PY1052 Problem Set 8 â Autumn 2004 Solutions

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m = 1.5 kg m = 2.0 kg1 2M = 4.0 kg(8) A board is balanced as is shown above. m 1 and m 2 are x 1 = 2.5 m andx 2 = 1.2 m to the left of the support point, which is just under the centre of theboard. How far is M from the support point?We have three conditions for the equilibrium:∑ ∑ ∑ Fx = 0 Fy = 0 τ = 0In this case, there are no horizontal forces, so we only have the last two of these equations.(i) ∑ F y = 0F support + m 1 g + m 2 g + Mg + m board g = 0where F support is the upward force of the support and m board is the mass of the board (whichwe do not know).(ii) ∑ τ = 0We can choose any point about which to take the torques. It is convenient in this case tochoose the centre of the board as the point about which we take our torques. The reason isthat if we choose any other point, we will get a torque equation that has three unknowns:F support , m board and X, where X is the distance from the support to the mass M. This meansthat we’ll have two equations with three unknowns, since we are not given the mass of theboard. If we take torques about the centre of the board, neither the weight of the board norF support exert torques, and the only unknown in our torque equation will be X:∑ τ = m1 gx 1 sin(90) + m 2 gx 2 sin(90) − MgX sin(90) = 0X = m 1x 1 +m 2 x 2= 0M(1.5 kg)(2.5 m)+(2.0 kg)(1.2 m)X = = 1.54 m4.0 kg(1b) A bowler throws a bowling ball of radius R = 11 cm along a lane. Theball slides on the lane with an initial speed v 0 = 8.5 m/s and initial angular speedω 0 = 0. The coefficient of kinetic friction between the ball and lane is µ k = 0.21.The kinetic friction force acting on the ball causes a linear acceleration of theball while also producing a torque that causes an angular acceleration of the ball.When the speed v has decreased enough and the angular speed ω has increasedenough, the ball stops sliding and then rolls smoothly. (a) What then is v interms of ω? (b) During the sliding, what are the ball’s translational and angularacceleration? (c) How far does the ball slide?(a) When the ball has slowed down enough to roll smoothly, the linear and angular velocitiesmust be related by the usual formula for smooth rolling: v = Rω.(b) The ball’s translational acceleration is given by∑F = ma−µ k N = −µ k mg = ma−µ k g = a
where N is the normal force and m is the mass of the bowling ball.acceleration is given byThe ball’s angular∑τ = Iαµ k NR = µ k mgR = [ 2 5 mR2 ]α5µ k g2R = αHere, τ = RF k sin φ, where F k is the force of kinetic friction and φ is the angle between theradius vector to the point where the force of kinetic friction acts and the direction of F k , whichis 90 ◦ .(c) We can find how far the ball slides by first finding v using a and finding ω using α, thenfinding the time the ball slides by setting v = Rω :v = v i + at= v i − µ k gt slideω = ω i + αt= 5µ kg2R t slide (ω i = 0)v i − µ k gt slide = R( 5µ kg2R t slide)= (µ k g + 5 2 µ kg)t slide = 7 2 µ kgt slidet slide = 2 v i7 µ k g = 2 8.5 m/s7 (0.21)(9.8 m/s 2 ) = 1.18 sThe distance the ball slides is then given byd = v i t + 1 2 at2= v i t − µ kg2 t2 slide= (8.5 m/s)(1.18 s) − (0.21)(9.8 m/s2 )(1.18 s) 2 = 8.60 m2Putty¡¢¡¢¡¢¢¡¢¡¢¡ ¢¡¢¡¢¡ ¢¡¢¡¢¡ ¢¡¢¡¢¡ ¢¡¢¡¢Rotation axis(2b) Two 2.00 kg balls are attached to the ends of a thin rod of negligible massthat is 50.0 cm long. The rod is free to rotate in a vertical plane without frictionabout a horizontal axis through its centre. With the rod initially horizontal, a50.0 g wad of putty drops onto one of the balls, hitting it with a speed of 3.00 m/sand sticking to it. (a) What is the angular speed of the system just after the puttyhits? (b) What is the ratio of the kinetic energy of the system after the collisionto that of the putty just before the collision? (c) Through what angle will thesystem rotate until it momentarily stops?
Page 2 and 3: PE i + KE i = PE f + KE fMgh i + 0
Page 5: L i = L fmv i r i = mv f r fv f = v
Page 9: ©h max¡ ¡¡¡¡ ¡ ©Therefore,

PY1052 Problem Set 8 â Autumn 2004 Solutions

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PY1052 Problem Set 8 â Autumn 2004 Solutions