A grinding 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