Rotation Ques 48
- A stone of mass $m$, tied to the end of a string, is whirled around in a horizontal circle. (Neglect the force due to gravity). 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 a constant, $r$ is the instantaneous radius of the circle. Then $n=\ldots \ldots$.
$(1993,1 M)$
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Answer:
Correct Answer: 48.-3
Solution:
- $m v r=k($ a constant $) \Rightarrow v=\frac{k}{m r}$
$$ \begin{aligned} T=\frac{m v^{2}}{r} & =\frac{m}{r} \quad \frac{k^{2}}{m r} \quad=\frac{k^{2}}{m} \cdot \frac{1}{r^{3}} \\ & =A r^{-3} \quad \text { where, } A=\frac{k^{2}}{m} \end{aligned} $$
Hence, $\quad n=-3$
BETA
