"A rod of mass 4.0 kg and length 1.5 m hangs from a hinge. The end of the rod is then given a 'kick' so that it is moving at a speed of 5 m/s. How high will the rod swing? Express you answer in terms of the angle the rod makes with the vertical.
KE(rot) = 0.5Iw2
I(rod, pivot at end) = (1/3)mL2
w = v/r
h = L-Lcosθ
The Attempt at a Solution
I have tried many, all yielding the incorrect answer. Posted solutions (answers only, no work shown) says that θ=64°
KEo = PEf
0.5mv2 = mgh
0.5mv2 = mg(L-Lcosθ)
This yields an answer of 81.39°
If I do:
0.5mv2 + 0.5Iw2 = mgh
" + " = mg(L-Lcosθ)
this also results in an incorrect answer. I don't remember what it was.
0.5Iw2 = mgh → (mg(L-Lcosθ))
This one gives me θ=44.23°
I've spent 9 hours on 10 homework problems, and have 3 finished. Please help me.