Question

A particles 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:

• A. $\sqrt{gl}$
• B. $\sqrt{2gl}$
• C. $\sqrt{3gl}$
• D. $\sqrt{5gl}$.

Answer 1

Answer: (C)

$\sqrt{3gl}$

At the tap of circle

T = O and $mg=\frac{m}{l}$

$2mgl=\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_T^2$

$V_T=\sqrt{gl}$

So, $4gl+gl=V^2$

$V=\sqrt{5gl}$

So minimum speed of partical $mgl=\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_{min}^2$

$V_{min}=\sqrt{3gl}$

Hence (C) option is correct