Question 301 of 308

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 is in the ratio $5:1$. The velocity of the bob at the highest position is ______ $m{s}^{-1}$. (Take $g=10m{s}^{-2}$)

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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,
$\dfrac{1}{2}m{v}_{1}^{2}=\dfrac{1}{2}m{v}_{2}^{2}+mg2l$
$\Rightarrow {v}_{1}^{2}={v}_{2}^{2}+4gl....\left(1\right)$
Now, ${T}_{\max }-mg=\dfrac{m{v}_{1}^{2}}{l}$
$\Rightarrow {T}_{\max }=mg+\dfrac{m{v}_{1}^{2}}{l}$
&{T}_{\min }+mg=\dfrac{m{v}_{2}^{2}}{l}
$\Rightarrow {T}_{\min }=\dfrac{m{v}_{2}^{2}}{l}-mg$
$\dfrac{{T}_{\max }}{{T}_{\min }}=\dfrac{5}{1}$
$\Rightarrow \dfrac{mg+\dfrac{m{v}_{1}^{2}}{l}}{\dfrac{m{v}_{2}^{2}}{l}-mg}=\dfrac{5}{1}$
$\Rightarrow mg+\dfrac{m{v}_{1}^{2}}{l}=\left[\dfrac{m{v}_{2}^{2}}{l}-mg\right]5$
$\Rightarrow mg+\dfrac{m}{l}\left[{v}_{2}^{2}+4gl\right]=\dfrac{5m{v}_{2}^{2}}{l}-5mg$
$\Rightarrow mg+\dfrac{m{v}_{2}^{2}}{l}+4mg=\dfrac{5m{v}_{2}^{2}}{l}-5mg$
$\Rightarrow 10mg=\dfrac{4m{v}_{2}^{2}}{l}$
${v}_{2}=\dfrac{10\times 10\times 1}{4}$
Thus, the velocity of the bob at the highest position is $5m{s}^{-1}.$