Question

A simple pendulum has a string of length l and 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 circular path about the point of suspension. The tension in the string at the lowest position of the bob is

(a) 3mg

(b) 4mg

(c) 5mg

(d) 6mg

Answer 1

Correct Answer is: (d) 6mg

At A, mv12/l = T1 + mg.

For v1 to be minimum, T1 = 0.

or v1 =√gl.

1/2 mv22 - 1/2 mv12 = mg x 2l

or mv22 / l mv12 / l + 4mg

At B, mv22 / l = T2 - mg

or mg + 4mg = T2 - mg or T2 = 6mg.

Applying the principle of conservation of energy between A and D,

1/2 mv 23 - 1/2 mv12 = mgl or mv32 / l = mv12 / l + 2mg = 3mg = T3.

The net force on the bob at D = √((3mg)2 + (mg)2) =√10mg.