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  • A small bob tied at one end of a string of the length 1m is decribing vertical circle so that the maximum and minimum lensions in the string are in the ratio 5:1. The velocity of the bob at the highest position is __________m/s (take g= 10 m/s2</s

A small bob tied at one end of a string of the length 1m is decribing vertical circle so that the maximum and minimum lensions in the string are in the ratio 5:1. The velocity of the bob at the highest position is __________m/s (take g= 10 m/s2)
Option: 1 4
Option: 2 5
Option: 3 6
Option: 4 7

Answers (1)

best_answer

Let the speed of bob at the lowest position be V_1 and at the highest position be V_2. Maximum tension is at the lowest position and minimum tension is at the highest position.

Now, using, conservation of mechanical energy, 

\begin{aligned} &\frac{1}{2} \mathrm{mv}_{1}^{2}=\frac{1}{2} \mathrm{mv}_{2}^{2}+\mathrm{mg} 2 l\\ &\Rightarrow \mathrm{v}_{1}^{2}=\mathrm{v}_{2}^{2}+4 g l \end{aligned}

\begin{array}{l} \text { Now } \mathrm{T}_{\max }-\mathrm{mg}=\frac{\mathrm{mv}_{1}^{2}}{l} \\ \Rightarrow \mathrm{T}_{\max }=\mathrm{mg}+\frac{\mathrm{mv}_{1}^{2}}{l} \\ \\ \& \mathrm{~T}_{\min }+\mathrm{mg}=\frac{\mathrm{mv}_{2}^{2}}{l} \\ \Rightarrow \mathrm{T}_{\min }=\frac{\mathrm{mv}_{2}^{2}}{l}-\mathrm{mg} \end{array}

So

\begin{array}{l} \frac{\mathrm{T}_{\max }}{\mathrm{T}_{\min }}=\frac{5}{1} \\ \Rightarrow \mathrm{mg}+\frac{\mathrm{mv}_{1}^{2}}{l } \end{array}

\begin{array}{l} \Rightarrow \mathrm{mg}+\frac{\mathrm{mv}_{1}^{2}}{l} =\left[\frac{\mathrm{mv}_{2}^{2}}{l}-\mathrm{mg}\right] 5 \\ \Rightarrow \mathrm{mg}+\frac{\mathrm{m}}{l}\left[\mathrm{v}_{2}^{2}+4 \mathrm{~g} l\right] =\frac{5 \mathrm{mv}_{2}^{2}}{l}-5 \mathrm{mg} \\ \Rightarrow \mathrm{mg}+\frac{\mathrm{mv}_{2}^{2}}{l}+4 \mathrm{mg} =\frac{5 \mathrm{mv}_{2}^{2}}{l}-5 \mathrm{mg} \\ \Rightarrow 10 \mathrm{mg}=\frac{4 \mathrm{mv}_{2}^{2}}{l} \end{array}

\begin{array}{l} \mathrm{v}_{2}^{2}=\frac{10 \times 10 \times 1}{4} \\ \Rightarrow \mathrm{v}_{2}^{2}=25 \\ \Rightarrow \mathrm{v}_{2}=5 \mathrm{~m} / \mathrm{s} \end{array}

Thus, the velocity of the bob at the highest position is 5 m/s. 

Posted by

avinash.dongre

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