physics 1710 —warm-up quiz

Physics 1710Physics 1710—Warm-up Quiz—Warm-up QuizTwo 2.0 kg disks, both of radius 0.10 m are sliding (without friction) and rolling, respectively, down an incline. Which will reach the bottom first?

1 2 3

37%

23%

40%

1. Rolling disk wins.

2. Sliding disk wins.

3. Tie.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

Solution:Solution:

Physics 1710Physics 1710—C—Chapter 11 Rotating hapter 11 Rotating BodiesBodies

Kinetic Energy of sliding disk:

K1 = ½ mv2 = mg(h-z); v =√[2g(h-z)]

Kinetic Energy of rolling disk:

K2 = ½ mv2+ ½ I ω2= mg(h-z)= ½ mv2+ ½ ( ½ mr2) (v/r)2

= 3/4 mv2; v =√[4/3 g(h-z)]

Slider wins!

Consider two spindles rolling down a Consider two spindles rolling down a ramp:ramp:

a

b

c

c

Which one will win and why?


No Talking!No Talking!Think!Think!

Confer!Confer!

Peer Instruction Peer Instruction TimeTime

Which one will win and why?Which one will win and why?


11′′ Lecture Lecture •The energy of rotation The energy of rotation

K = ½ I K = ½ I ⍵⍵ 2 2

• Torque (“twist”) is the vector product of the Torque (“twist”) is the vector product of the “moment” “moment” and a force. and a force. τ = τ = r x Fr x F

• τ τ = I = I ⍺ = I d⍵/dt⍺ = I d⍵/dt

•Angular momentum Angular momentum LL is the vector product of the is the vector product of the moment arm and the linear momentum. moment arm and the linear momentum. LL= = r x p.r x p.

• τ = d τ = d LL/dt/dt


Moment of Inertia—sphereMoment of Inertia—sphere

I = ∭ R I = ∭ R 2 2 ρ d ρ d VVN.B. : R N.B. : R 2 2 = r = r 22 – z – z 22

R R 2 2 = r = r 22 – (r cos – (r cos θ)θ) 22

=r =r 22 (1– cos (1– cos 2 2 θ)θ)

I = I = ∫∫0022ππ d dφφ∫∫00

ππ (1– cos (1– cos 2 2 θ)θ) sin sin θdθθdθ∫∫00aa r r 4 4 ρρdrdr

= [2= [2π][2- 1/3(2)][1/5 aπ][2- 1/3(2)][1/5 a55] ρ = ] ρ = [4[4π][2/3][1/5 aπ][2/3][1/5 a55] ] ρ ρ

==[4[4π/3][2/5 aπ/3][2/5 a55] ρ = 2/5 M a] ρ = 2/5 M a22

Rrr


Kinetic Energy of Rotation:Kinetic Energy of Rotation:

K = ½ K = ½ ΣΣi i mmi i vvi i 22

K = ½ K = ½ ΣΣi i mmi i (R(Rii ω ωii ) ) 22

K = ½ K = ½ ΣΣi i mmi i R R ii 22ωωii 22

For rigid body ωFor rigid body ωii = ω = ω

K = ½ [K = ½ [ΣΣi i mmi i R R ii 22] ω ] ω 22

K = ½ K = ½ I ω I ω 22

With I =ΣWith I =Σi i mmi i R R ii 22 = the moment of inertia. = the moment of inertia.


Why do round bodies roll down Why do round bodies roll down slopes?slopes?

The torque is the “twist.”

θ Fsinθ

θFg

F = m a = mr F = m a = mr dω/dtdω/dt

rF = rrF = rFFg g sinsinθθ = mr= mr22dω/dtdω/dt

Torque = r x F = I αTorque = r x F = I α

τ τ = = r x Fr x F,, ||τ τ || = = rFrFsinsinθθ


Torque and the Right Hand Rule:Torque and the Right Hand Rule:


rr

FF r r x x FF

X


Vector Product:Vector Product:

C = A x BC = A x B

CCxx = = AAyy B Bzz – A – Azz B Byy

Cyclically permute: (xyz), (yzx), (zxy)Cyclically permute: (xyz), (yzx), (zxy)

||CC| =| =√[√[CCxx22 + C + Cyy

22 + C + Czz22 ]]

= = AB sin AB sin θθ

Directed by RH Rule.Directed by RH Rule.


