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A block of mass 10 kg is in contact against the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be: (g = 10 m/s²)

1. √10 rad/s
2. 10/2π rad/s
3. 10 rad/s
4. 10 π rad/s

Correct answer: √10 rad/s

Solution

To keep the block stationary, the frictional force must balance the weight of the block. The normal force is provided by the centripetal force, which is mω²R. Using the condition for limiting friction, μN = mg, we get μmω²R = mg. Simplifying, ω = √(g/μR). Substituting g = 10 m/s², μ = 0.1, and R = 1 m, we get ω = √(10/0.1) = √100 = 10 rad/s.

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