The combination of an applied force and a constant frictional forceproduces a co

Question

The combination of an applied force and a constant frictional forceproduces a constant total torque of 35.8N·m on a wheel rotating about a fixed axis.The applied force acts for 6.08 s. Duringthis time the angular speed of the wheel increases from 0 to10.5 rad/s. The applied force is thenremoved, and the wheel comes to rest in 60.9 s. (a) Find the moment of inertia of thewheel.
1kg·m2

(b) Find the magnitude of the frictional torque.
2N·m

(c) Find the total number of revolutions of the wheel.
3 rev

I have found part a and part b but i can't find part c i don't knowhow to set up the equation please help thank you
(a) Find the moment of inertia of thewheel.
1kg·m2

(b) Find the magnitude of the frictional torque.
2N·m

(c) Find the total number of revolutions of the wheel.
3 rev

I have found part a and part b but i can't find part c i don't knowhow to set up the equation please help thank you

Explanation / Answer

        Given that thetotal torque is = 35.8 N.m         The applied forceacts for t1 = 6.08 s          Intialangular velocity is 1 = 0 and final angular velocity is2 = 10.5 rad/s ----------------------------------------------------------------------------------------------    The rorque is = I                           = I*(2 - 1) / t1                        I = *t1 / (2 - 1)                            =-----------kg.m2              If the applied force is removed then timetaken to stop is t2 = 60.9s                   Then the frictional torque is = I*(2 - 0) / t2                                                                = -------- N.m Then the total angular displacment is =(1/2)(2 - 1)*t12 / t1 +2*t2 + (1/2)(2 - 0)*t22 /t2                                                            =---------- rad

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