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Physics 201, Lecture 18

Today’s Topics

q  Rotational Dynamics

§  Torque q  Exercises and Applications

§  Rolling Motion

q  Hope you have previewed Chapter 10. (really!)

Review Angular Velocity And Angular Acceleration

q  Angular Velocity (ω) describes how fast an object rotstes, it has two components: §  Angular speed: and

§  direction of ω: + counter clockwise - clockwise

Ø  All particles of the rigid object have the same angular velocity q  Angular Acceleration (α):

and

Note: the similarity between (θ,ω,α) and (x, v, a)

ωave ≡ΔθΔt

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ω ≡Δt→ 0limΔθ

Δt=dθdt

αave ≡ΔωΔt

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α ≡Δt→ 0limΔω

Δt=dωdt

è Angular velocity ω is a vector! (define direction next page)

è Angular acceleration α is also a vector!

axis Review: Moment of Inertia

q  Moment of Inertia of an object about an axis

(unit of I : kgm2)

Ø  “I” depends on rotation axis, total mass, and mass distribution.

∑≡ 2iirmI :Inertia ofMoment

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another form: I ≡ r2dmwhole object∫

Moments Of Inertial Of Various Objects

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I = miri2∑ (= r2dm)∫

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Which Has Larger Moment of Inertia?

q  Which of the following configurations has larger I Ø  Central or side axis ?

(Trivial) Quick Quiz q  A force is acting on a rigid rod around a fixed axis. Ø  Which of the follow case(s) will not turn.

Ø  Effective Turning: Force, direction, acting point(action length).

F F F

F

Turning counter clockwise

Turning clockwise No Turn No Turn

axis

Torque: Effect of Force on Rotation q  Torque:

§  Magnitude: τ = Fsinφ r , depends on F , r, and sinφ §  Direction:

•  conventional: –  clockwise = “-”, –  counter-clockwise = “+”

•  More strictly: Right Hand Rule for cross-product

q  The angular acceleration of an object is proportional to the torque acting on it Στ = I α

I: Moment if inertia (Tuesday)

τ =r ×F

Lever Arm

q  τ ≡ Fsinφ r = F d

axis (pivot)

acting point

Lever Arm d = r sinφ

The lever arm, d, is the perpendicular distance from the axis of rotation to a line drawn from the direction of the force

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Alternative View of Torque q  τ ≡ Fsinφ r = F⊥ r

q  Only the perpendicular component of the force contributes to the torque

axis (pivot)

acting point

Summary: Two Views of Torque q  τ ≡ Fsinφ r = F⊥ r = F r⊥

F⊥ r F r⊥

axis (pivot)

acting point

Torque Has a Direction q  Torque is a vector. It has a magnitude and a direction. q  For fixed axis rotation, the direction of torque can be described by

a sign (+/-) in one dimension; (and by right hand rule in general)

Ø  Convention: Counter-clockwise : +, clockwise - .

τ > 0

τ < 0

τ > 0

τ > 0

The Acting point of Gravity: Center of Gravity

q  The force of gravity acting on an object must be considered in determining equilibrium q  In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at one point called center of gravity (cg) q  Effectively, assuming gravitation field is uniform, the CG of an object is the same as its CM (that is usually true at the Phy103 level)

xcg =ΣmixiΣmi

and ycg =ΣmiyiΣmi

See demo: finding CG

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pivot

L/2

θ

Quick Exercise: Calculating Torque q  As shown, a pencil is falling down under gravity. What is the

torque (about the pivot) by the gravity?

q  Answer: + Fgrav cosθ L/2 (note: why cosθ?)

Rotational Dynamics

Angular Linear

tαωω += 0atvv += 0

221

00 tt αωθθ ++=2

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00 attvxx ++=

θαωω Δ+= 220

2 xavv Δ+= 220

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Iτ

α =mFa =

Στ = I α Rotational Dynamics compared to 1-D Dynamics

KE = 12Iω 2 KE = 1

2mv2

Exercise: Pulley with Mass q  A crate of mass mcrate is hanging on a pulley of mass mpulley and

radius Rpulley as shown. What is the acceleration of the crate? Solution: 1: per FBD for the crate T-mcrateg = mcratea 2: for the pulley: τ = -TR = Ipulley α 3: connection: a=αR Solve: a = - mcrateg/(mcrate+Ipulley/Rpulley

2) Get T and α yourself after class

Please make sure you understand this

Combined Translational and Rotational Motion

q  Generally, the motion of an extended object is a combination of the translational motion of the CM and the rotation about the CM

KEtot = KEtrans + KErot = ½ MvCM2 + ½ Iω2

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Combined Translation and Rotation q  Combined translational and rotational motion.

