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.(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: Rotating rod Please Help

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