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  GR0177 #91
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Mechanics}Conservation of Energy

Conservation of energy gives  Mg(H) = \frac{1}{2}Mv^2 + \frac{1}{2}I \omega^2. v=R\omega, and thus the equation becomes

 Mg(H) = \frac{1}{2}Mv^2 + \frac{1}{2}I (v/R)^2.

Thus, I=2(MgH-\frac{1}{2}Mv^2)R^2/v^2. (Note that the height of the center of mass is the same at both the end and the start, thus the extra bit of the potential energy MgR cancels out. Thanks to the user keflavich for this correction.)

Given v^2 = 8gH/7, one plugs it into the equation above to get

I=2(MgH  -M4gH/7)R^2/(8gH/7)=2(3MgH/7)R^2/(8gH/7)=3MR^2/4, as in choice (B).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
Ning Bao
2008-02-01 08:11:02
Let R, m,h,g=1. Then you just have

Alternate Solution - Unverified
2011-10-08 15:23:01
Did this by process of elimination (not even looking at the PE versus KE).

(A) can't be true, since the cylinder is not of uniform density.

(D) can't be true, since this would only occur if all of the mass were concentrated in a thin ring at a distance R.

(E) can't be true, since the largest moment would be (D).

We are left with (B) and (C). Because a radial mass distribution that varies as \rho R should be less than something very close to (D), we are left with choice (B).
2012-02-09 18:27:24
The radial dependence is not given to be linear. I Think (B) or (C) remain until you look at energy.
2012-03-06 12:06:24
I was looking at the problem again. The elimination you did is the right place to start. Once you have it down to (B) and (C), look at the translational energy.
T = \frac{1}{2}Mv^2
= \frac{1}{2}MgH\frac{8}{7}
\g \frac{1}{2}MgH

since the translation takes up a little more than half of the original energy of the object, less than half is available for the rotational energy. This implies a smaller moment of inertia, hence choice (B)
Ning Bao
2008-02-01 08:11:02
Let R, m,h,g=1. Then you just have

2008-10-21 11:27:20
Nice - sure helps save time! Caveat: you really can't set R=1 and h=1 simultaneously, since they are both measures of length.

Here it works rather nicely, because H cancels out as you could tell from the answers.
Alternate Solution - Unverified
2007-10-31 03:28:11
You can make the calculation more easy by noting that the rotational inertia will look like

I = \alpha M R^2  ,

with \alpha some constant. Now, you just have to solve:

MgH =  \frac{1}{2} M v^2 (1 + \alpha)

This might save you some time. Because, 1.7 min..., man, that is not much time.
2019-10-22 14:57:59
BRILLIANT!!! I think this is the best and the most make sense alternative solution under this pose.
2007-10-30 18:38:31
A potentially useless note:

The solution given in (E) is found by NOT accounting for the translational kinetic energy.
2006-03-05 00:00:42
Yosun, there is a typo on your solution. When you express I, the term MV^2 has a factor of 2 implicitly.

Recall the conservation of energy equation, multiply both terms by 2 and the "only" term that has now a factor of 2 is MgH.

The correct expression should have I=(2 MgH - MV^2) (r/v)^2

Yosun, thank you very much for working out the problems. If you are a girl I send you a big kiss :). I obtained the pdf from the yahoo-group. It also has the typo.
Typo Alert!
2005-11-10 22:48:10
You don't need to consider MgR as part of the gravitational potential energy. When the cylinder reaches the bottom of the plane, its center of mass will remain at a distance R above the ground.
2005-11-10 22:54:19
keflavich: you're right that the center of mass stays at a height of R above the ground at both points. thanks for the correction.
Fixed Typos!

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