Question

A small mass attached to a string rotates on frictionless table top as shown. If the tension of the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will

• A. Remain constant.
• B. Increase by a factor of 2.
• C. Increase by a factor of 4.
• D. Decrease by a factor of 2.

Answer 1

Answer: (C)

Increase by a factor of 4.

$K.E.=\frac{L^2}{2I}$

$\because$ From angular momentum conservation about centre.
$L\to$ constant
$I=m{r}^2$
${K.E.}^{\prime }=\frac{L^2}{2\left(m{r}^{\prime 2}\right)}{r}^{\prime }=\frac{r}{2}$

${K.E.}^{\prime }=4K.E.$

$K.E.$ is increased by a factor of 4.