A uniform rod, pivoted at its upper end, hangs vertically. It is displaced by an angle of 60^ and subsequently released. Determine the magnitude of the force acting on a particle of mass dm located at the tip of the rod when it forms an angle of 37^ with the vertical.

Step-by-step solution

Let the mass of the uniform rod be M and its length be L . The moment of inertia of the rod about the pivot is I = ML^2 3 . When the rod is at an angle = 37^ with the vertical, the torque about the pivot is provided by the weight of the rod acting at its center of mass: = Mg L 2 37^ Using = I , the angular acceleration is: ML^2 3 = Mg L 2 ( 3 5 ) = 3g 2L 3 5 = 9g 10L = 0.9g L The tangential acceleration of the particle at the tip of the rod is: a_t = L = 0.9g To find the angular velocity at = 37^ , we use the conse