A uniform solid sphere of mass M and radius R rolls without slipping. A horizontal spring of force constant k is attached to the material point that is uppermost in the equilibrium position. For sufficiently small oscillations, the spring extension may be taken as twice the displacement x of the centre. What is the angular frequency?
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Question

A uniform solid sphere of mass M and radius R rolls without slipping. A horizontal spring of force constant k is attached to the material point that is uppermost in the equilibrium position. For sufficiently small oscillations, the spring extension may be taken as twice the displacement x of the centre. What is the angular frequency?

Options

G1 Physics · Accepted Answer

For a rolling solid sphere, Mₑ = M+I/R² = M+2M/5 = 7M/5. For a small centre displacement x, the top material point moves horizontally by approximately 2x. Hence spring extension is 2x. Spring potential energy: U = ½k(2x)² = 2kx². Writing U = ½kₑx², we obtain kₑ = 4k. Thus, ω² = 4k/(7M/5) = 20k/(7M). Therefore, ω = √[20k/(7M)]. Correct option: C.