A stone of mass m , tied to the end of a string, is whirled around in a horizontal circle. The length of the string is reduced gradually, keeping the angular momentum of the stone about the centre of

Question

A stone of mass m , tied to the end of a string, is whirled around in a horizontal circle. The length of the string is reduced gradually, keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by T = a r n , where 'a' is constant, r is the instantaneous radius of the circle. Then n is equal to

TestHub · Accepted Answer

Angular momentum of the stone about the centre is $L = m v r$ .

Since $L$ is constant, the velocity is $v = \frac{L}{m r}$ .

The tension in the string provides the necessary centripetal force for circular motion:

$T = \frac{m v ^{2}}{r}$

Substituting the value of $v$ :

$T = \frac{m}{r} (\frac{L}{m r})^{2}$

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

$T = (\frac{L ^{2}}{m}) r^{- 3}$

Comparing this with $T = a r^{n}$ , we get $a = \frac{L ^{2}}{m}$ and $n = - 3$ .

Answer: $- 3$

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