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Question

A particle of mass m begins to slide down a fixed smooth sphere from the top as shown. What is its acceleration when it breaks off the sphere?


A
2 g/3
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B
5g/3
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C
g
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D
g/3
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Solution

The correct option is B 5g/3
Suppose the particle of mass m slides down a smooth sphere of radius R, starting from rest at the top. Suppose the particle leaves the sphere at angle θ with vertical.
The particle slides off when the normal force is zero.
Using Newton's second law,
mv2r=mg(cosθ)

from WE theorem
mv22=mgr(1cosθ)

from simplification,

mgcosθ=2mg(1cosθ)
3cosθ=2
cosθ=23
Now tangential acceleration is gsinθ=g1(23)2=5g3

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