Forces of Rolling. 1) Ifobjectisrollingwith a com =0 (i.e.no netforces), then v com =ωr = constant (smooth roll)

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1 Physics 2101 Section 3 March 12 rd : Ch. 10 Announcements: Mid-grades posted in PAW Quiz today I will be at the March APS meeting the week of th. Prof. Rich Kurtz will help me. Class Website:

2 Forces of Rolling 1) Ifobjectisrollingwith a =0 (i.e.no netforces), then v =ωr = constant (smooth roll) if constant speed, it has no tendency to slide at point of contact no frictional forces a = 0 α = 0 τ = 0 f = 0 a = αr I α = τ net τ net = R f sinθ 2) If object is rolling with a 0 (i.e. there are net forces) and no slipping occurs, then α 0 τ 0 static friction needed ddto supply torque! kinetic friction - during sliding (a αr) static ti friction - smooth rolling (a = αr)

3 Rolling down a ramp : Energy considerations An object at rest, with mass m and radius r, roles from top of incline plane to bottom. What is v at bottom and a throughout. ( KE ) final 0 KE rot +trans,com 1 mv 2 2 COM ΔE mech = 0 ΔKE tot = ΔU [ ] ( ) init = ( 0 ) final ( mgh ) init I COMω 2 = mgl sinθ Using v = ωr h L sinθ = h θ Using v 2 = v aL v COM = 2gL sinθ 1+ I mr 2 Using 1 D kinematics (const accel) a v2 gsinθ = = 2L 1+ I mr 2 Speed at Bottom Constant Acceleration

4 Rolling down a ramp: acceleration Acceleration downwards Friction provides torque Static friction points upwards Yo Yo Newton s 2 nd Law Linear version x ˆ : f s F g sinθ = ma y ˆ : N F g cosθ = 0 Tension provides torque Here θ = 90 Angular version z ˆ : I α = τ = Rf s a = R 2 ( mgsinθ ma ) I Al Axle: R R 0 a = αr α = a R = Rf s I gsinθ a = 1+ I mr 2 g a = 1+ I 2 mr 0

5 a Rolling Down a Ramp Consider a round uniform body of mass M and radius rolling down an inclined plane of angle θ. We will calculate the acceleration a of the center of mass along the x-axis using Newton's second law for the translational and rotational motion. R Newton's second law for motion along the x-axis: f Mgsin θ = Ma (eq. 1) Newton's second law for rotation about the center of mass: s τ = Rf s = I a a α = We substitute α in the second equation and get Rfs = I R R a fs = I (eq. 2). We substitute f from equation 2 into equation 1 2 s R a g sinθ I = a = Mgsinθ Ma 2 R I 1+ 2 MR α

6 a a = g sinθ I 1+ 2 MR Cylinder Hoop 2 MR 2 I1 = I2 = MR 2 g sin θ g sin θ a1 = a2 = 1 + I / MR 1 + I / MR g sinθ g sinθ a 1 = a = MR /2MR 1 + MR / MR gsinθ gsinθ a1 = a2 = 1+ 1/ gsinθ gsinθ a1 = = (0.67) gsin θ a2 = = (0.5) gsinθ 3 2

7 Prob NON smooth rolling motion A bowler throws a bowling ball of radius R along a lane. The ball slides on the lane, with initial speed v,0 and initial angular speed ω 0 = 0. The coefficient of kinetic friction between the ball and the lane is μ k. The kinetic frictional force f k acting on the ball while producing a torque that causes an angular acceleration of the ball. When the speed v has decreased enough and the angular speed ω has increased enough, the ball stops sliding and then rolls smoothly. a) [After it stops sliding] What is the v in terms of ω? Smooth rolling means v =Rω b) During the sliding, what is the ball s linear acceleration? From 2 nd law: (linear) ˆ x : f k = ma ˆ y : N mg = 0 But f k = μ k N = μ k mg So a = f k m = μ k g c) During the sliding, what is the ball s angular acceleration? From 2 nd law: τ = Rf k ( ˆ z ) But f k = μ k N (angular) Iα( ˆ z ) = τ = Rf k ( ˆ z ) = μ k mg So Iα = Rf k = R( μ k mg) α = Rμ kmg I d) What is the speed of the ball when smooth rolling begins? When does v =Rω? v = v 0 + a t v = v 0 μ k gt From kinematics: ω = ω 0 + αt t = ω α = Iω Rμ k mg I v R v = v 0 μ k g Rμ k mg e) How long does the ball slide? v = t = v 0 v μ k g v 0 1+ ImR 2 ( )

