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A particle is whirled in a vertical circle of radius $1.0 \mathrm{~m}$ using a string with one end fixed. If the ratio of maximum to minimum tensions in the string is $5 / 3$, then the minimum velocity (in $\mathrm{m} / \mathrm{s}$ ) of the particle during circular motion is
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The correct answer is:
$10 \mathrm{~m} / \mathrm{s}$
Step1: Draw the rough diagram for given situation
Step2: Find ratio of maximum and minimum tension.
Given, $T_{\max }: T_{\min }=5:3$
At the lower most point,
$T_{\max }-\mathrm{mg}=\mathrm{mv2maxR} \ldots$..(i)
At the upper most point,
$\mathrm{mg}+T_{\min }=\mathrm{mv} 2 \min R \ldots$...(ii)
Using equation (i),(ii)
$\frac{T_{\max }}{T_{\min }}=\frac{\left(\frac{m v_{\max }^2+m g}{R}\right)}{\left(\frac{m v_{\min }^2}{R}-m g\right)} \ldots(iii)$
step3: Find minimum velocity of the particle.
Apply conservation of energy at lower most point and upper most point,
$\begin{aligned}& \frac{1}{2} m v_{\text {max }}^2=m g(2 R)+\frac{1}{2} m v_{\text {min }}^2 \\& \Rightarrow v_{\text {max }}^2=2 g(2 R)+v_{\text {min }}^2 \ldots \text { (iv) }\end{aligned}$
Using equation (iii) and (iv)
$\begin{aligned}& \frac{5}{3}=\frac{\frac{v_{\text {min }}^2+2 \cdot g \cdot 2 R}{R}+g}{\frac{v_{\text {min }}^2}{R}-g} \\& \Rightarrow v_{\min }=10 \mathrm{~m} / \mathrm{s}\end{aligned}$
Step2: Find ratio of maximum and minimum tension.
Given, $T_{\max }: T_{\min }=5:3$
At the lower most point,
$T_{\max }-\mathrm{mg}=\mathrm{mv2maxR} \ldots$..(i)
At the upper most point,
$\mathrm{mg}+T_{\min }=\mathrm{mv} 2 \min R \ldots$...(ii)
Using equation (i),(ii)
$\frac{T_{\max }}{T_{\min }}=\frac{\left(\frac{m v_{\max }^2+m g}{R}\right)}{\left(\frac{m v_{\min }^2}{R}-m g\right)} \ldots(iii)$
step3: Find minimum velocity of the particle.
Apply conservation of energy at lower most point and upper most point,
$\begin{aligned}& \frac{1}{2} m v_{\text {max }}^2=m g(2 R)+\frac{1}{2} m v_{\text {min }}^2 \\& \Rightarrow v_{\text {max }}^2=2 g(2 R)+v_{\text {min }}^2 \ldots \text { (iv) }\end{aligned}$
Using equation (iii) and (iv)
$\begin{aligned}& \frac{5}{3}=\frac{\frac{v_{\text {min }}^2+2 \cdot g \cdot 2 R}{R}+g}{\frac{v_{\text {min }}^2}{R}-g} \\& \Rightarrow v_{\min }=10 \mathrm{~m} / \mathrm{s}\end{aligned}$
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