A particle attached to a string of length $r$ is moving on vertical circular path continuously. If the speed…

1. $1: 1$
2. $1: 2$
3. $1: 7$
4. $1: \sqrt{7}$

Solution

The given situation is shown in the figure
At highest point $A$, $ v_A=\sqrt{7 g r} $ $T_d$ and $T_s$ are tensions in the string when particle is at highest position $(A)$ and lowest position $(B)$. $\therefore$ At point $A$. $ \begin{aligned} & \frac{m v_A^2}{r}=T_A+m g \\ & \Rightarrow \quad \frac{m(\sqrt{7 g r})^2}{r}=T_A+m g \\ & \Rightarrow \quad 7 m g=T_A+m g \\ & \Rightarrow \quad T_A=6 \mathrm{mg} \\ & \end{aligned} $ At point $B$ by the law of conservation of energy, change in potential energy $=$ Change in kinetic energy $ \begin{aligned} & m g(2 r)=\frac{1}{2} m v_{\bar{B}}^2-\frac{1}{2} m v_A^2 \\ \Rightarrow \quad & 2 m g r=\frac{1}{2} m v_B^2-\frac{1}{2} m(7 g r) \\ \Rightarrow \quad & 4 g r=v_B^2-7 g r \\ \Rightarrow \quad & v_s=\sqrt{11 g r} \end{aligned} $ At lowest point $B, T_n-\frac{m v_n^2}{r}=m g$ $ \begin{aligned} T_s-\frac{m}{r} \cdot 11 g r & =m g \\ \Rightarrow \quad T_n-11 m g & =m g \\ \Rightarrow \quad T_s & =12 m g \\ \frac{T_A}{T_s} & =\frac{6 m g}{12 m g}=\frac{1}{2} \\ \Rightarrow \quad T_A: T_s & =1: 2 \end{aligned} $

Asked in: AP EAMCET 2022 (06 Jul Shift 2)

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