A 100 g block of wood is placed on a turntable that is turning at 25 rpm. If the coefficient of friction between the block and the table is 0.25 how far away from the center can the block be placed?
(g = 9.81 m/s^2)
The turntable is turning at 25 rpm. When a block of wood weighing 100 g is placed on the turntable it experiences two horizontal forces. One is the centrifugal force due to the circular motion of the block about the center of the turntable and the other is the frictional force.
The block does not move from its position until the centrifugal force is greater than the frictional force. The mass of the block is 100 g and the gravitational attraction is 9.81 m/s^2. As the coefficient of friction is 0.25, the frictional force is equal to 0.1*9.81*0.25 = 0.24525 N. The centrifugal force on the block is equal to 0.1*w^2*R where R is the distance of the block from the center and w is the angular velocity of the block. As the turntable turns at 25 rpm, w = 2*pi*(25/60) = 2*(22/7)*(25/60) = 2.619 radians/second. The centrifugal force is 0.1*2.619^2*R.
Equating the frictional force and the centrifugal force gives the maximum distance from the center that the block can be placed.
0.1*2.619^2*R = 0.24525
=> R = 0.24525/0.6859
=> R = 0.0357 m
=> R = 3.57 cm
The block can be kept at a maximum distance of 3.57 cm away from the center of the turntable to ensure that it does not move when the turntable turns.