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GR0177 #26
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Mechanics$\Rightarrow$}Conservation of Energy

Conservation of energy gives, $E=MgL/2=1/2 Mv^2 +1/2I\omega^2$, where the potential energy is given about the center of mass. The moment of inertia is about the center of mass, too. The translational velocity is the velocity at the center of mass.

For a rod, one has $I=1/12 M L^2$. $v=\omega L/2$ Plug stuff in to get $MgL/2=1/2 M\omega^2/4 +1/2(1/12 M L^2)\omega^2=1/2(1/4+1/12)ML^2\omega^2 \Rightarrow g = (1/4+1/12)L\omega^2=(1/3)\omega^2 \Rightarrow \omega=\sqrt{3g/L}$. Now that one has the angular velocity, one can calculate the velocity at the endpoint; the rod rotates about its other end on the ground. $v=\omega \times r = L\omega = \sqrt{3gL}$, which is choice (C).

Alternate Solutions
 kroner2009-10-05 11:10:58 It's quicker to consider the motion about the pivot point instead of about the center of mass, since then the linear kinetic energy term drops out (since the pivot point is stationary). $E = MgL/2 = I\omega^2/2$ with $I = ML^2/3$. The velocity of the end point is $v = \omega L$. $MgL = ML^2\omega^2/3 = Mv^2/3$ $v = \sqrt{3gL}$.Reply to this comment wittensdog2009-09-13 16:15:22 So, there's an alternate method I used to solve this problem which makes no reference to moment of inertia at all, which I think is relatively quick and simple. First, recognize that if we set the potential energy to be zero at the height of the pivot point, we start out with all of the energy as potential energy, and by the time the rod falls over, it is entirely kinetic. The initial potential energy of an infinitesimal part of the rod is g*h*dm, where h ranges from zero to L. Replacing dm with (M/L)*dh, we get g*(M/L)*h*dh. This is a two second integral which results in (M*g*L)/2 for the total potential energy (the same as if you averaged the energy of a point mass M at height zero and L). Now, the kinetic energy of an infinitesimal portion of the rod when it is moving is (1/2)*dm*v^2. Again replacing dm with (M/L)*dh, and the velocity with h*w, where w is whatever the angular velocity of the rod is, we have (1/2)*(M/L)*w^2*h^2*dh. Again, this integral takes about two seconds and gives (M*L^2*w^2)/6. Now, if you equate this with the potential energy, and cancel the mass terms and a factor of two, you immediately get, g*L = (L*w)^2 / 3 from which it is easy to solve for L*w, or the velocity at the end of the rod, and you indeed get answer C. The telling of the method makes it seem long, but it really is only about a 30 second problem at the very most, if you know what to do.Reply to this comment
nepo
2016-02-02 18:32:48
From where did we get E = mg L / 2 ?
tweetiebird
2014-11-28 20:32:15
The author leaves off a L^2 term in one step
saforem2
2013-06-26 15:03:27
I'm confused as to why you have $v = \omega * r = \omega * \frac{L}{2}$ initially, but then when you solve for v in terms of $\omega$ you use $v = \omega * L$ instead of $v = \omega * \frac{L}{2}$
wtf_is_wrong?
2011-06-07 21:48:33
Is it wrong to simply use $\tau_{net} = I \alpha$ ? Where $\tau_{net} = mg L$ and $I \alpha$ = $\frac{1}{12} m L^{2}$ $\frac{v^2}{\frac{1}{4} L^{2}}$ . Which ends up giving you $v = \sqrt{3 g L}$.
his dudeness
2010-09-24 16:11:53
Fellas! What's with all the fancy formulas? There's a much easier way to do the problem. Imagine we take a mass and drop it from a height L/2. Its potential energy gets converted into potential energy and it hits the ground with velocity $\sqrt{gL}$. Now, in our case, the potential energy doesn't get converted purely into kinetic energy, because some of it has to go into rotational energy as well. It's pretty clear that the final velocity of any point on the rod will be less than $\sqrt{gL}$, so answer (A) is the only one that fits.

