Question

A particle of mass \(m\) tied to a string of length \(r\) is whirled in a vertical circle. Its minimum speed at the bottom is}

• A. [option in figure]
• B. [option in figure]
• C. [option in figure]
• D. [option in figure]
• E. [option in figure]

Hint:

For minimum speed in vertical circle: \[ v_{\text{top,min}}=\sqrt{gr} \] and then use energy conservation to find the bottom speed.

Answer 1

Answer: (E)

For complete vertical circular motion, minimum speed at the top must satisfy: \[ v_{\text{top}}=\sqrt{gr} \] Using conservation of mechanical energy between bottom and top: \[ \frac12 mv_b^2 = \frac12 mv_{\text{top}}^2 + 2mgr \] Substitute: \[ v_{\text{top}}^2=gr \] \[ \frac12 mv_b^2 = \frac12 m(gr) + 2mgr \] \[ v_b^2 = gr + 4gr = 5gr \] So the minimum speed at the bottom is: \[ v_b=\sqrt{5gr} \]
Hence, the correct answer is the option corresponding to: \[ \boxed{\sqrt{5gr}} \]