Answer to Question #117072 in Physics for MICHAEL

Question

A wheel of mass 6.0kg and radius of gyration 40cm is rotating at 300prm. Find it's moment of inertia and it's rotational KE

Answer 1

By definition, the moment of inertia of the body with the radius of gyration $R_g = 40 cm$ and mass $m = 6kg$ will be:

$$I = mR_g^2 = 6\cdot 0.4^2 = 0.96 kg\cdot m^2$$

The rotational KE:

$$E_{rot} = \dfrac12I\omega^2$$

where $\omega$ is the angular velocity. If the frequency of rotation is $\nu = 300 rpm = 5Hz$, then the angular velocity will be:

$$\omega = 2\pi\nu = 2\pi\cdot 5 \approx 31.4 rad/s$$

Thus, the rotational KE will be:

$$E_{rot} = \dfrac12\cdot 0.96\cdot 31.4^2 = 473.7 J$$

Answer. I = 0.96 kg*m^2, Erot = 473.7 J.