A 55-kg runner runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner’s velocity relative to the earth has magnitude 2.8 m/s. The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.20 rad/s relative to the earth. The radius of the turntable is 3.0 m, and its moment of inertia about the axis of rotation is Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can model the runner as a particle.)
Solution 98P This question is based on the concept of angular momentum conservation. We shall have to find the initial and final angular momenta of the system. Given , mass of the runner m = 55 kg r Speed of the runner v = 2.8 r/s Radius of the turntable r = 3.0 m Therefore, angular momentum of the runner, L r m × v r r r L r 55 kg × 2.8 m/s × 3.0 m 2 L r 462 kg.m /s Now, moment of inertia of the turntable I = 80 kg.m 2 t Its angular velocity = 0.2s rad/s 2 Its angular momentum L = 80 kg.mt× 0.20 rad/s L = 16 kg.m /s 2 t Since the turntable is moving in the opposite direction, hence total initial momentum of 2 2 2 the system is L = L i L = r62 kg.t /s 16 kg.m /s = 446 kg.m /s L = 446 kg.m /s…..(1) i Let the final angular velocity be . f Final angular momentum of the runner L = 55 kg × (3.00) m × …..(2) rf f 2 Final angular momentum of the turntable L tf = 80 kg.m × …..(3)f Adding equations (2) and (3), we shall get the final angular momentum of the system. L f L rf + L tf L = (55 kg × (3.00) m + 80 kg.m ) 2 f f 2 2 L f (495 kg.m + 80 kg.m ) f L = 575 kg.m …..(4) f f According the principle of conservation of angular momentum and equating equations (1) and (2), we get L = L i f 2 2 446 kg.m /s = 575 kg.m . f = 0.776 rad/s f Therefore, the final angular velocity of the system will be 0.776 rad/s.
