θ (t) ( θ 1 ) Δ θ = θ 2 s = θ r ω (t) = d θ (t) dt v = d θ dt r = ω r v = ω r α (t) = d ω (t) dt = d 2 θ (t) dt 2 a tot 2 = a r 2 + a t 2 = ω 2 r 2 + αr 2 a tot = a t + a r = a r ω ω r a t = α r ( ) Rota%onal Kinema%cs: ( ONLY IF α = constant) ω = ω 0 + αt θ = θ + ω 0 t + 1 2 αt
2 If rigid body = few particles I = m i r i If rigid body = too-many-to-count particles Sum Integral Parallel Axis Theorem I = I COM + Mh 2 Energy of rota%onal mo%on KE rot = 1 2 Iω 2 [ KE trans = 1 2 ] mv2
Example #3 A bicycle wheel has a radius of 0.33m and a rim of mass 1.2 kg. The wheel has 50 spokes, each with a mass 10g. What is the moment of iner%al about axis of rota%on? What is moment of iner%a about COM? I tot,com = I rim,center + 50I spoke What is I spoke (parallel axis)? Ispoke = Irod.com + Mh 2 = 1 12 ML2 + M ( 1 L 2 ) 2 = 1 3 ML2 = 1 0.01kg 3 ( )( 0.33m) 2 3.6 10 4 kg m 2 PuUng together I tot,com = I rim,center + 50I spoke = M wheel R 2 + 50I spoke = ( 1.2kg) ( 0.33m) 2 + 50(3.6 10 4 kg m 2 ) = 0.149kg m 2
Example #4 What is the kine%c energy of the earth s rota%on about its axis? Energy of rota%onal mo%on is found from: KE rot = 1 Iω 2 2 What is earth s moment of iner%a, I? Iearth = Isphere = 2 5 Mr 2 = 2 6 10 24 kg 5 ( )( 6.4 10 6 m) 2 1 10 38 kg m 2 What is earth s angular velocity, ω? From T = 2π ω Now plug n chug: KE rot = 1 1 10 38 kg m 2 7.3 10 5 rad /s 2 ω = 2π radians day = 2π ( 3600 24) = 7.3 10 5 rad /s ( ) 2 = 2.6 10 29 J
10 8: Rota;onal dynamics Ability of force to rotate body What causes something to rotate What does it depend on?? F Depends on distance from pivot point τ = r F τ = r F sinϑ Depends on angle between force and radius Moment of force about an axis = torque to twist τ depends on: r, F and θ Units of torque: N m (not Joule)
Define: Torque 10 8: Rota;onal Dynamics τ = r F Units of torque: N m vector = magnitude + direc;on τ = r F sinϑ vector τ is always perpendicular to both vectors r and F If F t is tangen%al component of force If r is moment arm τ = r F sinϑ ( ) = rf t ( ) = r sinϑ F = r F
10-9: Rotational dynamics - Newton s 2 nd Law Force about a point causes torque τ = r F τ net = F net = τ i = I α { F i } = m a τ net = τ i = r F i = i = ( 2 m i r ) i α = I α r m i a i i = r m i r i α i torque is posi%ve when the force tends to produce a counterclockwise rota%on about an axis is nega%ve when the force tends to produce a clockwise rota%on about an axis ( VECTOR! posi%ve if poin%ng in +z, nega%ve if poin%ng in z ) net torque and angular accelera%on are parallel
Example #1 A uniform disk, with mass M and radius R, is mounted on a fixed horizontal axle. A block of mass m hangs from a massless cord that is wrapped around the rim of the disk. a) Find the accelera%on of the disk, and the tension in the cord. b) Assume now that m = M. Ini%ally at rest, what is the speed of the block aeer dropping a distance d? Method 1: Using conserva%on of Energy 1 ( Mv 2 + 1 Iω 2 2 2 ) 0 + 0 final Subs%tu%ng 1 2 Mv2 + 1 2 ΔKE tot = ΔU [ ] ( ) init = ( 0) final ( Mgd) init ( 1 MR 2 ) v 2 R v = ωr & I = 1 2 MR 2 2 = Mgd v 2 + ( 1 2) ( v) 2 = 2gd v 2 ( 1+ 1 2) = 3 2 v2 = 2gd v = 4 3 gd
Method 2 Using force + torque z ˆ : τ net = r F T = M ( g a) = ( RT)( ˆ z ) ˆ y : T Mg = Ma a t = αr & I = 1 2 MR 2 τ = Iα ( RT)( ˆ z ) = ( Iα)( ˆ z ) R[ M ( g a) ] = ( 1 MR 2 ) a 2 R a( 1 M + M 2 ) = Mg Now use 1 D kinema%cs v 2 = v 02 + 2ad v 2 = 0 + 2d 2 3 g v = 4 3 gd a = Mg 3 2 M = 2 3 g ( )
Work and Rota;onal Kine;c Energy where τ is the torque doing the work W, and θ i and θ f are the body s angular posi%ons before and aeer the work is done, respec%vely. When τ is constant, The rate at which the work is done is the power
