A wooden plank OA rotates in vertical plane about horizontal axis through O with a constant angular velocity `omega = 3 `rad/s . As it passes the position `theta = 0 ` , a small mass m is placed upon it at a radial distance `r = 0*5` m . If the mass m starts sliding at ` theta = 37^(@)` , find the coefficient of friction between mass and the plane .

Explanation

On first sight appears that `37^(@)` is angle of repose. In the case of angle of repose, body remains in equilibrium but here small mass is moving in a particular path so it is not in equilibrium . It has some centripetal acceleration . In limiting case .

` mg sin theta - mu N = N omega^(2) r`
Also , ` N = mg cos theta `
` rArr g sin theta - mu g cos theta = omega^(2)r`
` rArr mu g cos theta = g sin theta - omega^(2)r`
` rArr mu xx 10 xx 4/5 = 10 xx 3/5 - 9 (0*5)`
` = 6 - 4*5`
` mu = 3/16 `