Q.
A rod of mass m and length L is pivoted at point O in a car whose acceleration towards left is a0. The rod is free to rotate in vertical plane. In equilibrium state, the rod remains horizontal when other end is suspended by a spring of force constant k. The time period of small oscillations of rod is T=2πD3.The value of D is ___________. (Given, k=20Nm−1,a0=10ms−2,m=1kg,L=1m
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answer is 5.
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Detailed Solution
In equilibrium, mgL2=kyL∴y=mg2k( initial extension of spring) During oscillations, consider a situation when the angular displacement of rod is θ as shown in the figure.ma0= Pseudo forceln the reference frame of car, restoring torque is.τ=−ma0L2sinθ+myL2cosθ−k(y−Lθ)Lcosθ Using Eq. (i), τ=−ma0L2θ+kL2θ (∵θ is small ) τ=−ma0L2+kL2θ∴ Iα=−ma0L2+kL⋅Lθα=−ma02+kLmL2/3θ=−3a02L+3kmθUsing the given values, we get α=−75θ∴ ω=75=53⇒T=2πω=2π53∴ D=5
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