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

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 10π^2 N

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 ω is given by: ω = 2π = 2π × (1)/(3) = (2π)/(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 = mω^2 r Substituting the values: f = 15 × ((2π)/(3))^2 × 1.5 f = 15 × (4π^2)/(9) × (3)/(2) f = 15 × (2π^2)/(3) f = 10π^2 N

At a children's park, a heavy rod is pivoted at its centre and...

Options

Correct answer

Step-by-step solution

More from Circular Motion