Question

Please show all work

A wheel is rotating freely at angular speed 600 rad/s on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with three times the rotational inertia of the first, is suddenly coupled to the same shaft. What is the angular speed of the resultant combination of the shaft and two wheels? What fraction of the original rotational kinetic energy is lost in the collision?

Answer 1

From law of conservation of angular momentum $L= I_{1}\omega _{1}=I_{2}\omega _{2}$

$\Rightarrow Ix600= \left ( I+3I \right )\omega$

  $\Rightarrow Ix600= 4I\omega$

  $\Rightarrow \omega =600/4 = 150rad/s$

From energy equation $\frac{\Delta K}{K_{i}}= \frac{K_{f}-K_{i}}{K_{i}}$

  $\Rightarrow \frac{\Delta K}{K_{i}}= \frac{1/2x4Ix150^{2}-1/2xI600^{2}}{1/2xIx600^{2}}$

The loss in kinetic energy $\Rightarrow \frac{\Delta K}{K_{i}}= \frac{270000}{360000}=\frac{3}{4}$