Tangent force on a rotating rod

AI Thread Summary
A problem involving a thin rod hinged at the bottom and falling is discussed, focusing on finding the tangential force exerted by the lower part of the rod on the upper part. The angular acceleration was determined to be 3/2 g/L Sin(theta), and the moment of inertia was calculated using the parallel axis theorem. A suggestion was made to consider the center of mass of the segment to simplify the problem. Ultimately, the original poster successfully found the solution after several hours of effort. The discussion highlights the complexities of rotational dynamics and the importance of analyzing forces and accelerations in such systems.
abomination5
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Homework Statement


I've been working on this problem for several hours with no luck. Any help would be appreciated. (It's not homework)

A thin rod of mass M and length L is hinged at the bottom, and almost balanced vertically. It starts to fall. Find the tangential component of the force that the part of the rod below r exerts on the part of the rod above r.

(Theta is the angle between the rod and the vertical, r is the distance on the rod from the hinge)


Homework Equations



SOLN: F = Mg(L-r)(3r-L)/(4 L^2) * Sin(theta)

d x F = I *(angular acceleration)


The Attempt at a Solution



I've already determined the angular acceleration to be

3/2 g/L Sin(theta)

using the Lagrangian.

I've calculated the moment of inertia to be:

1/3 M (L-r)/L [(L-r)^2 + 3*r^2]

using the parallel axis theorem.

I've tried for several hours to get the final equation for the force but I haven't been able to do so yet... maybe there is a mistake in the moment of interia?
 
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abomination5 said:

The Attempt at a Solution



I've already determined the angular acceleration to be

3/2 g/L Sin(theta)
Good.

using the Lagrangian.

I've calculated the moment of inertia to be:

1/3 M (L-r)/L [(L-r)^2 + 3*r^2]
Rather than go this route, focus on the center of mass of the piece in question. What is its acceleration? What must be the net force on it?
 
Doc Al,

Thanks for your help. I was able to get the solution this morning. I won't spoil the rest of the details for anyone who wants to try it.
 
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