Explanation:
Mass of the diskshaped grindstone, m = 1.1 kg
Radius of disk, r = 0.09 m
Angular velocity, [tex]\omega=73.3\ rad/s[/tex]
Time, t = 42.4 s
We need to find the frictional torque exerted on the grindstone. Torque in the rotational kinematics is given by :
[tex]\tau=I\alpha[/tex]
I is moment of inertia of disk, [tex]I=\dfrac{mr^2}{2}[/tex]
[tex]\tau=\dfrac{mr^2\alpha }{2}\\\\\tau=\dfrac{1.1\times (0.09)^2\times 73.3 }{2\times 42.4}\\\\\tau=7.7\times 10^{-3}\ N-m[/tex]
So, the frictional torque exerted on the grindstone is [tex]7.7\times 10^{-3}\ N-m[/tex].
