Question

9) A flywheel initially rotating at 1200 rpm stops in 4.00 min when only friction acts. If an additional torque of 300 N-m is applied, it stops in 1.00 min. (a) What is the rotational inertia of the wheel? (b) What is the frictional torque?

Answer 1

Solution:

Initial angular velocity of the flywheel,

Time taken to stop the flywheel when friction acts,

Final angular velocity of the flywheel when it stops,

Apply to find angular acceleration, due to frictional torque

Negative sign indicates that the frictional torque acts in the opposite direction to the motion of flywheel.

If an additional torque of 300 N-m is applied the flywheel stops in 1 min.

Apply to find total angular acceleration due to frictional torque and additional torque.

Negative sign indicates that the frictional torque and additional torque act in the opposite direction to the motion of flywheel.

Additional torque,

Let the rotational inertia of the wheel be .

Angular acceleration due to additional torque, can be derived from equation

Now,

(a) The rotational inertia of the wheel,

(b) Frictional torque is given by,

  

So, the frictional torque is 98.73 Nm.