Question

A small bob tied at one end of a thin string of length $1m$ 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 $g=10m/s^2)$

Answer 1

Answer: 5

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,

$\frac{1}{2}m{v}_1^2=\frac{1}{2}m{v}_2^2+mg2l$
$\Rightarrow v_1{}^2=v_2{}^2+4gl\dots (1)$
Now $T_{\max }-mg=\frac{m{v}_i^2}{l}$
$\Rightarrow T_{\max }=mg+\frac{m{v}_1^2}{l}$
$\&T_{\min }+mg=\frac{m{v}_2^2}{l}$
$\Rightarrow T_{\min }=\frac{m{v}_2^2}{l}-mg$
$\frac{T_{\max }}{T_{\min }}=\frac{5}{1}$
$\Rightarrow \frac{mg+\frac{m{v}_1^2}{l}}{\frac{m{v}_2^2}{l}-mg}=\frac{5}{1}$
$\Rightarrow mg+\frac{m{v}_1^2}{l}=\left[\frac{m{v}_2^2}{l}-mg\right]5$
$\Rightarrow mg+\frac{m}{l}\left[v_2^2+4gl\right]=\frac{5m{v}_2^2}{l}-5mg$
$\Rightarrow mg+\frac{m{v}_2^2}{l}+4mg=\frac{5m{v}_2^2}{l}-5mg$
$\Rightarrow 10mg=\frac{4m{v}_2^2}{l}$
$v_2{}^2=\frac{10\times 10\times 1}{4}$
$\Rightarrow v_2^2=25\Rightarrow v_2=5m/s$
Thus, velocity of bob at highest position is $5m/s$