Questions
A solid cylinder of radius r and mass M rests on a curved path of radius R as shown figure. When displaced through a small angle as shown and left to itself, it executes simple harmonic motion. Then the time period of oscillation is (Assume that the cylinder rolls without slipping)
detailed solution
Correct option is A
Restoring torque acting on a cylinder after a small displacement θ , about an axis passing through point of contact O with the curved path, τ=−Mgrsinθ, for small angles sinθ≈θTherefore, τ=−Mgrθ………….(1)Angular acceleration of the cylinder is α=d2ϕdt2 From the diagram, rϕ=(R−r)θ ϕ=(R−r)θr Consequently, α=d2dt2(R−r)θr α=(R−r)rd2θdt2……………………. (2) Moment of inertia of the cylinder about a point of contact is I=32Mr2 Using the definition of torque τ=Iα and substituting equation (2) in it, τ=32Mr2(R−r)rd2θdt2 Substituting equation (1) in the above relation, −Mgrθ=32Mr2(R−r)rd2θdt2 −Mgθ=32M(R−r)d2θdt2 −gθ=32(R−r)d2θdt2 This equation can be written as 0=d2θdt2+23(R−r)gθ Where d2θdt2=α, Therefore −23(R−r)gθ=α Comparing this with the equation of motion for angular simple harmonic motion, α=−ω2θ , ω2=2g3(R−r) Hence angular frequency is ω=2g3(R−r)2πT=2g3(R−r), where ω=2πT T=2π3(R−r)2gTalk to our academic expert!
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A disc of radius R and mass M is pivoted at the rim and is set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be
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