A grinding wheel, with a mass of 22.9 kg and a radius of 22.6 cm, is a uniform c

Question

A grinding wheel, with a mass of 22.9 kg and a radius of 22.6 cm, is a uniform cylindrical disk. (a) Find the rotational inertia of the wheel about its central axis. Correct: Your answer is correct. kg·m2 (b) When the grinding wheel's motor is turned off, friction causes the wheel to slow from 1200 rpm to rest in 59.8 s. What torque must the motor provide to accelerate the wheel from rest to 1200 rpm in 3.95 s? Assume that the frictional torque is the same regardless of whether the motor is on or off. Incorrect: Your answer is incorrect. N·m

Explanation / Answer

The rotational inertia of the wheel about its central axis is 0.5 M R R, where M is the mass of the wheel and R is the radius of the wheel.

0.5 x 22.9 x 0.226 x 0.226 = 0.5848 kg.m^2.

Torque = Rotational Inertia x angular acceleration.

In the problem the initial angular speed = 1200 rev/ minute = 20 rev/s.

Final angular speed is zero.

Change in angular speed = -20 rev/ s

Angular acceleration = change in angular speed / time.

= -20/ 59.8

= -0.33 revolutions / s^2

This is equal but opposite to the frictional torque.

This torque will oppose the rotation when the motor is on.

The final angular acceleration needed is 20/ 3.95 = 5.06 revolutions / s^2

Therefore we have to give an acceleration of (5.06 + 0.33)

= 5.39 revolutions / s^2

= 33.86 radian/ s^2 (Since one revolution = 2pi radian)

Torque needed is I x acceleration = 0.5848 x 33.86 = 19.80 Nm.

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