Question

A particle tied to a string of negligible weight and length $L$ is rotated in a horizontal circular path with constant angular velocity having time period $T$. If the string length is shortened by $L/2$ while the particle is in motion, the time period is:

• A. $4T$
• B. $2T$
• C. $T/2$
• D. $T/4$

Answer 1

The correct option is D $T/4$

According to the law of conservation of angular momentum,
$m{v}_1{r}_2=m{v}_2{r}_2$

or, ${\omega }_1{r}_1^2={\omega }_2{r}_2^2$

or $\frac{2\pi }{T_1}{r}_1^2=\frac{2\pi }{T_2}{r}_2^2$
or, $T_2=T_1\frac{r_2^2}{r_1^2}=T\left(\frac{r^2/4}{r^2}\right)$ [Length in shortened by half]

$=\frac{T}{4}$
So, Time period will be one fourth of the initial time period.