Rod attached to rotating vertical shaft

[M.Jenniskens]

1. Homework Statement

This is my first time posting on Physics Forums. I would appreciate your help with a question I'm doing in preparation for the CAP university prize exam in February. The following question is from the 1996 CAP exam.

Rod attached by frictionless pin to vertical rotating shaft. At what angle will rod hang in steady state? What will be the speed of the rod's center when angle has increased to 30 degrees (rod starts from rest)?

Please find attached a complete copy of the question and my attempt at its solution.
I think My problem lies in the fact that I am using forces that perhaps are not present in such a system.


2. Homework Equations



3. The Attempt at a Solution

Please find a copy of my solution attached

!Thanks for your help!

 

[Simon Bridge]

Welcome to PF;
I cannot read docx files.
Please summarize your reasoning.

This is basically a modification of the swing-ball problem.
In an inertial frame, there are two forces only. They should add to give a net centripetal force.

 

[M.Jenniskens]

the rotating axis is parallel to the y-axis
rod is attached at it's end to the rotating vertical shaft by a pin.
i say there's a force, n, acting on the rod. A force of n(cos(theta)) in the x direction and a force of n(sin(theta)) in the y direction. force of gravity is acting in the downward direction.
i found angular velocity from rpm.
i state that torque is equal to length of rod (5 m) multiplied by the force in the x direction.
torque is equal to 1/2 the rods moment of inertia multiplied by it angular velocity. solved for theta. found it to be 38.17 degrees in steady state.

velocity=(angular velocity)(radius) where radius is Lsin(theta).
found velocity at time when rod makes angle of 30 degrees to be 0.982 m/s


I have attached my solutions in pdf format. maybe you will be able to read them now so you can see my full solution.

!Thanks!

 

[M.Jenniskens]

here is the pdf file. I don't think it attached in my other post.

 

[Simon Bridge]

i say there's a force, n, acting on the rod. A force of n(cos(theta)) in the x direction and a force of n(sin(theta)) in the y direction. force of gravity is acting in the downward direction.

Just a niggle - the n force does not always have a component in the x direction does it? I mean, the rod is rotating about the y-axis (you can just decide that - it's no problem) so sometimes the n force has a component in the z direction right?

That to one side ... I cannot see your reasoning at each step - i.e. where on the rod does the force of gravity act? Doesn't it matter?
It looks like you are doing kinetic energy and arc-length calculations but I cannot see why.

I'd check that you are using the correct value/relationship for r.

BTW: thanks for the pdf.
docx seems not to like to play with non-MS software.

 

[M.Jenniskens]

okay. Thanks for the response. I'm still not sure what to do. I think lagrangian mechanics would be a better method for going about this problem but I'm not sure how to even start the problem using lagrangian mechanics. In the current method i think I've made a lot of errors with the forces. I'm thinking now that the force of gravity should be in the center of the rod but I'm not sure.

 

[Simon Bridge]

Where on the rod does the force of gravity act?