Rotating Man

Classical Mechanics Level 2

A hollow, vertical cylinder is rotating around its axis. A person stuck on the inner surface of the cylinder rotates with the cylinder without anything to rest his/her feet on. The person rotates at the same rotational velocity as the cylinder. The coefficient of static friction between the person and the surface of the cylinder is 0.4. If the cylinder has a radius of 3 meters, what is the minimum velocity at which the cylinder needs to rotate for the person to not fall down?

Remember, the linear velocity is asked, not rotational velocity.

g = 9.81 m/s^2


The answer is 8.58.

Siddharth Gopujkar by Siddharth Gopujkar , 30, India

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1 solution

The centrifugal force acting on the man is m v 2 r \dfrac{mv^2}{r} where m , r m, r and v v are the mass of the man, radius of cross section of the cylinder, and linear velocity of the man required respectively. Therefore the friction force acting on him is μ s m v 2 r \dfrac{\mu_smv^2}{r} where μ s \mu_s is the coefficient of static friction. Since this force balances the weight of the man, we can write μ s m v 2 r = m g \dfrac{\mu_smv^2}{r}=mg , or v = g r μ s v=\sqrt {\dfrac{gr}{\mu_s}} . Substituting the values of the quantities, we get v = 9.81 × 3 0.4 = 8.57787... v=\sqrt {\dfrac{9.81\times 3}{0.4}}=\boxed {8.57787...}

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