Question

The bob of a simple pendulum at rest position is given a velocity $V$ in horizontal direction so that the bob describes vertical circle of radius equal to length of pendulum $l$ . If the tension in string is $4$ times weight of bob when the string is horizontal, the velocity of bob when it is crossing highest point of vertical circle is:

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

Answer 1

The correct option is D $\sqrt{2gl}$

At pt B.

$T=\frac{m{v}^2}{L}andT=4mg.$

$4\frac{mg}{m}L=v^2$

$V=\sqrt{4gL}$

Now WET between B and C.

$mgL=\frac{1}{2}m{v}^2-\frac{m}{2}{v}_0^2$

$mgL=2mgL-\frac{m}{2}{v}_0^2$

$v_0=\sqrt{2gL}$