Question

A small bob tied at one end of a thin string of length $1 m$ is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio $5: 1$ . The velocity of the bob at the height position is $m / s$___
(Take $\left. g =10 m / s ^{2}\right)$

Answer 1

Let the speed of bob at lowest position be $v_{1}$ and at the highest position be $v_{2}$. Maximum tension is at lowest position and minimum tension is at the highest position. Now, using, conservation of mechanical energy,

$\frac{1}{2} mv _{1}^{2}=\frac{1}{2} mv _{2}^{2}+ mg 2 l$
$\Rightarrow v _{1}{ }^{2}= v _{2}{ }^{2}+4 g l \quad \ldots (1)$
Now $T _{\max }- mg =\frac{ mv _{ i }^{2}}{l}$
$\Rightarrow T _{\max }= mg +\frac{ mv _{1}^{2}}{l}$
$\& T _{\min }+ mg =\frac{ mv _{2}^{2}}{l}$
$\Rightarrow T _{\min }=\frac{ mv _{2}^{2}}{l}- mg$
$\frac{ T _{\max }}{ T _{\min }}=\frac{5}{1}$
$\Rightarrow \frac{ mg +\frac{ mv _{1}^{2}}{l}}{\frac{ mv _{2}^{2}}{l}- mg }=\frac{5}{1}$
$\Rightarrow mg +\frac{ mv _{1}^{2}}{l}=\left[\frac{ mv _{2}^{2}}{l}- mg \right] 5$
$\Rightarrow mg +\frac{ m }{l}\left[ v _{2}^{2}+4 g l\right]=\frac{5 mv _{2}^{2}}{l}-5 mg$
$\Rightarrow mg +\frac{ mv _{2}^{2}}{l}+4 mg =\frac{5 mv _{2}^{2}}{l}-5 mg$
$\Rightarrow 10 mg =\frac{4 mv _{2}^{2}}{l}$
$v _{2}{ }^{2}=\frac{10 \times 10 \times 1}{4}$
$\Rightarrow v _{2}^{2}=25 \Rightarrow v _{2}=5 m / s$
Thus, velocity of bob at highest position is $5 m / s$