A uniform disc of mass M and radius R rolls without slipping on a rough horizontal surface. A horizontal spring of force constant k is attached to the material point that is uppermost in the equilibrium configuration. For small oscillations, the spring extension is approximately 2x when the disc centre moves by x. What is the angular frequency?
Options

Question

A uniform disc of mass M and radius R rolls without slipping on a rough horizontal surface. A horizontal spring of force constant k is attached to the material point that is uppermost in the equilibrium configuration. For small oscillations, the spring extension is approximately 2x when the disc centre moves by x. What is the angular frequency?

Options

G1 Physics · Accepted Answer

For a disc, I = MR²/2. Thus, Mₑ = M+I/R² = 3M/2. Because the spring extension is 2x, the effective spring constant with respect to x is kₑ = 4k. Hence, ω² = 4k/(3M/2) = 8k/(3M). Therefore, ω = √[8k/(3M)]. Correct option: C.