A horizontal rod of mass m = 20 kg and length L = 5 meters is fixed on the left end to a horizontal, frictionless axis. The other end is held by a thread attached to the ceiling. The rod's moment of inertia is mL2/3.(a) What is the force that the axis exerts on the rod? N(ai) What is the CW torque caused by the mass of the rod about the axis? N*mWhat is the moment of inertia of the rod when it rotates about the axis? kg*m2The thread is burned by a match. Determine the following for the time immediately after the thread breaks (while the rod is still horizontal).(b) The initial angular acceleration of the rod.(Hint: Solve using a force diagram and net torque equation.) rad/sec2(c) The initial linear acceleration of the rod's center of mass. m/s2(d) The force that the axis exerts on the rod. N

The problem can be solved in the following steps:

(a) The force that the axis exerts on the rod can be found by considering the forces acting on the rod. The weight of the rod acts downwards at its center of mass, which is at a distance L/2 from the axis. This force is balanced by the tension in the thread and the force exerted by the axis. Since the rod is in equilibrium, the sum of the forces is zero. Therefore, the force exerted by the axis is equal to the weight of the rod, which is m*g = 20 kg * 9.8 m/s^2 = 196 N.

The CW torque caused by the mass of the rod about the axis is the force of gravity acting at the center of mass times the distance from the axis, which is (mg)(L/2) = (20 kg * 9.8 m/s^2) * (5 m / 2) = 490 N*m.

The moment of inertia of the rod when it rotates about the axis is given as mL^2/3 = 20 kg * (5 m)^2 / 3 = 166.67 kg*m^2.

(b) When the thread is burned, the only force acting on the rod is its weight, which causes a torque about the axis. The net torque is equal to the moment of inertia times the angular acceleration. Therefore, the initial angular acceleration of the rod is the net torque divided by the moment of inertia, which is (mgL/2) / (mL^2/3) = 3g / (2*L) = 3 * 9.8 m/s^2 / (2 * 5 m) = 2.94 rad/s^2.

(c) The initial linear acceleration of the rod's center of mass is the angular acceleration times the distance from the axis, which is (angular acceleration) * (L/2) = 2.94 rad/s^2 * 5 m / 2 = 7.35 m/s^2.

(d) After the thread breaks, the force that the axis exerts on the rod is the component of the rod's weight that acts perpendicular to the rod, which is mgcos(theta), where theta is the angle between the rod and the vertical. Since the rod is horizontal immediately after the thread breaks, theta is 90 degrees and cos(theta) is zero. Therefore, the force that the axis exerts on the rod is zero.