Question
The coefficient of friction between a certain brass block and alarge revolving turntable is u=0.27. How far from the axis ofrotation can the block be placed before it slides off the turntableif the turntable rotates at a constant rate of 33 1/3 rev/min (sothat it requires time T= 60/33.33 seconds to complete onerevolution)?
I was trying to use the formula v= circumference/t However, I'm confused about how to determine theradius.
I was trying to use the formula v= circumference/t However, I'm confused about how to determine theradius.
Explanation / Answer
The coefficient of friction is = 0.27 The angular speed of the turntable is w = 33 (1/3)rev/min = (100/3) rev/min = (100/3) * (2/60) rad/s = 3.49rad/s The speed of the turntable is v = r * w --------------(1) We know from the relation v = ( * r * g)1/2 --------------(2) From equations (1) and (2),we get r * w = ( * r * g)1/2 or (r * w)2 = * r * g or r2 * w2 = * r * g or r = ( * g/w2) Here,g = 9.8 m/s2 Substituting the values in equation (3),we get r = (0.27 * 9.8/(3.49)2) or r = 0.2172 m Therefore,the distance from the axis of rotation can the blockbe placed before it slides off the turntable is r = 0.2172 m. Substituting the values in equation (3),we get r = (0.27 * 9.8/(3.49)2) or r = 0.2172 m Therefore,the distance from the axis of rotation can the blockbe placed before it slides off the turntable is r = 0.2172 m.