Question

A uniform flexible chain of length 3/2 m rests on fixed smooth sphere of radius r=2/pi m such that one end A of chain is on the top of the sphere while the other end b is hanging freely. chain is held stationary by a horizontal thread PA. Calculate the acceleration of chain when the horizontal string PA is burnt

Answer 1

The acceleration of the chain is a = 7.58 ms-2

Explanation:

• The length of chain is in contact with sphere and lower 0.5 m hanging freely.
• When the thread is burnt, chain starts to accelerate.
• Let the magnitude of acceleration be a and mass per unit length of chain be λ.

Mass of this elemental length = λ(Rdθ)

Consider the tangential forces

(T + dT) + λRg sinθdθ-T = λRadθ

Or  dT = λR(a – g sinθ)dθ

To solve equation integrate it with 0 to π/2 .

ʃ dT = λRʃ(a – g sinθ)dθ

Here θ = 0 means at top point A, tension is zero and tension is assumed to be T1 at θ=π/2.

Therefore T1 = λR(a π/2 – g)……..(1)

Consider top l m length and lower 0.5 m length.

0.5λg - T1 = 0.5 λa……….(2)

λR(a π/2 – g) = 0.5λg - 0.5 λa

From equations 1 and 2 we get

a = (4+π/3π)g = 7.58 ms-2

The acceleration of the chain is a = 7.58 ms-2