Question:
A particle attached to a string of length \(r\) is moving on a vertical circular path continuously. If the speed of the particle at the highest point is
\[
\sqrt{7gr},
\]
then the ratio of the respective tensions in the string holding it at the highest and the lowest points is
A particle attached to a string of length \(r\) is moving on a vertical circular path continuously. If the speed of the particle at the highest point is
\[
\sqrt{7gr},
\]
then the ratio of the respective tensions in the string holding it at the highest and the lowest points is
Show Hint
In vertical circular motion, use centripetal force equations separately at the top and bottom points, and use conservation of mechanical energy to relate the speeds.
Updated On: Jun 26, 2026
- \(1:1\)
- \(1:2\)
- \(1:7\)
- \(1:\sqrt{7}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Find tension at the highest point.
At the highest point of vertical circular motion, both tension and weight act towards the center.
Therefore, \[ T_H+mg=\frac{mv_H^2}{r} \] Given, \[ v_H=\sqrt{7gr} \] So, \[ v_H^2=7gr \] Substitute: \[ T_H+mg=\frac{m(7gr)}{r} \] \[ T_H+mg=7mg \] \[ T_H=6mg \]
Step 2: Find speed at the lowest point using energy conservation.
The particle descends through a height \[ 2r \] from the highest point to the lowest point.
Using conservation of mechanical energy: \[ \frac{1}{2}mv_L^2 = \frac{1}{2}mv_H^2+mg(2r) \] Substitute \[ v_H^2=7gr \] \[ \frac{1}{2}mv_L^2 = \frac{1}{2}m(7gr)+2mgr \] \[ \frac{1}{2}mv_L^2 = \frac{7}{2}mgr+2mgr \] \[ \frac{1}{2}mv_L^2 = \frac{11}{2}mgr \] Thus, \[ v_L^2=11gr \]
Step 3: Find tension at the lowest point.
At the lowest point, \[ T_L-mg=\frac{mv_L^2}{r} \] Substitute \[ v_L^2=11gr \] \[ T_L-mg=\frac{m(11gr)}{r} \] \[ T_L-mg=11mg \] \[ T_L=12mg \]
Step 4: Find the ratio of tensions.
\[ T_H:T_L = 6mg:12mg \] \[ =1:2 \]
Step 5: Final conclusion.
Hence, the required ratio is \[ \boxed{1:2} \]
At the highest point of vertical circular motion, both tension and weight act towards the center.
Therefore, \[ T_H+mg=\frac{mv_H^2}{r} \] Given, \[ v_H=\sqrt{7gr} \] So, \[ v_H^2=7gr \] Substitute: \[ T_H+mg=\frac{m(7gr)}{r} \] \[ T_H+mg=7mg \] \[ T_H=6mg \]
Step 2: Find speed at the lowest point using energy conservation.
The particle descends through a height \[ 2r \] from the highest point to the lowest point.
Using conservation of mechanical energy: \[ \frac{1}{2}mv_L^2 = \frac{1}{2}mv_H^2+mg(2r) \] Substitute \[ v_H^2=7gr \] \[ \frac{1}{2}mv_L^2 = \frac{1}{2}m(7gr)+2mgr \] \[ \frac{1}{2}mv_L^2 = \frac{7}{2}mgr+2mgr \] \[ \frac{1}{2}mv_L^2 = \frac{11}{2}mgr \] Thus, \[ v_L^2=11gr \]
Step 3: Find tension at the lowest point.
At the lowest point, \[ T_L-mg=\frac{mv_L^2}{r} \] Substitute \[ v_L^2=11gr \] \[ T_L-mg=\frac{m(11gr)}{r} \] \[ T_L-mg=11mg \] \[ T_L=12mg \]
Step 4: Find the ratio of tensions.
\[ T_H:T_L = 6mg:12mg \] \[ =1:2 \]
Step 5: Final conclusion.
Hence, the required ratio is \[ \boxed{1:2} \]
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