- #1

Asrai

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*A uniform rod of length l is freely pivoted at one end. It is intially held horizontally and then released from rest.*

Write down expressions for the angular momentum L(theta) of the rod and the torque T(theta) about the pivot when the rod is inclined at an angle theta to the horizontal.

Hence show that the angular velocity at this instant is given by:

theta[double-dot]^2 = (3*g*sin(theta))/l

Write down expressions for the angular momentum L(theta) of the rod and the torque T(theta) about the pivot when the rod is inclined at an angle theta to the horizontal.

Hence show that the angular velocity at this instant is given by:

theta[double-dot]^2 = (3*g*sin(theta))/l

I apologise in advance for the horribleness of my equation; I'll figure out how to use that Latex thing in a little while=)

Anyway, I think I've solved the first two bits of the problem; for the angular moment I get L = (1/3)*m*l^2*theta[dot], using the angular velocity and the moment of inertia of the rod.

Differentiating gives the torque N = (1/3)*m*l^2*theta[double-dot] and the torque is also given by N = l*m*g*sin(theta) because of the gravitational force acting on the rod. Equating those two expressions for N and solving for theta[double-dot] gives me:

(3*g*sin(theta))/l

which is the result that I'm supposed to get for the squared angular velocity. Now, I have the feeling that I went wrong

*somewhere*, I just can't figure out where! What have I done wrong?