A large drum having radius R is spinning around its axis with angular velocity ω, as shown in figure. The minimum value of ω so that a body of mass M remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass M as μ, is

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(2) √g/(μ R). Idea: the wall pushes inward to supply the centripetal force, and friction against that same wall holds the body up. Two equations, and M cancels. Radially: N=Mω²R. Vertically, at the slowest spin that works, friction is at its limit: μ N=Mg. Substituting, μ Mω²R=Mg ωₘᵢₙ=√g/(μ R). Sanity check on the direction of each dependence: a slipperier wall (smaller μ) or a tighter drum (smaller R) both demand a faster spin, and both appear in the denominator, as they should.

A large drum having radius R is spinning around its axis with angular velocity ω, as shown in figure. The minimum value of ω so that a body of mass M remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass M as μ, is

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