Question

A particle is rotated in a vertical circle by connecting it to a string of length l and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is

Options

• \[\sqrt{gl}\]
   
• \[\sqrt{2gl}\]
   
• \[\sqrt{3gl}\]
   
• \[\sqrt{5gl}\]
   

Answer 1

\[\sqrt{3gl}\]

Suppose that one end of an extensible string is attached to a mass m, while the other end is fixed. The mass moves with a velocity v in a vertical circle of radius R. At some instant, the string makes an angle θ with the vertical as shown in the figure.

For a complete circle, the minimum velocity at L must be \[v_L = \sqrt{5gl}\] . 

Applying the law of conservation of energy, we have:
Total energy at M = total energy at L

\[\text{ i .e }  . , \frac{1}{2}m {v_M}^2 + mgl = \frac{1}{2}m {v_L}^2 \]
\[ \Rightarrow \frac{1}{2}m {v_M}^2 = \frac{1}{2}m {v_L}^2 - mgl\]
\[\text{ Using }  v_L \geq \sqrt{5gl}, \text{ we have} : \]
\[\frac{1}{2}m {v_M}^2 \geq \frac{1}{2}m(5gl) - mgl\]
\[ \therefore v_M = \sqrt{3gl}\]