Vector Product:Vector Product:A x B = - B x AA x B = - B x A

A x ( B + C ) = A x B + A x CA x ( B + C ) = A x B + A x Cd/dt d/dt ( ( A x BA x B ) = d ) = d AA /dt /dt x B + A x x B + A x dd B B/dt/dt

i x i = j x j = k x k = 0i x i = j x j = k x k = 0i x j = - j x i = ki x j = - j x i = k

j x k = - k x j = ij x k = - k x j = ik x i = - i x k = jk x i = - i x k = j

Torque Bar:Torque Bar:


rr

FF

τ τ == r r x x FFττ

A B C

Teeter-totter:Teeter-totter:


τ τ == r r x x FF

FF22

FF11

Where should the fulcrum be place to balance the teeter-totter?

Where should the fulcrum be place to balance the teeter-totter?

A B C

6%

86%

8%

A.A.

B.B.

C.C.

81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120


Torque LadderTorque Ladder

Which way will the torque ladder move?

?



A B C

3%

27%

70%

A.A. Clockwise Clockwise B.B. CounterclockwiseCounterclockwiseC.C. Will stay balancedWill stay balanced

81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120


Torque LadderTorque Ladder


?

r

r sin θ


Second Law of MotionSecond Law of Motion

L = r L = r xx pp is the “angular momentum.”

FF = m = m aa

Or Or F F = d= dpp/dt/dt

Then:Then:

r x r x F F = d (= d (r r xx pp)/dt)/dt

Torque = Torque = ττ = d = d LL//dtdt


Angular Momentum:Angular Momentum:

L = r x pL = r x p

The angular momentum is the The angular momentum is the vector vector productproduct of the moment arm and the linear of the moment arm and the linear momentum.momentum.

∑ ∑ T = T = d d LL/dt/dt

The net torque is equal to the time rate of The net torque is equal to the time rate of change in the angular momentum.change in the angular momentum.


Angular Momentum:Angular Momentum:

Proof:Proof:∑ ∑ T = T = r x r x ∑∑FF = = r x r x d d pp/dt/dt

AndAndd d LL/dt = d( /dt = d( r x pr x p) /dt ) /dt

= d = d rr/dt/dt x p + r x x p + r x d d pp/dt./dt.

But But p = p = m d m d rr/dt/dt , , therefore therefore d d rr/dt/dt x p = 0 x p = 0

d d LL/dt = /dt = r x r x d d pp/dt/dtAnd thusAnd thus

∑ ∑ T = T = d d LL/dt./dt.


Second Law of MotionSecond Law of Motion

L = L = constant meansconstant means angular momentum si conserved.

Torque = Torque = ττ = d = d LL//dtdt

If If ττ = 0, then = 0, then

LL is a constant. is a constant.


Rotating Platform Rotating Platform DDemonstrationemonstration


Analysis:Analysis:

•Why does an ice skater increase her angular Why does an ice skater increase her angular velocity without the benefit of a torque?velocity without the benefit of a torque?

L = r x pL = r x p= r x = r x ( m ( m vv))

= = r x r x ( m ( m r x r x ⍵⍵))

LLii = m = mii r rii 2 2 ⍵ ⍵

LLzz = (∑ = (∑ii m mii r rii 2 2 ) ⍵ ) ⍵

LLzz = I ⍵; & ⍵ = L = I ⍵; & ⍵ = Lzz / I / I

Therefore, a decrease in I ( by reducing r) will Therefore, a decrease in I ( by reducing r) will result in an increase in ⍵.result in an increase in ⍵.


Summary:Summary:

•The total Kinetic energy of a rotating system is The total Kinetic energy of a rotating system is the sum of the rotational energy about the Center the sum of the rotational energy about the Center of Mass and the translational KE of the CM. of Mass and the translational KE of the CM.

K = ½ IK = ½ ICMCM ⍵ ⍵ 22 + + ½ MR½ MR 2 2 ⍵ ⍵ 22

ττ = r x F = r x F


Summary:Summary:


•Angular momentum Angular momentum LL is the vector product of is the vector product of the moment arm and the linear momentum.the moment arm and the linear momentum.

L = r L = r x x pp

• The net externally applied torque is equal to The net externally applied torque is equal to the time rate of change in the angular the time rate of change in the angular

momentum.momentum.

∑ ∑ ττzz = d L = d Lzz /dt = I /dt = Izz ⍺ ⍺

physics 1710 —warm-up quiz

Documents

r x fwhere

x d bdti x

x cddt

d r x pdttorque

r x d pdtandd

r x fteetertotter

r x pthe angular momentum

b x aa x b c