Ø  At point A (top). vrot = Rω to the right. vtop = vCM + Rω Ø  At point B (bottom). vrot = Rω to the left. vbottom = vCM - Rω

Ø  At arbitrary point C

CM: moving linearly (1-D)

Everything on the wheel rotation about CM

R

v = vCM +vrot

vCM

A

B

C

vrot

vrot

vCM

v

Rolling/Sliding Conditions q  Recall: vbottom = vCM - Rω q  Depending relative size of Vcm and Rω, there can be three

classes of rolling /sliding conditions.

vcm

ω

vbottom = vcm - Rω =0 vcm = Rω

vcm

ω

vbottom = vcm – Rω<0 vcm < Rω

vcm

ω

vbottom = vcm – Rω>0 vcm > Rω

“spinning” “pure rolling” (rolling w/o slipping)

“sliding”

“Pure” Rolling Motion (Rolling without Slipping)

q  Rolling motion refers to a form of combined translational and rotational motion.

Ø  Condition for rolling w/o slipping:

vCM = ωR and aCM = αR

CM: moving linearly (1-D)

wheel: rotation about CM

No slipping on road i.e. vbottom =0

R vCM

Quick Quiz and Demo q  Consider a wheel in pure rolling without slipping. After a full

resolution (i.e. in period T), how far the CM moves?

q  R, 2R, πR, 2πR, other

See Demo

CM: moving linearly (1-D)

R vCM = Rω=R2π/T ΔxCM =VCM*T =2πR

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Trajectory of a Point on the Rim of a Pure-Rolling Wheel Quick Quiz: Rolling Without Slipping

q  Consider a wheel rolling down a (not smooth) hill without slipping.

How many (external) forces are acting on the wheel? 2, 3, more than 3, other q  In the process, the work down by friction is Positive, negative, zero q  Now consider a wheel rolling on a (flat but non smooth)

horizontal plane without slipping. How many external forces are acting on it?

2, 3, more than 3, other Why? See next slide

See Demo

fs

mg

N

mg

N

Exercise: Rolling w/o Slipping Down a Slope

q  A uniform disc (or wheel, or sphere) of mass M , radius R, and moment of inertia I is rolling down a slope without slipping as shown. Calculate its CM acceleration.

q  Solution: Ø  Step 1: FBD as shown Ø  Step 2: Set up axis as shown Ø  Step 3: Dynamics for CM (x direction): mgsinθ – fs = maCM Ø  Step 4: Dynamics for rotation: -fsR =- Iα Ø  Step 5: rolling w/o slipping: Rα=aCM Ø  Solve for unknowns:

fs

mg

N

θ

x

aCM =gsinθ

1+ ImR2

, fs =mgsinθmR2

I+1

Results Discussion: Rolling Down a Slope (w/o slipping)

Consider a wheel rolling down a flat (but not smooth) slope. Ø  On a slope, a friction is necessary to keep it from slipping Ø  For same mR2, the larger the I, the slower it moves. Ø  Spheres (or wheels or discs) of the same shape and mass

distribution roll at the same speed regardless of their size and mass. Ø  On a horizontal flat surface, the friction reduces to zero and the

rolling can go forever even when the surface is not smooth!

fs

mg

N

θ

x aCM =gsinθ

1+ ImR2

, fs =mgsinθmR2

I+1

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Moments Of Inertial Of Various Objects

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I = miri2∑ (= r2dm)∫

See demo for rolling of wheels with different I

q  General Work: W = F · Δs = Fin_direction_of_Δs Δs q  For an object rotating about a fixed axis, all mass elements are

moving in the tangential direction : W = Ft Δs = rFt Δs/r = τ Δθ P = W/Δt = τ Δθ/Δt = τω Ø  Rotational version of Work-Energy theorem

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if

rot

IIKEW

ωωθτ −=Δ

Δ=

Work and Power by Torque (Self Reading) Δs

Δs

Δθ