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. \(1:1\)
• B. \(1:2\)
• C. \(1:7\)
• D. \(1:\sqrt{7}\)

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.

Answer 1

The Correct Option is B

Step 1: Apply Newton's second law at the highest point of the circular path.
At the highest point, both the tension $T_H$ in the string and the weight $mg$ of the particle act downward, both pointing toward the center of the circle. The net centripetal force provides the circular motion: \[ T_H + mg = \frac{mv_H^2}{r} \] At the top of the circle, the center is below the particle, so both forces add up toward the center.
Step 2: Calculate the tension at the highest point.
Given that $v_H = \sqrt{7gr}$, so $v_H^2 = 7gr$. Substituting: \[ T_H + mg = \frac{m(7gr)}{r} = 7mg \] \[ T_H = 7mg - mg = 6mg \]
Step 3: Use conservation of mechanical energy to find speed at the lowest point.
As the particle moves from the highest point to the lowest point, it descends a height of $2r$ (the diameter of the circle). By conservation of energy: \[ \frac{1}{2}mv_L^2 = \frac{1}{2}mv_H^2 + mg(2r) \] \[ v_L^2 = v_H^2 + 4gr = 7gr + 4gr = 11gr \] The particle gains kinetic energy as it descends, so $v_L > v_H$, which is physically correct.
Step 4: Apply Newton's second law at the lowest point.
At the lowest point, the tension $T_L$ acts upward (toward the center) and the weight $mg$ acts downward (away from the center). The net centripetal force is: \[ T_L - mg = \frac{mv_L^2}{r} \] Substituting $v_L^2 = 11gr$: \[ T_L - mg = \frac{m(11gr)}{r} = 11mg \] \[ T_L = 11mg + mg = 12mg \]
Step 5: Find the ratio of tensions.
\[ \frac{T_H}{T_L} = \frac{6mg}{12mg} = \frac{1}{2} \] So the ratio is $T_H : T_L = 1 : 2$.
Step 6: Interpret the physical meaning of the result.
The tension at the lowest point is always greater than at the highest point in vertical circular motion. At the bottom, the string must not only support the weight but also provide additional centripetal force, while at the top, gravity itself helps provide the centripetal force. The ratio 1:2 shows the lowest point string tension is twice the highest point tension. \[ \boxed{T_H : T_L = 1 : 2} \]