Question

A torque of 7.4 N • m is applied to a wheel with a moment of inertia of 0.092 kg . cm?

. What is the resulting

angular acceleration?

Answer 1

$$\tau=I\alpha,$$ $$\alpha=\dfrac{\tau}{I}=\dfrac{7.4\ N\times m}{0.092\ kg\times m^2}=80.4\ \dfrac{rad}{s^2}.$$