Summary: Newton s 2 nd law for rota;on τ net = τ i = I α F net = { F i } = m a Summary: Work and Rota%onal Kine%c Energy NET Work done ON system W = ΔKE rot ( ) = 1 I ω 2 2 2 f ω i W = ΔKE trans = 1 m v 2 2 v 2 f i ( ) Rota%onal work, fixed axis rota%on (if torque is const) W = θ 2 θ 1 τ net dθ ( ) = τ θ f θ i x 2 W = Fdx 1 D motion x 1 Power, fixed axis rota%on P = dw dt = d ( dt τθ ) = τω P = dw dt = d dt F ( x ) = F v
Example: Block and Pulley Accelera;ng In the figure one block has a mass M, the other has mass m, and the pulley (I), which is mounted in horizontal frictionless bearings, has a radius r. When released from rest, the heavier block falls a distance d in time t. (a) What is the magnitude of the block's acceleration? (b) What is the tension in the part of the cord that supports the heavier block? (c) What is the tension in the part of the cord that supports the lighter block? (d) What is the magnitude of the pulley's angular acceleration? Force about a point causes torque τ = F Basic Equations Mg-T 2 = Ma T 1 mg = ma T 1 T 2 Torque causes rota%on τ net = τ i Trans rota%onal rela%onship a T = rα = I α ( T 2 T 1 )r = Iα = Ia r Ia r 2 + M + m ( )a = M m ( )g a = ( M m)g I r + M m 2 + ( ) Check: I=0 Check: M=m Check: m=0
a Example: Torque or Kine;c Energy Massless cord wrapped around a pulley of radius R and mass M W (fric%onless surface/bearings) and I W =1/2M W R 2. What is angular accelera%on, α, of pulley (disc)? α into board 1) What are forces on m 1? 2) What are forces on m 2? x ˆ : T 1 m 1 gsinθ = m 1 a y ˆ : N m 1 gcosθ = 0 T 1 = m 1 ( a + gsinθ) 3) What are torques about wheel? τ net = τ 1 + τ 2 τ net = RT 1 sin 90 NOTE: T 1 & T 2 are NOT equal ( ) + RT 2 sin( 90 ) = R( T 1 T 2 )(+ z ˆ ) NOTE: angular accelera%on vector is in nega%ve z direc%on y ˆ : T 2 m 2 g = m 2 a T 2 = m 2 ( g a) 4) Solve for α? I W α = R T 2 here I W α = ( 1 M 2 W R 2 )α = R m 2 g a a t = αr 1 ( M 2 W R2 )α = Rg m 2 m 1 sinθ α = 2g R [( ) ( T 1 )] ( ( )) m 1 ( a + gsinθ) ( ) R( αr) ( m 2 + m 1 ) [ ( )] m 2 m 1 sinθ M w + 2 m 1 + m 2 ( ) a) If M w = 0, same as problem 5 43 b) If θ =90º, same as problem 11 55 (HW # 9)
Problem 90 (Ch 10): A rigid body is made of three iden%cal thin rods, each with a length L=0.600 m, fastened together in the form of a leoer H. The body is free to rotate about the axis shown through on of the rods. The body is allowed to fall rota%ng around the axis. What is the angular velocity when it is ver%cal? Can't do Torque--use conservation of energy! ΔK ω = ΔU g = Iω 2 Iω 2 2 = mg L 2 + mgl 2 = 4mL2 3 ω 2 = mg 3L 2 ω = g 9L 4L I 1 = 0 why? I 2 = ml2 12 + M L 2 2 I 3 = ml 2 I = 4mL2 3 = ml2 3
Chapter 11 Rolling, Torque, and Angular Momentum In this chapter we will cover the following topics: -Rolling of circular objects and its relationship with friction -Redefinition of torque as a vector to describe rotational problems that are more complicated than the rotation of a rigid body about a fixed axis -Angular momentum of single particles and systems of particles -Newton s second law for rotational motion -
What Is Rolling? Understanding rolling with wheels Wheel moving forward with constant speed v com s = θr displacement: transla%on rota%on These rela%onships define smooth rolling mo%on ( ) v com = ds dt = d θr dt a com = dv com dt = d ( ωr ) dt = ωr = αr Only if NO SLIDING [smooth rolling] At point P (point of contact), wheel does not move
Understanding rolling with wheels II All points on wheel move with same ω. All points on outer rim move with same linear speed v = v com. v = ω r All points on wheel move to the right with same linear velocity v com as center of wheel Combina%on of pure rota%on and pure transla%on Note at point P: vector sum of velocity = 0 (point of sta;onary contact) at point T: vector sum of velocity = 2v com (top moves twice as fast as com)
Kine;c Energy of Rolling 1 2 I comω 2 + 1 2 Mv com 2 = KE tot Note: rota%on about COM and transla%on of COM combine for total KE Remember: v com =ωr