1. Tardigrade
  2. Question
  3. Physics
  4. A stone of mass 1 kg tied to a light inextensible string of length L=(5/3) m, is whirling in a vertical circle of radius L. If the ratio of the maximum and minimum tension in the string is 4 and g=10 m / s 2, the speed of stone at the highest point of the circle is

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Q. A stone of mass $1\, kg$ tied to a light inextensible string of length $L=\frac{5}{3} m$, is whirling in a vertical circle of radius $L$. If the ratio of the maximum and minimum tension in the string is $4$ and $g=10\, m / s ^{2},$ the speed of stone at the highest point of the circle is

Work, Energy and Power

Solution:

In the vertical circular motion, the tension in the string is maximum at $A$ and minimum at $B$.
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$T_{A}=\frac{m v_{A}^{2}}{L}+m g=T_{\max }$
$T_{B}=\frac{m v_{B}^{2}}{L}-m g=T_{\min }$
$\frac{T_{\max }}{T_{\min }}=4=\frac{\frac{m v_{A}^{2}}{L}+m g}{\frac{m v_{B}^{2}}{L}-m g}$
or $ \frac{v_{A}^{2}+g L}{v_{B}^{2}-g L}=\frac{4}{1} \ldots$(i)
But $v_{A}^{2}=v_{B}^{2}+2 g(2 L)=v_{B}^{2}+4 g L$
(from the law of conservation of energy)
From (i) $\frac{v_{B}^{2}+5 g L}{v_{B}^{2}-g L}=\frac{4}{1}$
On solving, we get
$v_{B}=\sqrt{3 g L}=\sqrt{3 \times 10 \times \frac{5}{3}}$
$=\sqrt{50}=5 \sqrt{2}\, m / s$