A point mass is tied to one end of a cord whose other end passes through a vertical hollow tube, caught in one hand. The point mass is being rotated in a horizontal circle of radius $2m$ with a speed of $4m/s$. The cord is then pulled down so that the radius of the circle reduces to $1m$. Compute the new linear and angular velocities of the point mass and also the ratio of kinetic energies in the initial and final states.

The force on the point mass due to cord is radial and hence the torque about the center of rotation is zero. Therefore, the angular momentum must remain constant as the cord is shortened. Let mass of the particle be $m$ let it rotate initially in circle of radius $r_1$ with linear velocity $v_1$ and angular velocity ${\omega }_1$. Further let the corresponding quantities in the final state be radius $r_2$, linear velocity $v_2$ and angular velocity ${\omega }_2$.

Since, Initial angular momentum = Final angular momentum

Therefore, $I_1{\omega }_1=I_2{\omega }_2\Rightarrow m{r_1}^2\frac{v_1}{r_1}=m{r_2}^2\frac{v_2}{r_2}\Rightarrow r_1{v}_1=r_2{v}_2$

Therefore, $v_2=\frac{r_1}{r_2}{v}_1=\frac{2}{1}\times 4=8m/s$and ${\omega }_2=\frac{v_2}{r_2}=\frac{8}{1}=8 rad/s$.

$\frac{Final K.E.}{Initial K.E.}=\frac{{\displaystyle \frac{1}{2}}{I}_2{{\omega }_2}^2}{{\displaystyle \frac{1}{2}}{I}_1{{\omega }_1}^2}=\frac{m{r_2}^2\times {\left[{\displaystyle \frac{v_2}{r_2}}\right]}^2}{m{r_1}^2\times {\left[{\displaystyle \frac{v_1}{r_1}}\right]}^2}=\frac{{v_2}^2}{{v_1}^2}=\frac{{\left(8\right)}^2}{{\left(4\right)}^2}=4$