Answer to Question #108519 in Physics for Aaishah

Question

If a constant torque of 500Nm turns a wheel of moment of inertia 100kgm² about an axis through its center find the gain in anguler velocity in 2 seconds

Answer 1

J=500Nm - torque

I=100kgm² - moment of inertia

$\Delta$t=2s - time interval

$\alpha$ - angular acceleration

$\Delta\omega$ - ? - gain in anguler velocity

$J=I\alpha \\ \alpha=\frac{J}{I}\\ \alpha=\frac{500}{100}=5 \ rad/s^2\\ \Delta\omega=\alpha \Delta t\\ \Delta\omega=5*2=10\ rad/s$

Answer: $\Delta\omega=10\ rad/s$