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Rolling Down a Ramp with No FricOn

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Rolling Down a Ramp with No Fric�On → → → → → → → Fnet = ∑Fi = ma τ net = τ i = r× F = r× m a = m r× (r×α )= Iα { } ∑ € € Work and Rotational Kinetic Energy W = ΔKE rot W = ΔKE trans NET Work done ON system 1 2 2 1 2 2 = 2 I(ω f −ω i ) = 2 m(v f − vi ) θ 2 Rotational work, x2 W = ∫ τ net dθ fixed axis rotation€ θ1 € W = ∫ Fdx 1− D motion x1 (if torque is const) = τ (θ f −θi ) dW d € dW d Power, P = = (τθ) = τω P = = (F • x ) = F •v fixed axis rotation€ dt dt dt dt € € We can compare linear variables with rotational variables x θ v ω s = rθ a α v = rω Δt Δt T a = rα F τ T m I The same can be done for work and energy: € € For translational€ systems For rotational systems W = F ⋅ x W = τ ⋅θ 1 1 KE = mv 2 KE = Iω 2 2 2 € € All points on wheel move All points on wheel Combination of “pure with same ω. All points on move to the right with same rotation” and “pure outer rim move with same linear velocity vcom as center translation” linear speed v = v . of wheel com v = ω ×r Note at point P: vector sum of velocity = 0 (point of stationary contact) at point T: vector sum of velocity = 2v (top moves twice as fast as com) € com 1 2 1 2 2 Icomω + 2 Mvcom = KEtot Note: rotation about COM and translation of COM combine for total KE Remember: v =ωr € € € com Example: Rolling down a ramp with no fricon Object, with mass m and radius r, roles from top of incline plane to bottom. What is v, a, and Δt at bottom h ΔE = 0 mech L sinθ = h AT BOTTOM ΔKE tot = −ΔU θ KE − 0 = − 0 − mgh ( rot +trans,COM ) final ( )init [( ) final ( )init ] 1 2 1 2 € Using 1-D kinematics mv + I ω = mgL sinθ 2 € COM 2 com 2 2 v = v0 + 2aL 2 v2 gsin 1 2 1 v θ mv + I = mgL sinθ a = = 2 COM 2 COM r 2L I € 1 + 2 2 I mr v m + COM = 2mgL sinθ COM r2 AND 2L I 2gL sinθ t = 1 + v = 2 COM I gsinθ mr 1 + mr 2 AT BOTTOM € € Compare Different Objects Assuming same work done (same change in U), objects with larger rotational inertial have larger 2gL sinθ vCOM = KE and during rolling, their KE is smaller. Icom rot trans 1 + I mr 2 KE = KE + KE = KE 1 + com tot trans rot trans mr 2 Roll a hoop, disk, and solid sphere down a ramp - what wins? € Rotational Fraction of Energy in Object Inertia, I Translation Rotation € com 2 large Moment of inertia Hoop 1mr 0.5 0.5 slowest 1 2 → Disk 2 mr 0.67 0.33 2L I small Δt = 1 + bottom gsinθ mr 2 2 2 Sphere 5 mr 0.71 0.29 € sliding block 0 1 0 fastest (no friction) € Question A ring and a solid disc, both with radius r and mass m, are released from rest at the top of a ramp. Which one gets to the bottom first? 1. Solid disc 2. Ring (hoop) 3. both reach boom at same me Question #2 Two solid disks of equal mass, but different radii, are released from rest at the top of a ramp. Which one arrives at the bottom first? 1. The smaller radius disk. The larger radius disk. 2. Using 1-D kinematics 3. Both arrive at the same time. 2 2 v = v0 + 2aL v2 gsinθ a = = 2L I 1 + mr 2 The equation for the speed of the a disk at the bottom of the ramp is 4 AND 3 gl sinθ Notice, it does not depend on the radius or the mass of 2L I the disk!! t = 1 + gsinθ mr 2 € RollingRollin Downg Down a RRampamp with a Frictional Force Consider a round uniform body of mass M and radius R acom rolling down an inclined plane of angle θ. We will calculate the acceleration acom of the center of mass along the x-axis using Newton's second law for the translational and rotational motion. Newton's second law for motion along the x-axis: fs − Mg sinθ = Macom (eq. 1) Newton's second law for rotation about the center of mass: τ = Rfs = Icomα a a α = − com We substitute α in the second equation and get Rf = −I com → R s com R a f = −I com (eq. 2). We substitute f from equation 2 into equation 1→ s com R2 s a g sinθ com acom = − −Icom 2 − Mg sinθ = Macom I R 1+ com MR2 g sinθ | a |= com I a 1+ com com MR2 Cylinder Hoop MR2 I = I = MR2 1 2 2 g sinθ g sinθ a1 = 2 a2 = 2 1+ I1 / MR 1+ I2 / MR g sinθ g sinθ a = a = 1 1+ MR2 / 2MR2 2 1+ MR2 / MR2 g sinθ g sinθ a = a = 1 1+1/ 2 2 1+1 2g sinθ g sinθ a = = (0.67)g sinθ a = = (0.5)g sinθ 1 3 2 2 Sample Problem A solid cylinder starts from rest at the upper end of the track as shown. What is the angular speed of the cylinder about its center when it is at the top of the loop? Sample Problem #2 A A solid cylinder of radius 10 cm and mass 12 kg starts from rest and rolls without slipping a distance of 6 m down a house roof that is inclined at 30º. B Where does it hit? Forces of Rolling 1) If object is rolling with acom=0 (i.e. no net forces), then vcom=ωR = constant (smooth roll) …if constant speed, it has no tendency to slide at point of contact - no frictional forces acom = 0 α = 0 τ = 0 f = 0 acom = αR Icomα = τnet τnet = |R||f|sinθ 2) If object is rolling with acom≠ 0 (i.e. there are net forces) and no slipping occurs, then α ≠ 0 ⇒ τ ≠ 0 … static friction needed to supply torque ! kinetic friction - during sliding (acom ≠ αR) static friction - smooth rolling (acom = αR) Rolling Down a Ramp: Acceleraon by fs - Acceleration downwards Yo-Yo - Friction provides torque - Static friction points upwards Newton’s 2nd Law Linear version xˆ : fs − Fg sinθ = −macom yˆ : N − Fg cosθ = 0 - Tension provides Angular version torque zˆ : Icomα = τ = Rfs 2 R (mgsinθ − macom ) - Here = 90 a = θ ° → a = αR com com Icom - Axle: R ⇒ R0 a Rf € α = com = s R I com g € € a gsinθ com = 2 a = 1+ Icom mR0 com 1+ I mR2 € com € € A yo-yo has a rotational inertia of Icom and mass of m. Its axle Problem radius is R0 and string’s length is h. The yo-yo is thrown so that its initial speed down the string is v0. a) How long does it take to reach the end of the string? 1-D kinematics given acom 1 2 −h = Δy = −v0 t − 2 acom t ⇒ solve for t (quadradic equation) b) As it reaches the end of the string, what is its total KE? Conservation of mechanical energy 2 1 2 1 vcom,0 € KE f = KE i +U = 2 mvcom,0 + 2 Icom + mgh R0 c) As it reaches the end of the string, what is its linear speed? 1-D kinematics given acom € −vcom = −v0 − acom t ⇒ solve for vcom d) As it reaches the end of the string, what is its translational KE? Knowing |vcom| 1 2 € KEtrans = mvcom 2 e) As it reaches the end of the string, what is its angular speed? g a = Knowing |v | com 2 com vcom 1+ Icom mR0 ω = € R downwards f) As it reaches the end of the string, what is its rotational KE? 1 Two ways: KE = I ω 2 or KE = K − KE €rot 2 com rot Ef ,tot trans € € Which way will it roll?? 1 1 3 Problem 11‐13 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 vcom,0 and initial angular speed ω0 = 0. The coefficient of kinetic friction between the ball and the lane is µk. The kinetic frictional force fk acting on the ball while producing a torque that causes an angular acceleration of the ball. When the speed vcom 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 vcom in terms of ω ? Smooth rolling means vcom =Rω b) During the sliding, what is the ball’s linear acceleration? nd ˆ f N a f m From 2 law: x : − fk = macom But k = µ k So com = − k (linear) yˆ : N − mg = 0 = µ k mg = −µ k g c) During the sliding, what is the ball’s angular acceleration? nd ˆ Iα = Rfk = R(µ k mg) From 2€ law: τ = Rfk (−z ) €But f k = µ k N So€ (angular) I α (−zˆ ) = τ = Rf (−zˆ ) = µ mg Rµ k mg k k α = I d) What is the speed of the ball when smooth rolling begins? v I com When does€ v =Rω ? vcom€= v0 + acom t vcom = v0 − µ k gt R com € v = v − µ g t com 0 k From kinematics: ω = ω 0 + α t ω Iω Rµ k mg = α = Rµ mg k v e) How long does the ball slide? v = 0 € com (1+ I mR2 ) v − v € t = 0 com µ k g € € .
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