At a children's park, a heavy rod is pivoted at its centre and forced to rotate about the pivot, maintaining a horizontal orientation. As shown in the figure, two kids hold the rod near its ends and rotate along with it. Assume the mass of each kid is 15 kg , the distance between the points where the two kids hold the rod is 3.0 m , and the rod rotates at a rate of 20 revolutions per minute. Determine the force of fr

Step-by-step solution

Given: Mass of each kid, m = 15 kg Distance between the two kids is 3.0 m , so the radius of the circular path for each kid is r = 3.0 2 = 1.5 m . Frequency of revolution, = 20 rev/min = 20 60 rev/s = 1 3 rev/s . The angular velocity is given by: = 2 = 2 1 3 = 2 3 rad/s The force of friction exerted by the rod on the kid provides the necessary centripetal force to keep the kid moving in the circular path. Therefore, the force of friction f is: f = m ^2 r Substituting the values: f = 15 ( 2 3 )^2 1.5 f = 15 4 ^2 9 3