A block of mass 0.5 kg is pressed against the inner vertical wall of a cylindrical drum of radius 4 m. The drum rotates about its vertical axis. The minimum angular speed required so that the block does not slide downward is 5 rad·s⁻¹. Taking g = 10 m·s⁻², what is the coefficient of friction between the block and the drum wall?
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Question

A block of mass 0.5 kg is pressed against the inner vertical wall of a cylindrical drum of radius 4 m. The drum rotates about its vertical axis. The minimum angular speed required so that the block does not slide downward is 5 rad·s⁻¹. Taking g = 10 m·s⁻², what is the coefficient of friction between the block and the drum wall?

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G1 Physics · Accepted Answer

The normal force provided by the rotating wall supplies the centripetal force: N = mω²R. The block is just prevented from sliding downward at the minimum angular speed, so the maximum static friction balances its weight: f(max) = mg. Also, f(max) = μN. Therefore, mg = μmω²R. Cancel m: μ = g ÷ (ω²R). Substitute g = 10 m·s⁻², ω = 5 rad·s⁻¹, R = 4 m. Thus, μ = 10 ÷ [5²×4] = 10 ÷ 100 = 0.1. Correct option: A.