We need to calculate the angular acceleration of the disk first.
The disk starts at 217 rpm and stops, so its final angular speed is zero, therefore the variation of angular speed is
[tex]\Delta \omega = 0 - 217 rpm =-217 rpm = -22.7 rad/s[/tex]
The time interval in which the disk stops is
[tex]\Delta t=2.25 min=135 s[/tex]
Therefore, the angular acceleration is
[tex]\alpha = \frac{\Delta \omega}{\Delta t}= \frac{-22.7 rad/s}{135 s}=-0.17 rad/s^2 [/tex]
Now, we can find the torque needed to stop the disk by using the equivalent of Netwon's second law for the rotational motions:
[tex]\tau = I \alpha[/tex]
where [tex]\tau [/tex] is the torque, [tex]I[/tex] is the moment of inertia of the object and [tex]\alpha[/tex] is the angular acceleration.
The moment of inertia for a disk is [tex]I= \frac{1}{2} mr^2 [/tex], therefore substituting this into the formula and by using the data of the exercise, we can find the torque:
[tex]\tau = ( \frac{1}{2} mr^2)\alpha = \frac{1}{2}(24.1 kg)(1.83m)^2 (-0.17rad/s^2)=-6.79kgm^2s^{-2} [/tex]
where the negative sign means the torque is directed against the direction of the rotation, so anti-clockwise.
