Find h in terms of angle with the vertical (rotational KE)

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Homework Statement



"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.


Homework Equations



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.

If:

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.
 

Answers and Replies

  • #2
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0.5mv2 = mgh

0.5mv2 + 0.5Iw2 = mgh

0.5Iw2 = mgh → (mg(L-Lcosθ))
1st equation is not correct because, besides KE of the Centre of mass, there is also KE of rotation about the centre of mass - in this case it would be 0.5Iw^2 where I would be (1/12)mL^2.

The 2nd equation is not correct because, as I said already, if one considers the KE of the centre of mass then one must add the KE of rotation ABOUT THE CENTRE OF MASS and so the moment of inertia will be 1/12)mL^2.

The 3rd is the correct one because it says the KE of rotation of the body about its end with the corresponding correct moment of inertia. So your mistake must be in calculating h. Your original PE is zero and hence you are taking the position of the Centre of Mass as the zero reference for PE. Hence I think that h = (L/2) - (L/2)costheta.
 
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