As always, the lesson about the Physics GRE is: never plug and chug unless you absolutely have to!
 his dudeness2010-10-10 13:14:20 Whoah, my original answer was totally off base, please disregard it! (I guess the key here is that the end of the rod is moving twice as fast as the middle of the rod). In retrospect, it seems like the best solution is conservation of energy about the pivot point. You get: $Mg\frac{L}{2} = \frac{1}{2} I w^2$, where $I=\frac{1}{3}ML^2$. Plug and chug to get the right answer.
kroner
2009-10-05 11:10:58
It's quicker to consider the motion about the pivot point instead of about the center of mass, since then the linear kinetic energy term drops out (since the pivot point is stationary).

$E = MgL/2 = I\omega^2/2$ with $I = ML^2/3$. The velocity of the end point is $v = \omega L$.

$MgL = ML^2\omega^2/3 = Mv^2/3$
$v = \sqrt{3gL}$.
 kroner2009-10-05 11:25:51 If you get hit on the head and forget all the moment of inertia formulas, the basic ones are pretty easy to derive from scratch. For a rod: $I = \int \lambda r dr$ where $\lambda$ is the linear density, in this case $\lambda = M/L$. About the center $I = \int_{-L/2}^{L/2} \lambda r dr$. About the end $I = \int_0^L \lambda r dr$.
 kroner2009-10-05 11:28:41 Ha ha, sorry I managed to screw that up. The integrals should be $\int \lambda r^2 dr$. I guess I would fail.
 kiselev2010-11-02 18:13:28 Hi! It says "With what speed does the FREE end of the rod strike the ground?" This means that your answer must be multiply by 2, so you get (D) \sqr{12gL} The first solution is just terrible... the center of mass doesnt have translational kinetic energy. However it seems that ETS made the same mistake and considers that the answer is (C)...Unless they said what is the center of mass velocity when strikes the surface I dont see how this is right
 aayush30092012-09-16 09:23:19 On the same lines, i have found a really quick method. The PE at the top most point of the rod is M.g.L Now for KE when it hits the ground, consider the rod to be of length 2L and of mass 2M. Now the point of rotation becomes the center of mass of this extended rod. With the new mass and length, MoI, I = (2M)(2L)^2/12. = 2/3 ML^2. Equate MgL = Iw^2 where w=v/L, you get v = (3gL)^1/2 !!
 nvp102014-04-02 06:21:47 Thanks! this is the best solution. Maybe it would also be useful to say why I is what it is: $I=I_CM+Mr_CM^2=\frac{Ml^2}{12}+M\frac{l}{2}^2=ML^2/3$ due to the central axis theorem.
 nvp102014-04-02 06:25:00 Sorry, I got the latex wrong. here is what i mean: I=I_{CM}+Mr_{CM}^2=\frac{Ml^2}{12}+M\frac{l}{2}^2=ML^2/3
 nvp102014-04-02 06:26:19 and again: $I=I_{CM}+Mr_{CM}^2=\frac{Ml^2}{12}+M\frac{l}{2}^2=ML^2/3$
 RusFortunat2016-08-11 22:05:48 this is a correct solution. disregard OP one
 NervousWreck2017-03-28 11:00:04 also disregard people trying to find a mistake in this solution. this is the fastest one.
wittensdog
2009-09-13 16:15:22
So, there's an alternate method I used to solve this problem which makes no reference to moment of inertia at all, which I think is relatively quick and simple.

First, recognize that if we set the potential energy to be zero at the height of the pivot point, we start out with all of the energy as potential energy, and by the time the rod falls over, it is entirely kinetic.

The initial potential energy of an infinitesimal part of the rod is g*h*dm, where h ranges from zero to L. Replacing dm with (M/L)*dh, we get g*(M/L)*h*dh. This is a two second integral which results in (M*g*L)/2 for the total potential energy (the same as if you averaged the energy of a point mass M at height zero and L).

