A torque $\tau =20\text{ Nm}$ is applied to a wheel having a moment of inertia $I=4{\text{ kg m}}^2$. What is the angular acceleration $\alpha$ of the wheel?

Solution

1. The relation between torque and angular acceleration is $\tau =I\alpha$.
2. Rearrange to solve for angular acceleration: $\alpha =\frac{\tau }{I}$.
3. Substitute $\tau =20\text{ Nm}$ and $I=4{\text{ kg m}}^2$ to get $$\begin{array}{rl}\alpha & =\frac{20}{4}\\ & =5{\text{ rad/s}}^2\end{array}$$.
4. Hence the answer is (B).