Question

A stone of mass $m$, tied to the end of a string, is whirled in a horizontal circle in zero gravity. The length of the string is gradually decreased without exerting any external torque. At an instant when the radius of the circle is r, the tension in the string is $T=A{r}^{-n}$, where $A$ is a constant. Find the value of n (integer only)

Answer 1

Let, initial and final radius of the horizontal circle be $R$ and $r$ and initial and final velocity of the stone be $v$ and $v^{\prime }$respectively.

Since there is no external torque acting on the system, hence angular momentum is conserved.

$L_i=L_f$

$mvR=m{v}^{{}^{\prime }}r........\left(1\right)$

Let, $mvR=K\left[K\text{is a constant}\right]$

From (1) we get,

$v^{\prime }=\frac{K}{mr}$

Now, Tension $T$ in the string is,

$T=\frac{m(v^{\prime }{)}^2}{r}(\because g=0)$

$T=\frac{m}{r}\frac{K^2}{m^2{r}^2}=\frac{K^2}{m}{r}^{-3}$

Given expression is $T=A{r}^{-n}$

$\therefore A=\frac{K^2}{m}\&n=-3$

Therefore, answer is $n=3$.

Why this question ? This question involves a law of conservation of angular momentum where there is no external torque on the object.