Answer to Question #156169 in Physics for Marie

Question

A wheel of mass 50kg has a radius of 0.4m. It is making 480rpm. What a the torque necessary to bring it to rest in 40 seconds.

Answer 1

Let's first find the angular acceleration of the wheel:

$\alpha=\dfrac{\omega_f-\omega_i}{t},$ $\alpha=\dfrac{0-480\ \dfrac{rev}{min}\cdot\dfrac{1\ min}{60\ s}\cdot\dfrac{2\pi\ rad}{1\ rev}}{40\ s}=-50.3\ \dfrac{rad}{s^2}.$

The sign minus means that the wheel decelerates.

Let's find the moment of inertia of the wheel:

$I=mr^2=50\ kg\cdot(0.4\ m)^2=8\ kg\cdot m^2.$

Finally, we can find the torque necessary to bring the wheel to rest:

$\tau=\alpha I=-50.3\ \dfrac{rad}{s^2}\cdot 8\ kg\cdot m^2=-402.4\ N\cdot m.$

The sign minus means that the torque directed in the opposite direction to the rotation of the wheel.

Answer:

$\tau=-402.4\ N\cdot m.$