Question

The flywheel of an engine has moment of inertia 2.50kg.m^2 about its rotation axis.What constant torque is required to bring it up to an angular speed of 400rev/min in 8s,starting from rest

Answer 1

According the the law of rotational movement, the torque is

$$\tau = I\varepsilon$$

where $I = 2.5\space kg\cdot m^2$ and $\varepsilon$ is the angular acceleration.

By definition, the angular acceleration is:

$$\varepsilon = \dfrac{\omega - \omega_0}{t} = \dfrac{\omega}{t}$$

where $\omega = 2\pi \cdot 400 = 800\pi \dfrac{rad}{min} = \dfrac{40}{3}\pi\space \dfrac{rad}{s}$ is the final angular velocity and $\omega_0 = 0$ is the intial angular velocity (since starts from rest). Here $t = 8s$ is the time of velocity increasing.

Substituting angular acceleration, obtain:

$$\tau = I\varepsilon = \dfrac{I\omega}{t}\\ \tau = \dfrac{2.5\cdot \dfrac{40}{3}\pi}{8}\approx 13.1\space N\cdot m$$

Answer. $13.1\space N\cdot m$.