Calculating Resultant Force at a Pivoted Rod | Uniform Rod in Vertical Plane

[fkf]

Homework Statement

A uniform rod of length 1.6 m and mass 20 kg is free to rotate about a frictionless pivot at one end in a vertical plane. The rod is released from rest in the horizontal position. The acceleration of gravity is 9.80 m/s2.

Homework Equations

What is the resultant force in the pivot immediately after the release?

The Attempt at a Solution

I know that the Moment of Inertia around such a rod is 1/3*m*L^2 and that the tourqe around that rod is L/2*mg, this gives me the angular acceleration 9.1875. I assume that the normal force in this situation is zero?

That would only give me the tangental force F_t = m*r_cm*a left (since F_n = m*r*w^2, but w = 0), which gives me the answer 147 N, but the answer should be 49 N

 

[tiny-tim]

hi fkf!

That would only give me the tangental force F_t = m*r_cm*a left (since F_n = m*r*w^2, but w = 0), which gives me the answer 147 N, but the answer should be 49 N

hint: 49 + 147 = 196

 

[haruspex]

I assume that the normal force in this situation is zero?

What are you calling the normal force here? What is the linear acceleration of the rod's centre?