Answer
$1.28\ kg.m^2$
Work Step by Step
Given;
Torque on a wheel $\tau =32\ N.m$
Angular acceleration $\alpha = 25\ rad/s^2$
Newton's second law relation between torque and angular acceleration is $\tau = I\alpha$. We rearrange the formula and solve for $I$:
$\tau = I\alpha$
$I = \frac{\tau}{\alpha} $
$I = \frac{32\ N.m}{25\ rad/s^2} $
$I =1.28\ kg.m^2$
