Question

A wheel is rotating freely at angular speed 800 rev/min on a shaft whose rotational inertia is negligible. A sec- ond 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) The final angular speed can be found from conservation of angular momentum:

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

(b) The fraction of lost kinetic energy:

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