Question

A point of mass is tied to one end of a cord whose other end is passes through a vertical hollow tube, caught in one hand. The point mass is being rotated in a horizontal circle of radius $2\text{m}$ with speed of $4\text{m/s}$. The cord is then pulled down so that the radius of the circle reduces to $1\text{m}$. Find the ratio of final kinetic energy and initial kinetic energy.

• A. $1:1$
• B. $3:1$
• C. $2:1$
• D. $4:1$

Answer 1

The correct option is D $4:1$

Here, the force on the point mass due to cord is radial hence the torque about the centre of rotation is zero.

Therefore, the angular momentum will remain constant as the cord is shortened.

Let initially,
$r_1=2\text{m}$
$v_1=$ linear velocity $=4\text{m/s}$
${\omega }_1=$ angular velocity
Finally, when $r_2=1\text{m}$
$v_2=$ linear velocity
${\omega }_2=$ Final angular velocity

From angular momentum conservation,

Initial angular momentum = Final angular momentum

$I_1{\omega }_1=I_2{\omega }_2$

MOI of hollow ring $I=m{r}^2$

Hence, $m{r}_1^2\times \frac{v_1}{r_1}=m{r}_2^2\times \left(\frac{v_2}{v_1}\right)[\because v=\omega r]$

$\Rightarrow r_1{v}_1=r_2{v}_2$

$\Rightarrow v_2=\left(\frac{r_1}{r_2}\right)\times v_1$

$\Rightarrow v_2=\left(\frac{2}{1}\right)\times 4=8\text{m/s}$

Initial KE $=\frac{1}{2}{I}_1{\omega }_1^2=\frac{1}{2}\times m{r}_1^2{\left(\frac{v_1}{r_1}\right)}^2$
$=\frac{1}{2}m{v}_1^2$

Final KE $=\frac{1}{2}{I}_2{\omega }_2^2=\frac{1}{2}\times m{r}_2^2{\left(\frac{v_2}{r_2}\right)}^2$

$=\frac{1}{2}m{v}_2^2$

$\frac{\text{Final KE}}{\text{Initial KE}}=\frac{\frac{1}{2}m{v}_2^2}{\frac{1}{2}m{v}_1^2}$

$=\frac{v_2^2}{v_1^2}=\frac{8^2}{4^2}$

$\frac{\text{Final KE}}{\text{Initial KE}}=\frac{4}{1}$

Hence, $\left(D\right)$ is the correct answer.