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.
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
Wfriction =(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
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