance the painter can move to the right without tipping the board off?



If its just balancing on “B”, then FA = 0 the only forces on the beam are:

FB

Mg mg

Using FTOT = 0: FB = Mg + mg This does not tell us x

x



Find net torque around pivot B: (or any other place)

FB

Mg mg

(FB ) = 0 since lever arm is 0

(Mg ) = Mgd1

d1 d2

(mg ) = -mgd2

Total torque = 0 = Mgd1 -mgd2

So d2 = Md1 /m and you can use d1 to find x



Painter standing at the support B.

FA FB

Mg mg

Find total torqueabout this axisD

d

(FA) = - FAD

(Mg) = Mgd

(FB) = 0 (since distance is 0)

(mg) = 0 (since distance is 0)

Total torque = 0 = Mgd -FAD

So FA = Mgd /D


MORE EXAMPLES (bar and weights suspended by the string):Find net torque around this (or any other) place

(m1g) = 0 since lever arm is 0

x

T

Mgm2g

m1g


(Mg ) = -Mg L/2


L/2

T

Mgm2g

m1g


(Mg ) = -Mg L/2


x

T

Mgm2g

m1g

(T ) = T x


(Mg ) = -Mg L/2


L

T

Mgm2g

m1g

(T ) = T x

(m2g ) = -m2g L

All torques sum to 0: Tx = MgL/2 + m2gL So x = (MgL/2 + m2gL) / T





Moment of Inertia & Rotational KE

Textbook Sections 9.4 - 9.5:


Torque and Stability

Center of mass outside of base:

--> unstable

Center of mass over base:

--> stable




Moments of Inertia of Common Objects

Hollow cylinder or hoop about central axis

I = MR2

Solid cylinder or disk about central axis

I = MR2/2

Solid sphere about center

I = 2MR2/5

Uniform rod about center

I = ML2/12

Uniform rod about end

I = ML2/3





The picture below shows two different dumbbell shaped objects. Object A has two balls of mass m separated by a distance 2L, and object B has two balls of mass 2m separated by a distance L. Which of the objects has the largest moment of inertia for rotations around the x-axis?

1. A 2. B 3. Same

CORRECT

x

2LL

m

m

2m

2m

A B

I = mL2 + mL2

= 2mL2

I = 2m(L/2)2 + 2m(L/2)2

= mL2


Rotational Kinetic Energy

Translational kinetic energy:

KEtrnas = 1/2 MV2cm

Rotational kinetic energy:

KErot = 1/2 I2

Rotation plus translation:

KEtotal = KEtrans + KErot = 1/2 MV2cm + 1/2 I2


Angular MomentumAngular Momentum

Textbook Section 9.6


See text: chapters 8-9

See Table 8.1

Define Angular MomentumDefine Angular Momentum

MomentumMomentum Angular MomentumAngular Momentum

p = mV L = I

conserved if Fext = 0 conserved if ext =0

Vector Vector!

units: kg-m/s units: kg-m2/s


Chapter 9, Pre-flightsChapter 9, Pre-flights

You are sitting on a freely rotating bar-stool with your arms stretched out and a heavy glass mug in each hand. Your friend gives you a twist and you start rotating around a vertical axis though the center of the stool. You can assume that the bearing the stool turns on is frictionless, and that there is no net external torque present once you have started spinning. You now pull your arms and hands (and mugs) close to your body.



What happens to your angular momentum as you pull in your arms? 1. it increases 2. it decreases 3. it stays the same

L1 L2

This is like the spinning skater example in the book. Since the net external torque is zero (the movement of the arms and hands involve internal torques), the angular momentum does not change.

CORRECT



1 2

I2 I1

L L

What happens to your angular velocity as you pull in your arms? 1. it increases 2. it decreases 3. it stays the same

as with the skater example given in the book....as you pull your arms in toward the rotational axis, the moment of inertia decreases, and the angular velocity increases.

CORRECT

My friends and I spent a good half hour doing this once, and I can say...based on a great deal of nausea, that the angular velocity does increase.



What happens to your kinetic energy as you pull in your arms? 1. it increases 2. it decreases 3. it stays the same

CORRECT

Your angular velocity increases and moment of inertia decreases, but angular velocity is squared, so KE will increase with increasing angular velocity

1 2

I2 I1

L L

K 12

2I 12

2 2

II 1

22

IL (using L = I )


Two different spinning disks have the same angular momentum, but disk 2 has a larger moment of inertia than disk 1. Which one has the biggest kinetic energy ?

(a) disk 1 (b) disk 2

Spinning disksSpinning disks


K 12

2I 12

2 2

II

If they have the same L, the one with the smallest I will have the biggest kinetic energy.

L I1 1

disk 2

L I2 2

disk 1I1 < I2

12

2

IL (using L = I )


Preflights: Turning the bike wheelPreflights: Turning the bike wheel

A student sits on a barstool holding a bike wheel. The wheel is initially spinning CCW in the horizontal plane (as viewed from above). She now turns the bike wheel over. What happens?

1. She starts to spin CCW.2. She starts to spin CW.3. Nothing

CORRECT


Turning the bike wheel...Turning the bike wheel...

Since there is no net external torque acting on the student-stool system, angular momentum is conserved. Remenber, L has a direction as well as a magnitude!

Initially: LLINI = LLW,I

Finally: LLFIN = LLW,F + LLS

LLW,F

LLS

LLW,I LLW,I = LLW,F + LLS


Rotation Summary Rotation Summary (with comparison to 1-d linear motion)(with comparison to 1-d linear motion)

Angular Linear

constant

t 0

0 021

2t t

constanta

v v at 0

x x v t at 0 021

2

See text: chapters 8-9

See Table 8.1

maF I

)2/(2

1 22 mpmvKEtrans ILIKErot 2/

2

1 22

L vm p

UB, Phy101: Chapter 9, Pg 1 Physics 101: Chapter 9 l Today’s lecture will cover Textbook Sections 9.1 - 9.6.

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