Question

A stone of mass $1\text{kg}$ tied to a light string of length $l=\frac{10}{3}\text{m}$ is whirled in a circular path in a vertical plane. If the ratio of the maximum to minimum tension in the string is $4$, find the speed of the stone at the highest point.

• A. $20\text{m/s}$
• B. $10\text{m/s}$
• C. $5\text{m/s}$
  
• D. $1\text{m/s}$

Answer 1

The correct option is B $10\text{m/s}$

Let the velocity at the highest point be $V_P$
Let the velocity at the lowest point be $V_R$

From conservation of mechanical energy,$\frac{1}{2}m{V_p}^2+mg\left(2l\right)=\frac{1}{2}m{V_R}^2$

${V_p}^2+4gl={V_R}^2$

At the lowest point, $T$ is maximum,

Using Newton's second law at the lowest point we have,

$T_{max}-mg=\frac{{m{V}_R}^2}{l}$

$\Rightarrow T_{max}=mg+\frac{m{V_R}^2}{l}$

At the highest point, $T$ is minimum.

Using Newton's second law at the highest point we have,

$T_{min}+mg=\frac{{m{V}_P}^2}{l}$

$\Rightarrow T_{min}=\frac{m{V_P}^2}{l}-mg$

$\because \frac{T_{max}}{T_{min}}=4\therefore \frac{\frac{m{V_R}^2}{l}+mg}{\frac{m{V}_p^2}{l}-mg}=4\Rightarrow \frac{{V_R}^2+gl}{{V_P}^2-gl}=4$

We know that

${V_R}^2={V_P}^2+4gl\Rightarrow \frac{{V_P}^2+5gl}{{V_P}^2-gl}=4\Rightarrow 3{V_P}^2=9gl$

$\Rightarrow V_P=\sqrt{3gl}=\sqrt{3\times 10\times \frac{10}{3}}=10\text{m/s}$

Why this Question ? Key point- Always remember that in a verticle circular motion, the tension is maximum at the lowest point and minimum at the highest point.