8 Torque & Angular Momentum ω Conservation of: Linear momentum (closed, isolated system) KE (elastic collisions) Mechanical energy (only conservative forces) Total energy Angular momentum (closed, isolated system)

9 Angular Momentum (single particle) A Linear momentum when particle passes through point it has linear momentum: p = mv Angular momentum with respect to point O the angular momentum is defined as: = r p = m r v ( ) where r is the position vector of the particle with respect to O. with ponents in x y plane Units kg m 2 /s ponents in z direction only! ( to x y plane) just as in τ, has meaning only with respect to a specified point (axis)

10 Newton s 2 nd Law in Angular Form (single particle) Relationship between force and linear momentum Newton s 2 nd Law in linear form F net = F i = ma = d p dt Relationship between torque and angular momentum τ net = τ i = d dt The vector sum of all the torques acting on a particle is equal to the time rate change of the angular momentum of that particle. τ net = d dt has no meaning unless the net torque τ net, and the rotational momentum, are defined with respect to the same origin

11 Example #1 A cannon ball, with mass m, is shot directly upward with velocity v 0. What is the angular momentum and torque of the ball about a point a distance x 0 from the cannon? The only force on the in flight cannon ball is gravity: The vector position and velocity of the in flight ball is: x ˆ : x(t) = x 0 + v 0,x t = x 0 y ˆ : y(t) = y 0 + v 0,y t + 1 at 2 = v 2 0 t 1 gt 2 2 F g = mgˆ j r = ( x 0 )ˆ i + ( v 0 t 1 gt 2 2 )ˆ j v = ( v 0 gt)ˆ j The vector angular momentum about the man of the in flight ball is: i ˆ ˆ j k ˆ l = m r v = m x 0 v 0 t 1 gt = mx 0 ( v 0 gt)ˆ k 0 v 0 gt 0 The vector torque about the man of the in flight ball is: ˆ ˆ ˆ τ = r F i j k = x 0 v 0 t 1 gt 2 0 = x 2 0 mgk ˆ 0 mg 0 Check τ = d l dt = d [ mx 0 ( v 0 gt)ˆ k ] dt = mx 0 g k ˆk

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14 Total angular momentum Newton s 2 nd Law in Angular Form System of particles L = l 1 + l 2 + l l n = Relationship between torque and angular momentum for system of particles n l i i=1 τ net = d L dt τ net represents the net external torque.

15 Example #2 (Problem 11 27) #1 Two particles are moving as shown. a) What is their total moment of inertia about point O? b) What is their total angular momentum about point O? c) What is their net torque about point O? #2 a) The total moment of inertia about point O is found from: I tot = I i = m i r i = (6.5 kg)(1.5 m) + (3.1 kg)(2.8 m) = 38.9 kg m b) The total angular momentum about point 0 is: r L = = ri r tot l i m i ( r i v i ) = m 1 ( r 1 v 1 sinθ 1 )( ẑ) z) + m 2 ( r 2 v 2 sinθ 2 )(+ẑ) )(+z) L tot = [ (6.5 kg)(1.5 m)(2.2 m /s) + (3.1 kg)(2.8 m)(3.6 m /s)]ˆ z = 9.8 kg m 2 /s z ˆ c) The net torque about point 0 is: dl τ tot net = = 0 closed, isolated system tot dt

16 Angular Momentum of a Rigid Body Rotating about a Fixed axis dl I α d ω τ net = Iα dt = = I dt L = Iω Example #3 (Problem 11 39) Three particles are fastened to each other with massless string and rotate around point O. What is the total angular momentum about point O? I tot = md ( ) 2 + m( 2d) 2 + m( 3d) 2 = 14mdd 2 L 2 tot = Iω = 14mωd ẑz [kg m 2 /s]

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