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.