Question

Calculate the rotational kinetic energy of a 9-kg motorcycle wheel if its angular velocity is 137 rad/s and its inner radius is 0.280 m and outer radius 0.330 m.

Answer 1

Answer:

The rotational kinetic energy = 7909.73 Joules

Explanation:

The mass of the motorcycle, m = 9kg

The angular velocity, w = 137 rad/s

Inner radius, r = 0.280 m

Outer radius, R = 0.330 m

The rotational kinetic energy is calculated as:

$KE_{\text{rot}}=\frac{m}{4}(r^2+R^2)w^2$

Substitute m = 9, r = 0.280, R = 0.330, and w = 137 rad/s into the formula above

$\begin{gathered} KE_{\text{rot}}=\frac{9}{4}(0.280^2+0.33^2)137^2 \\ KE_{\text{rot}}=\frac{9}{4}(0.1873)(18769) \\ KE_{\text{rot}}=7909.73\text{ }Joules\text{ } \end{gathered}$