A block of mass 5 kg is in contact with the inner wall of a hollow cylindrical drum of radius 2 m . The coefficient of static friction between the block and the inner wall is 0.5 . The cylinder is vertical and rotates about its axis at an angular velocity of 2 10 rad s ⁻¹ . What is the magnitude of the frictional force acting on the block? (Take g = 10 m s ⁻² )

Step-by-step solution

The normal force acting on the block provides the necessary centripetal force: N = m ^2 R N = 5 (2 10 )^2 2 = 5 40 2 = 400 N The maximum possible static friction (limiting friction) is: f_ max = N = 0.5 400 = 200 N The downward gravitational force on the block is its weight: W = mg = 5 10 = 50 N Since the required upward force to keep the block stationary ( 50 N ) is less than the maximum available static friction ( 200 N ), the block will not slip. Static friction is self-adjusting and will only act with the magni