PY1052 Problem Set 8 – Autumn 2004 Solutions

L i = L fmv i r i = mv f r fv f = v ir ir fv f = (3.50 m/s) 0.5000.400= 4.38 m/s(d) We can now find the mass M 2 using the same approach as in part (a)T − (M 1 + M 2 )g = 0mv 2− (M 1 + M 2 )gr= 0M 2 = mv2rg − M 1M 2 =(0.200 kg)(4.38 m/s)2(0.400 m)(9.8 m/s 2 )M 2 = 0.479 kg− 0.500 kg(7) A wheel with a moment of inertia I is rotating freely at an angular speed of800 rev/min on an axle with negligible moment of inertia. A second wheel withmoment of inertia 2I that is initially at rest is then suddenly coupled to the sameaxle, so that it also rotates. (a) What is the angular speed of the combinationof the two wheels rotating about the axle? (b) What fraction of the originalrotational kinetic energy is lost?(a) The key is that the angular momentum of the two wheels is conserved when they arecoupled:L before = L afterI 1i ω 1i + I 2i ω 2i = I 1f ω 1f + I 2f ω 2fIω 1i + 2I(0) = (I + 2I)ω fω f = ω 1i3(b) The kinetic energy that is lost is given by∆KE = KE before − KE after= 267 rev/min= ( 1 2 I 1ω 2 1i + 1 2 I 2(0) 2 ) − 1 2 I fω 2 f= 1 2 Iω2 1i − 1 2 (3I)ω2 fand the fraction that is lost is given by∆KEKE i=12 Iω2 1i − 1 2 (3I)ω2 f12 Iω2 1i= ω2 1i − 3ω 2 fω 2 1i= (800 rev/min)2 − 3(267 rev/min) 2(800 rev/min) 2= 0.6666% of the initial kinetic energy is lost when the wheels are coupled.