PLEASE I NEED HELP WITH PHYSICS, IVE BEEN TRYING ON THIS QUESTION FOR DAYS

Question

A long, uniform 5 kg rod of length 1.5 m is supported at the left end by a horizontal axis into the page and perpendicular to the rod, as shown. The right end is connected to the ceiling by a thin vertical thread so that the rod is horizontal.

a) Determine the magnitude of the force exerted on the rod by the axis. The acceleration of gravity is 9.8 m/s 2 , and the moment of inertia of the rod about the axis at the end of the rod is M ℓ2/3. Answer in units of N.

b) The thread is then burned by a match. Find the angular acceleration of the rod about the axis immediately after the thread breaks. Answer in units of rad/s 2 .

c) Find the translational acceleration of the center of mass of the rod immediately after the thread breaks. Answer in units of m/s 2 .

d) Find the force exerted on the end of the rod by the axis immediately after the thread breaks. Answer in units of N.

e) The rod rotates about the axis and swings down from the horizontal position. When the rod is at an angle θ = 35◦ with the horizontal, find the angular velocity of the rod. Answer in units of rad/s.

I got 24.525 for part A and it was marked correct

I got 9.81 for part B

I got 7.358 for part C

I got 12.263 for part D

Answer 1

a. The translational forces on the rod are its 5kg x 9.8 m/s² = 49 N in the center of the rod and the two supports which equally share the load. So the force on the rod by the axis is 49/2 = 24.5 N.

b. You would use the formula L = Iα where L is the torque (W x l/2) = 49 N x 1.5/2m = 36.75 Nm.

Moment of inertia is given by 2/3 x M x l = 2/3 x 5kg x 1.5 m = 5.0 kg-m.

α = 36.75 N-m / 5.0 kg-m = 7.35 rad /s²

c. Make use of the formula a = α x r = 7.35 rad/s² x 0.75m = 5.51 m/s².

d. I am not 100% sure of this one, but I think it would be the centripetal force. This would be ma_c = m x v² / r = 5kg x 5.51² / 0.75 m = 202 N.

e. I think this can be solved using energy considerations. The height of the rod when horizontal relative to the point when it makes an angle of 35 degrees is given by

H = l/2 sinθ = 0.430 m. The potential energy is mgH = 5kg x 9.81m/s^2 x 0.430 m = 21.1 j.

At 35 degrees, the kinetic energy is equal to the potential energy, so 1/2 x 5kg x v^2 = 21.1j, so v = sqrt ( 2 x 21.1j/ 5kg) = 2.90 m/s. This can be converted to angular velocity by ω = v/r, so ω = 2.90 m/s / 0.75 m = 3.99 rad/s.

This was a challenging problem. Check all the math and reasoning. Please let me know the results when you know.