VA Chapter 11 Homework x C A Counterweight of Mass X A Count

VA Chapter 11 Homework x C A Counterweight of Mass X A Counterweight of Mass X C dvanced physics questio X C fi www.webassign.ne 13683435# t/web/Student/Assignment-Responses/submit?dep Q10 11. points SerPSE9 P.035 My Notes Ask Your Teacher A uniform cylindrical turntable of radius 1.70 m and mass 32.3 kg rotates counterclockwise in a horizontal plane with an initial angular speed of 4m rad/s. The fixed turntable bearing lump of clay of mass 2.43 kg and is frictionless. A negligible size is dropped onto the turntable from a small distance above it and immediately sticks to the turntable at a point 1.60 m to the east of the axis. (a) Find the final angular speed of the clay and turntable magnitude rad/s direction Selec (b) Is mechanical energy of the turntable-clay system constant in this process? Yes No What, if any, is the change in internal energy? (c) Is momentum of the system constant in this process? Yes No What, if any, is the amount of impulse imparted by the bearing? magnitude direction Selec v 12. C+ -l 1 points SerPSE9 P.043. My Notes Ask Your Teacher 4/18/2016

Solution

a) MI = 0.5 * m * r² = 0.5 * 32.3 * 1.7² = 46.67 kg.m²

L = 46.67 * 4 pi = 586.47 kg.m²/s

Final MI = 46.67 + (2.43 * 1.6²) = 52.89 kg.m²

w_final = 586.47 / 52.89

= 11.1 rad/s

b) The inelastic colliding and sticking of the lump of clay will not allow mechanical energy conservation. And there will be a loss in internal energy

c) The linear momentum of the system is not conserved and the impulse imparted by the bearing is just the change in linear momentum which corresponds to the momentum gained by the lump of clay as it sticks to the rotating turntable

impulse = m v\' = m r w\'

= 2.43 * 1.6 * 11.1

= 43.16 kg.m/s