Q. An object of mass m is attached to light string which passes through a hollow tube. The object is set into rotation in a horizontal circle of radius , $r_{1}.$If the string is pulled shortening the radius to $r_{2}$,the ratio of new kinetic energy to the original kinetic energy is
A) $\left( \frac{r_{2}}{r_{1}} \right)^{2}$
B) $\left( \frac{r_{1}}{r_{2}} \right)^{2}$
C) $\frac{r_{1}}{r_{2}}$
D) $\frac{r_{2}}{r_{1\ \ \ }}$
Explanation
(b) Kinetic energy $K = \frac{1}{2}mr^{2}\omega^{2}$
$ie,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K \propto r^{2}$
The ratio of new kinetic energy to the original KE is given
$\frac{K_{2}}{K_{1}} = \left( \frac{r_{2}}{r_{1}} \right)^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.