Question

A wheel is rotating freely at angular speed

800 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertia of the

first, is suddenly coupled to the same shaft. (a) What is the angular

speed of the resultant combination of the shaft and two wheels?

(b) What fraction of the original rotational kinetic energy is lost?

Answer 1

(a) By conservation of angular momentum:

$$L_i=L_1=L_1+L_2=L_f,\\\space\\ I_1\omega_i=(I_1+I_2)\omega_f,\\\space\\ \omega_2=\omega_1\dfrac{I_1}{I_1+I_2}=\omega_1\dfrac{I_1}{I_1+2I_1}=\dfrac{\omega_1}3,\\\space\\ \omega_2=\frac{800}3=267\text{ rev/min}.$$

(b) The initial, final, and fraction of lost kinetic energy:

$$KE_i=\dfrac12I_1\omega_1^2,\\\space\\ KE_f=\dfrac12I_1\omega_2^2+\dfrac12I_2\omega_2^2,\\\space\\ \epsilon=1-\dfrac{KE_f}{KE_i}=1-\dfrac{\omega_2^2+2\omega_2^2}{\omega_1^2}=0.67.$$