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

In summary, the correct equation to use for this problem is 0.5Iw^2 = mgh, where I is (1/12)mL^2 and h is equal to (L/2) - (L/2)cos(theta). This yields an answer of approximately 64° for theta.
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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.
 
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  • #2
ebeckwith said:
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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1. What is rotational kinetic energy?

Rotational kinetic energy is the energy an object possesses due to its rotational motion. It is dependent on the mass, velocity, and radius of the rotating object.

2. How is rotational kinetic energy related to the angle with the vertical?

The angle with the vertical is a measure of the orientation of the rotating object. The greater the angle, the more energy is required to rotate the object, resulting in an increase in its rotational kinetic energy.

3. How do you find the value of h in terms of angle with the vertical for rotational kinetic energy?

To find the value of h in terms of angle with the vertical, you can use the formula h = r(1-cosθ), where h is the height of the object, r is the radius of rotation, and θ is the angle with the vertical.

4. Can you calculate the rotational kinetic energy of an object without knowing the angle with the vertical?

Yes, it is possible to calculate the rotational kinetic energy of an object without knowing the angle with the vertical. You can use the formula KE = 1/2Iω^2, where I is the moment of inertia and ω is the angular velocity of the object.

5. How does the value of h change as the angle with the vertical increases?

As the angle with the vertical increases, the value of h also increases. This is because the height of the object is directly proportional to the angle with the vertical, according to the formula h = r(1-cosθ).

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