Agrinding wheel is in the form of a uniform solid disk of radius 7.00 cm and mass 2.00 kg. it starts from rest and accelerates uniformly under the action of the constant torque of 0.600 n ? m that the motor exerts on the wheel. how long does the wheel take to reach its final operating speed of 1200 rev/ min? (note that moment of inertia of a disk is given by i = 0.5*mr2)

It will take 1.03 sec for the wheel to reach its final operating speed.

Solution:
First we calculate for the angular acceleration a from the torque formula which is the product of the moment of inertia i and the angular acceleration a:
     torque = ia
     a = torque / i

We know that the moment of inertia i of a solid disk is i = 0.5*mr^2, where radius r is in meters:
     a = torque / 0.5 * mr^2

Now, we substitute the values to the angular acceleration a equation:
     a = 0.600 Nm / (0.5 * 2.00 kg * (0.07m)^2)
        = 122.45 radians/s^2

We convert the final angular velocity Wf, which is the final operating speed 1200 rev/min, to rad/s: 
     Wf = 1200 rev/min * 2π/60 = 125.66 rad/s

For rotational motion,
     Wf = Wi + at 
     125.66 rad/s  = 0 + (122.45 rad/s^2)t 

The time t is therefore
     t = 125.66 / 122.45t
       = 1.03 sec

This is the answer: body 1 + engine 2

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