Question

A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +58 N·m is applied to the wheel for 16 s, giving the wheel an angular velocity of +575 rev/min. The external torque is then removed, and the wheel comes to rest 120 s later. (Include the sign in your answers.)

1) Find the moment of inertia of the wheel.

2) Find the frictional torque, which is assumed to be constant.

Answer 1

the applied torque on the wheel T(tau)=Iw /t

58 =I(575)(pie)/30) /16 =575(3.14)( I ) /30(16) here I=moment of inertia w=angular velocity

the moment of inertia=I=15.42 kg)m²

when the wheel gains 575rev/min torque is removed the frictional torque will act on it to stop rotation

applying work energy theorem work done by frictional torque = its initial kinetic energy at frictional torque acts

W_friction =(1/2)(Iw²)=0.5(15.42)(575(3.14) /30 )²   =27928.88J

but Work done by friction =(T(tau)(1/2)(wt) =T(0.5)(575(3.14/30)(120)=3611T

so T(3611)=27928.88 finally torque =27928.88/3611=7.73Nm