How is the kinetic energy of a falling rod on a frictionless surface calculated?

[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 don't know where the (Sin^2theta + 1/3) comes from


#2)


i don't 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

 

[Doc Al]

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

T= 1/2 I w^2

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).