Question

A stone of mass \(1\text{ kg}\) tied to a light inextensible string of length \(L = \frac{5}{3}\text{ m}\) is rotating in a circular path of radius \(L\) in a vertical plane. If the ratio of maximum tension in the string to the minimum tension in the string is 3 , the speed of the stone at the highest point of the circle is ( \(g =\) acceleration due to gravity)

• A. \(\sqrt{gL}\)
• B. \(\sqrt{2gL}\)
• C. \(\sqrt{4gL}\)
• D. \(\sqrt{8gL}\)

Hint:

Use energy + tension formulas together for vertical circular motion.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For an object moving in a vertical circle, tension is maximum at the lowest point and minimum at the highest point.
We must use Newton's second law for circular motion and the conservation of mechanical energy.
Step 2: Key Formulas or Approach:
Tension at the lowest point: \( T_{\text{max}} = \frac{mv_L^2}{L} + mg \).
Tension at the highest point: \( T_{\text{min}} = \frac{mv_H^2}{L} - mg \).
Conservation of energy: \( \frac{1}{2}mv_L^2 = \frac{1}{2}mv_H^2 + mg(2L) \).
Step 3: Detailed Explanation:
Let \( v_L \) be the speed at the lowest point and \( v_H \) be the speed at the highest point.
From the conservation of energy, we can find the relation between \( v_L \) and \( v_H \):
\[ \frac{1}{2}mv_L^2 - \frac{1}{2}mv_H^2 = 2mgL \] \[ v_L^2 = v_H^2 + 4gL \] It is given that the ratio of maximum tension to minimum tension is 3:
\[ \frac{T_{\text{max}}}{T_{\text{min}}} = 3 \] Substitute the expressions for tension:
\[ \frac{\frac{mv_L^2}{L} + mg}{\frac{mv_H^2}{L} - mg} = 3 \] Cancel mass \( m \) from numerator and denominator and multiply by \( L \):
\[ \frac{v_L^2 + gL}{v_H^2 - gL} = 3 \] Cross-multiply to solve for \( v_L^2 \):
\[ v_L^2 + gL = 3(v_H^2 - gL) \] \[ v_L^2 + gL = 3v_H^2 - 3gL \] \[ v_L^2 = 3v_H^2 - 4gL \] Now, equate the two expressions obtained for \( v_L^2 \):
\[ v_H^2 + 4gL = 3v_H^2 - 4gL \] Rearrange the terms to solve for \( v_H^2 \):
\[ 4gL + 4gL = 3v_H^2 - v_H^2 \] \[ 8gL = 2v_H^2 \] \[ v_H^2 = 4gL \] Taking the square root, we get the speed at the highest point:
\[ v_H = \sqrt{4gL} \] The mass \( m=1\text{ kg} \) and length \( L=\frac{5}{3}\text{ m} \) are extra information not needed to find the speed in terms of \( g \) and \( L \).
Step 4: Final Answer:
The speed of the stone at the highest point is \( \sqrt{4gL} \).