GREPhysics.NET: GR0177 #26

GR0177 #26

Problem

GREPhysics.NET Official Solution

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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

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 $v = \omega L$ . $MgL = ML^2\omega^2/3 = Mv^2/3$ $v = \sqrt{3gL}$ . Reply to this comment
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 ghdm, where h ranges from zero to L. Replacing dm with (M/L)dh, we get g(M/L)hdh. This is a two second integral which results in (MgL)/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)dmv^2. Again replacing dm with (M/L)dh, and the velocity with hw, where w is whatever the angular velocity of the rod is, we have (1/2)(M/L)w^2h^2dh. Again, this integral takes about two seconds and gives (ML^2w^2)/6. Now, if you equate this with the potential energy, and cancel the mass terms and a factor of two, you immediately get, gL = (Lw)^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

Comments

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 $v = \omega L$ .

$MgL = ML^2\omega^2/3 = Mv^2/3$
$v = \sqrt{3gL}$ .

kroner
2009-10-05 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 $I = \int_0^L \lambda r dr$ .

kroner
2009-10-05 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.

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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.

wittensdog
2009-10-09 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.

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AER
2009-04-02 16:23:28

How do we know there's no force being exerted on the rod by the pivot?

kroner
2009-10-05 11:17:47

There is force being exerted on the rod by the pivot. Yosun's solution uses conservation of energy rather than forces.

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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

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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

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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.

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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:

$1/12 ML^2 + M(L/2)^2 = 1/3 ML^2$

nitin
2006-11-14 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.

Reply to this comment

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?

herrphysik
2006-09-22 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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