1. Feb 5, 2006

### dowjonez

A thin homogeneous rod hsa length L and mass M. Initially the rod is vertical at rest on a frictionless surface. The rod is given a slight push and begins to fall in the gravitational field.

1) write an expression for the kinetic energy T of the rod as functions of M,L,g, theta and d theta/dt

2) if y is the distance the center of mass drops. find d y/dt as a function of L, g and theta

#1)i know the Kinetic energy of the rod is given by the equation

T= 1/2 I w^2

where w = omega = d theta/dt
and I = M(L/2)^2

so T = 1/2 M(L/2)^2 d theta/dt^2

T = 1/8 M L^2 d theta/dt^2

The answer is T = 1/8 M L^2 d theta/dt^2 (Sin^2theta + 1/3)

i dont know where the (Sin^2theta + 1/3) comes from

#2)

i dont even know where to start on this one

but the answer is Y prime = sintheta sqrt ( (3gL(1-costheta) / 4-3cos^2theta)

any help would be greatly appreciated

2. Feb 5, 2006

### Staff: Mentor

That would be true if the rod were purely rotating, but it's also translating: the total KE is the sum of the translational KE (of the cm) and the rotational KE (about the cm).