Question

A simple pendulum of length $L$ carries a bob of mass $m$. When the bob is at its lowest position, it is given the minimum horizontal speed necessary for it to move in a vertical circle about the point of suspension. When the string is horizontal the net force on the bob is:

• A. $\sqrt{10}mg$
• B. $\sqrt{5}mg$
• C. $4mg$
• D. $1mg$

Answer 1

The correct option is A $\sqrt{10}mg$

V=minimum horizontal speed= $\sqrt{5Lg}$

using energy conservation,

$\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=mgL\therefore \frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2-mgL=\frac{1}{2}m5gL-mgL\therefore v^2=\frac{3gL}{2}\times 2=3gLT=\frac{m{v}^2}{L}=3mg\therefore force=\sqrt{{\left(3mg\right)}^2+{\left(mg\right)}^2}=\sqrt{10}mg$