Uniform rod can pivot about a horizontal, fr

[synchronous]

Homework Statement

The thin uniform rod in the figure below has length 2.5 m and can pivot about a horizontal, frictionless pin through one end. It is released from rest at angle θ = 40° above the horizontal. Use the principle of conservation of energy to determine the angular speed of the rod as it passes through the horizontal position.

Homework Equations

E(mech, final) = E(mech, initial)
K(final) + U(final) = K(initial) + U(initial)
K(rotational) + K(translational) + U(final) = 0 + U(initial)
.5mr^2w^2 + .5m(wr/sin 40)^2 + mgh(final) = mgh(initial)
w = vsin(theta)/r

The Attempt at a Solution

I used both rotational and translational energy (not sure why translational energy applies though) for final kinetic energy. After masses cancel out,

.5(2.5)^2(w)^2 + .5w^2(2.5)^2/(sin(40))^2 + 9.81(2.5) = 9.8(4.1)
3.125w^2 + 7.56w^2 + 24.5 = 40.32
w = 1.22 rad/s

 

[LowlyPion]

I used both rotational and translational energy (not sure why translational energy applies though) for final kinetic energy. After masses cancel out,

.5(2.5)^2(w)^2 + .5w^2(2.5)^2/(sin(40))^2 + 9.81(2.5) = 9.8(4.1)
3.125w^2 + 7.56w^2 + 24.5 = 40.32
w = 1.22 rad/s

The gravitational potential is going to be the height of the center of mass isn't it? And doesn't that lay half way along the rod at the initial angle?

Won't just the gravitational potential then be the kinetic energy - 1/2*Iω²

 

[synchronous]

Rock on! Thanks for the assistance. I was thinking of the problem as a particle at the end of a massless rod. Finding the center of mass and using that in my calculations led to a correct answer of 2.75 rad/sec.