Question

A student holds a bike wheel and starts it spinning with an initial angular speed of 7.0 rotations per second. The wheel is subject to some friction, so it gradually slows down. In the 10.0 s period following the inital spin, the bike wheel undergoes 57.5 complete rotations. Assuming the frictional torque remains constant, how much more time At, will it take the bike wheel to come to a complete stop? At, = The bike wheel has a mass of 0.625 kg and a radius of 0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque Tf that was acting on the spinning wheel. N.m

Answer 1

W = 7 rotation/ second = 7x22 = 140 = 43.982 rad/sec t-los o= 57.5x2n 361-283 rad. o- wt + tat² 361.283 = 43.982 x 10 + $ xd*(10) 361. 283 - 439.82 + 504 Q- 361283-439.82 →-|-5707 radse w6=0 w = 43.982 rad/s d = 1.5707 1. t= wf-w =0-43.982 » 28 seconds. o 1.5707 28-10 = 18 seconds more. st So P= Id = Mg2a I = 482 P = 0.625*(0.315)? x 1.5707 T 0.0974 Nm Scanned with CamScanner