Question

A bob of mass m tied to one end of a light string of length 50 cm whirled in a vertical circle. The minimum velocity with which it has to move so that the string remains taut, when it reaches the top most point of the circle is $(g=10\text{ ms}^{-2})$

• A. $10\text{ ms}^{-1}$
• B. $8\text{ ms}^{-1}$
• C. $15\text{ ms}^{-1}$
• D. $5\text{ ms}^{-1}$
• E. $12\text{ ms}^{-1}$

Hint:

Logic Tip: Memorize the critical velocities for vertical circles: Top: $\sqrt{gr}$ Horizontal (midpoint): $\sqrt{3gr}$ Bottom: $\sqrt{5gr}$ If the calculated $v_{top} = \sqrt{10 \times 0.5} = \sqrt{5} \approx 2.23$ is not in the options, the question is asking for the initial launching velocity at the bottom!

Answer 1

The Correct Option is D

Concept:
For an object moving in a vertical circle on a string, the critical condition for the string to just remain taut at the highest point (top) is that the tension $T \ge 0$. The minimum velocity at the \textit{top} to achieve this is $v_{top} = \sqrt{gr}$. By applying the law of conservation of mechanical energy, the minimum velocity required at the \textit{bottom} to ensure the bob reaches the top and satisfies this condition is $v_{bottom} = \sqrt{5gr}$. \textit{(Note: Based on the provided options, the question implies finding the necessary minimum velocity at the bottom to achieve the taut condition at the top.)}
Step 1: Identify the given parameters.
Length of the string (radius), $r = 50\text{ cm} = 0.5\text{ m}$ Acceleration due to gravity, $g = 10\text{ ms}^{-2}$
Step 2: Calculate the minimum required velocity at the bottom.
Use the standard formula for critical velocity at the lowest point of a vertical circle: $$v_{bottom} = \sqrt{5gr}$$ Substitute the known values: $$v_{bottom} = \sqrt{5 \times 10 \times 0.5}$$ $$v_{bottom} = \sqrt{5 \times 5}$$ $$v_{bottom} = \sqrt{25} = 5\text{ ms}^{-1}$$