Moments Of Inertial Of Various ·...

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  • 1

    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):


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

    ωave ≡ΔθΔt

    ω ≡Δt→ 0limΔθΔt =


    αave ≡ΔωΔt

    α ≡Δt→ 0limΔωΔt =


    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

    another form: I ≡ r2dmwhole object∫

    Moments Of Inertial Of Various Objects

    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


    Turning counter clockwise

    Turning clockwise No Turn No Turn


    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

  • 3

    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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    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 += 0 atvv += 02


    00 tt αωθθ ++=2


    00 attvxx ++=

    θαωω Δ+= 2202 xavv Δ+= 220


    α =mFa =

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

    KE = 12Iω 2 KE =


    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/Rpulley2) 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


    v = vCM +vrot









    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.



    vbottom = vcm - Rω =0 vcm = Rω



    vbottom = vcm – Rω0 vcm > Rω

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


    “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






    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:






    aCM =gsinθ

    1+ ImR2

    , fs =mgsinθmR2


    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!





    x aCM =gsinθ

    1+ ImR2

    , fs =mgsinθmR2


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    Moments Of Inertial Of Various Objects

    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






    ωωθτ −=Δ


    Work and Power by Torque (Self Reading) Δs