Spring Constant
A thin uniform rod has mass M = 0.5 kg and length L= 0.37 m. It has a pivot at one end and is at rest on a compressed spring as shown in (A). The rod is released from an angle θ1= 55.0o, and moves through its horizontal position at (B) and up to (C) where it stops with θ2 = 111.0o, and then falls back down. Friction at the pivot is negligible. Calculate the speed of the CM at (B).

(in m/s)

how can I solve it?
I tried this :


ENERGY(B) = ENERGY(C)
0.5mV^2 + 0.5Iω^2 = mgh
0.5m(L(sin(θ2))^2/3 + 1)V^2 = 0.5mLg(cos(θ2) - cos(θ3))

V^2 = Lg(cos(θ2) - cos(θ3))/(L(sin(θ2))^2/3 + 1)

θ2 = 90
θ3 = 111
L = 0.37 m
g = 9.81 m/s^2

V^2 = 1.158
V = 1.076 m/s <=============

but it's wrong !

I don't think we can make sense of the question without a diagram, I can't work out where the angles are measured from.

Your equations are also confusing, can you go through and make sure they are typed out legibly?

My first thought is to make sure you are analysing the motion of the center of mass of the rod. I'm afraid I can't check for you because I can't read your equations very well.

If you can explain the question better I'm sure we can work out where the error is, you have done well to post your incorrect solution!