Question

A solid cylinder of mass, $m$, and radius, $r$, is rotating at an angular velocity, $\omega$ when a non-rotating cylinder of double the mass and equal radius drops onto the cylinder.
By what factor does this change the rotational kinetic energy with regard to its initial rotational kinetic energy $K_{rot}$?

• A. $K_{rot}^{\prime }=3K_{rot}$
• B. $K_{rot}^{\prime }=\frac{1}{3}{K}_{rot}$
• C. $K_{rot}^{\prime }=9K_{rot}$
• D. $K_{rot}^{\prime }=\frac{1}{9}{K}_{rot}$
• E. $K_{rot}^{\prime }=K_{rot}$

Answer 1

Answer: (B)

$K_{rot}^{\prime }=\frac{1}{3}{K}_{rot}$

Moment of inertia of the solid cylinder $I=\frac{1}{2}m{r}^2$

Thus initial kinetic energy of the system $K_{rot}=\frac{1}{2}I{w}^2=\frac{1}{4}m{r}^2{w}^2$

Final moment of inertia of the system $I^{\prime }=\frac{1}{2}m{r}^2+\frac{1}{2}\left(2m\right)r^2=\frac{3}{2}m{r}^2$

Let the final angular velocity of the system be $w^{\prime }$.

Using conservation of angular momentum : $Iw=I^{\prime }{w}^{\prime }$

$\therefore$ $\frac{1}{2}m{r}^{2}w=$ $\frac{3}{2}m{r}^2{w}^{\prime }$ $⟹w^{\prime }=\frac{w}{3}$

Thus final kinetic energy of the system $K_{rot}^{\prime }=\frac{1}{2}{I}^{\prime }(w^{\prime }{)}^2=\frac{1}{12}m{r}^2{w}^2$

$⟹$ $K_{rot}^{\prime }=\frac{K_{rot}}{3}$