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← IIT-JEE 2008 · Paper 2
PhysicsSingle correct+3 / -1Question 27 of 66

IIT-JEE 2008 · Paper 2, Physics question 27

A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A is just sufficient to make it reach the point B. The angle θ\theta at which the speed of the bob is half of that at A, satisfies

A vertical circle of radius L: the bob at the lowest point A has horizontal velocity V; B is the top of the circle. The string makes angle θ with the vertical (measured from the downward vertical through the centre) at a general position.
Choose one option
Answer and worked solution

Official answer: D

Just reaching the top on a string needs V2=5gLV^2=5gL.

Half speed. v2=V24=54gLv^2=\frac{V^2}4=\frac54gL. Energy conservation from A:
v2=V2−2gL(1−cos⁡θ)  ⟹  54=5−2(1−cos⁡θ)  ⟹  cos⁡θ=−78.v^2=V^2-2gL(1-\cos\theta)\implies\frac54=5-2(1-\cos\theta)\implies\cos\theta=-\frac78.
θ≈151°\theta\approx151°, between 3π4\frac{3\pi}4 (135°) and π\pi.

(D)

Topics:Motion in a vertical circle
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