Now, the kinetic energy of an infinitesimal portion of the rod when it is moving is (1/2)*dm*v^2. Again replacing dm with (M/L)*dh, and the velocity with h*w, where w is whatever the angular velocity of the rod is, we have (1/2)*(M/L)*w^2*h^2*dh. Again, this integral takes about two seconds and gives (M*L^2*w^2)/6. Now, if you equate this with the potential energy, and cancel the mass terms and a factor of two, you immediately get,

g*L = (L*w)^2 / 3

from which it is easy to solve for L*w, or the velocity at the end of the rod, and you indeed get answer C. The telling of the method makes it seem long, but it really is only about a 30 second problem at the very most, if you know what to do.
 wittensdog2009-10-09 12:56:01 Of course, what I'm really doing here in disguise is just deriving the moment of inertia for a rod about one end. Integrals of the form r^2 * dm come up so often in calculations of angular momentum and rotational energy that they were just given their own name as the moment of inertia. So I guess you can just blindly integrate as in my original method, or integrate to get the moment of inertia of a rod about its end, either way you're pretty much doing the same thing. Or of course you could always just memorize the moment of inertia of a rod about its end.
 bcomnes2011-09-22 16:35:28 I like this solution a lot. I really need to overcome my aversion dealing with inertia. This way looks long, but it is super simple.
AER
2009-04-02 16:23:28
How do we know there's no force being exerted on the rod by the pivot?
 kroner2009-10-05 11:17:47 There is force being exerted on the rod by the pivot. Yosun's solution uses conservation of energy rather than forces.
anmuhich
2009-03-14 10:58:08
For those of us who don't have these mathematical theorem's memorized (like me) and don't want to memorize them, another way to find the moment of inertia for a rod spinning about one end is to take half the moment of inertia of two of the same rods put together spinning about their center. This makes intuitive sense and you could come up with it on your own. So if you work it out:

Irod= (1/2)(1/12) Mnew*Lnew*Lnew

with Mnew=2*M and Lnew= 2*L

plugging in

Irod= (1/2)(1/12)(2M)(4L*L) = 1/3 M*L*L

Then continue with the rest. This way
mfidanis1
2009-01-29 16:27:41
it is called Steiner Theorem.
$I_c_m = I + Md^2$ where $I_c_m$ is the moment of inertia of center mass and d the distance between the two axes
michealmas
2006-12-28 16:56:20
Ok, I don't understand why she is using the

$\frac{1}{2}M v^2$

term. The intial potential energy is all transferred to rotational kinetic energy in the final state. Once we get rot. kin. E, THEN we change it to translational energy of the tip of the rod. So:

Use the proper equation for moment of inertia of a rod rotating at its end: ($I=\frac{1}{3} M L^{2}$) and say:

$E_{pot} = E_{kin} \Rightarrow$

$M g \frac{L}{2} = \frac{1}{2} I \omega^{2}$

(using $\frac{L}{2}$ instead of $L$ in the LHS because that is the center of mass. This is essentially the $M g h$ grav. pot. E term from newbie (or 'Yosunist') mechanics.)

Then plug in $I=\frac{1}{3} M L^{2}$, and solve for $\omega$ to get:

$\omega = \sqrt{\frac{3g}{L}}$

and finish it as Yosun did.
senatez
2006-10-29 10:30:53
I think it is easier to just use parallel axis theorem to derive the moment of inertia around the end of a rod. It will be:

$1/12 ML^2 + M(L/2)^2 = 1/3 ML^2$
 nitin2006-11-14 01:04:03 Well, what Yosun wrote: $E=Mg(\frac{L}{2})=\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$ actually comes from the so-called parallel-axis theorem. Nevertheless, you're right, and it is much simpler your way.
naama99
2006-09-18 20:22:39
How come the moment of inertia of the rod is I=1/12 MR^2 instead of 1/3 MR^2?
 herrphysik2006-09-22 20:38:51 naama99: $1/12ML^2$ is the moment of inertia of a rod about the center of mass of the rod (axis through the middle), which is what yosun is using in her solution. Granted this problem is much easier and faster to just use $I=1/3ML^2$ (axis at the end), however ETS doesn't supply this equation on their info sheet, only the first one. But had you the latter memorized, that would be the one to use. Without it though one is forced to use this more involved translational + rotational kinetic energy of the CM business.

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