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
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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
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