6. system of particles and rotational motion — Physics STD 11 Science — Question

To find the ratio of kinetic energies in the two cases Solution$:$

When hollow cylinder slides without rolling, it posses only transnational Kinetic energy $K_{r}=\frac{1}{2} m v^{2}$

When it rolls slipping, it posses both type of Kinetic energy

$K_{N}=\frac{1}{2} m v^{2}\left[1+\frac{K^{2}}{R^{2}}\right]$

Therefore

$\frac{K_{r}}{K_{N}}=\frac{\frac{1}{2} m v^{2}}{\frac{1}{2} m v^{2}\left[1+\frac{K^{2}}{R^{2}}\right]}$

and for hollow cylinder $\frac{K^{2}}{R^{2}}=1$

So the above equation becomes,

$\Longrightarrow \frac{K_{r}}{K_{N}}=\frac{1}{2}$

Hence the ratio of kinetic energies in the two cases is $1: 2$