Question

A simple pendulum has a string of length $l$ and bob of mass $m$. When the bob is at its lower position, it is given the maximum horizontal speed necessary for it to move in a circular path about the point of suspension. The tension in the string at the lowest position of the bob is

• A. $mg$
• B. $3mg$
• C. $\sqrt{10}mg$
• D. $4mg$

Answer 1

Answer: (A), (D)

The correct options are
A $mg$
D $4mg$

Lets A is the topmost point of the circle and B be the lowest point of the circle.

Let $v_2$ and $v_1$ be the velocities at B and A respectively.
Applying the principle of conservation of energy between A and B.

$\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2=mg2l$

or $\frac{m{v}_2^2}{l}=\frac{m{v}_1^2}{l}+4mg$ .....(1)

At B, $\frac{m{v}_2^2}{l}=T-mg$ .....(2)

At A. $\frac{m{v}_1^2}{l}=mg$ ....(3)

Solving (1), (2) and (3)

or $mg+4mg=T-mg$ or $T=6mg$