Question

A stone of mass 1kg is tied to a light string of length l=10m is whirling in a circular path in the vertical plane. If the ratio of maximum to minimum tensions in the string is 3, find the speeds of the stone at the lowest and highest points.

Answer 1

We will firstly draw the free body diagram of the particle at lowest and highest position during circular motion. We will form equations of energy using all the forces at lower and highest points. From the law of conservation of energy we know that the energy in any motion is conserved. And hence use this concept and will obtain the answer.

Formula used:
$\text{centripetal force}=\dfrac{m{{v}^{2}}}{r}$

Complete step by step answer:
Let the tension at the highest point be denoted by$={{T}_{H}}$.
Let the tension at the highest point be denoted by$={{T}_{L}}$.
Velocity at highest point be v.
Velocity at lowest point be u.
We know that Tension at the highest point and force mg on stone is in the same direction. So let’s see energy at its highest point.
${{T}_{H}}+mg=\dfrac{m{{v}^{2}}}{r}$
Since Tension at lowest point and force mg when stone is whirl is in the opposite direction. So we will find energy at the lowest point.
${{T}_{L}}-mg=\dfrac{m{{u}^{2}}}{r}$
We will use the law of energy of conservation.
$\dfrac{1}{2}m{{u}^{2}}=\dfrac{1}{2}m{{v}^{2}}+mg\left( 2r \right)$
We are given a ratio of lowest and maximum tension.
$\dfrac{{{T}_{L}}}{{{T}_{H}}}=\dfrac{3}{1}\text{ }......(1) \\\$
Now,

& {{T}_{L}}=\dfrac{m{{v}^{2}}}{r}+mg \\\ & {{T}_{H}}=\dfrac{m{{u}^{2}}}{r}-mg \\\ \end{aligned}$$ We will substitute the value in equation (1) $$\begin{aligned} & \dfrac{\dfrac{m{{v}^{2}}}{r}+mg}{\dfrac{m{{u}^{2}}}{r}-mg}=3 \\\ \end{aligned}$$ Solving the equation, $$\begin{aligned} & {{u}^{2}}={{v}^{2}}+4gr.......(2) \\\ & \dfrac{{{v}^{2}}+4gr+gr}{{{v}^{2}}-gr}=3 \\\ & \dfrac{{{v}^{2}}+5gr}{{{v}^{2}}-gr}=3 \\\ & {{v}^{2}}+5gr=3{{v}^{2}}-3gr \\\ & 2{{v}^{2}}=8gr \\\ & {{v}^{2}}=4gr \\\ & v=\sqrt{4\times 10\times 10}=20m/\sec \\\ \end{aligned}$$ Now putting value of $v$ in equation (2) $$\begin{aligned} & {{u}^{2}}={{v}^{2}}+4gr=400+400=800 \\\ & u=20\sqrt{2}m/\sec \\\ \end{aligned}$$ Therefore speed of stone at highest point and lowest point is $20m/\sec \text{ and }20\sqrt{2}m/\sec .$ **Note:** When a body is rotated in a vertical circle, there are three forces acting upon the mass at all times; gravity, tension and centripetal force. Centripetal force tends to move the mass away from the centre whereas the tension tends to pull the mass towards the centre. The centripetal force and tension are acting in opposite directions at all times, so they balance each other out leaving out gravity that determines the tension depending on the position of the mass. When the mass is at the lowest position, the centripetal force acts downward (away from centre) while the tension acts upward (towards the centre). But since the force of gravity is always acting downward, the tension is loaded with two downward forces. When the mass is at the highest position, the centripetal force acts upward (away from centre) while the tension acts downward (towards the centre). But since the force of gravity is again acting downward, the Centripetal force is loaded with two downward forces.