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Question

# A uniform disc of mass m and radius r is pivoted at point p and is free to rotate in vertical plane. The centre C of disc is initially in horizontal position with p. if it is released from this position then it's angular acceleration when the line PC is inclined to the horizontal at an angle q

A
α=2gcosθ6R
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B
α=5gcosθ5R
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C
α=4gcosθ3R
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D
α=2gcosθ3R
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Solution

## The correct option is D α=2gcosθ3Rwe need to find the torque on the disc about its point P when it is at angle with the horizontal.now by the equation we can sayτ=F×R×sin(90−θ)plug in values in itτ=mg×R×cosθnow we can useτ=Iαnow the moment of inertia of disc about point PI=32mR2now plug in all values in above equationmgRcosθ=32mR2×aso angular acceleration is given asα=2gcosθ3Rso above is the angular acceleration of disc at given positionHence,option (D) is correct answer.

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