Question

A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ________ m/s. (Take g=10 m/s²)

Hint:

In vertical circular motion, the difference between maximum and minimum tension ($T_B - T_T$) is always $6mg$ regardless of the speed, provided the string remains taut.

Answer 1

Correct Answer: 5

Step 1: $T_{max}$ is at the bottom: $T_B = \frac{mv_B^2}{L} + mg$.
Step 2: $T_{min}$ is at the top: $T_T = \frac{mv_T^2}{L} - mg$.
Step 3: By energy conservation: $v_B^2 = v_T^2 + 4gL$.
Step 4: $T_B = \frac{m(v_T^2 + 4gL)}{L} + mg = \frac{mv_T^2}{L} + 5mg$.
Step 5: Given $\frac{T_B}{T_T} = 5 \Rightarrow \frac{mv_T^2/L + 5mg}{mv_T^2/L - mg} = 5$.
Step 6: $\frac{mv_T^2}{L} + 5mg = 5\frac{mv_T^2}{L} - 5mg \Rightarrow 10mg = 4\frac{mv_T^2}{L}$.
Step 7: $v_T^2 = \frac{10gL}{4} = \frac{10 \times 10 \times 1}{4} = 25 \Rightarrow v_T = 5$ m/s.