GREPhysics.NET: GR9277 #74

GR9277 #74

Problem

GREPhysics.NET Official Solution

\prob{74}
Two circular hoops, X and Y, are hanging on nails in a wall. The mass of X is four times that of Y, and the diamter of X is also four times that of Y. If the period of small oscillations of X is T, the period of small oscillations of Y is

T
T/2
T/4
T/8
T/16

Mechanics $\Rightarrow$ }Small Oscillations

The small oscillations of the hoop has the same frequency as that of a simple pendulum. Thus, $\omega^2 = \frac{g}{l}$ . However, in this case, $l$ is the distance from the center of mass to the oscillation point---which is just the radius of the loop.

Since $\omega = 2\pi/T$ , the period $T\propto \sqrt{\frac{r}{g}} \propto \sqrt{r}$ does not depend on mass.

Since $r_x = 4 r_y$ . The ratio of periods is $T_X/T_Y ~ T/\sqrt{r_y} ~ T/\sqrt{r_x/4} ~ 2$ . Thus, the period of Y is just $T/2$ . (Note, the technique of leaving out constants requires that $~$ 's are used instead of $=$ 's. Practice a few times with this technique, as this will save time on the actual exam.)

Alternate Solutions

99percent
2008-11-05 06:52:16

Assuming both the hoops to be point masses concentrated at their respective center of mass, they can be considered as simple pendulums. For small oscillations, time period of a simple pendulum $=\sqrt{\frac{l}{g}}$ .
Thus, the time period is proportional to the $\sqrt{l}$ .
Therefore, the time period of Y is $T/\sqrt{4} = T/2$

Bingo..!!

Reply to this comment

Comments

gravity
2010-11-04 23:36:45

You can do it this way - and this is the way I did it first time around - but you might not get the same physical pendulum on the test and I'm noticing that I'm using this equation a lot of the time on practice tests, so I suggest knowing how to do it this way too:

The period for a physical pendulum is (memorize this):

T = 2 $\pi$ $\sqrt{ I / m g r_{cm}}$

Where $r_{cm}$ is the center of mass.

The moment of Inertia is in the form of mr $^2$ (don't care about the constant in front). The mass and one r in the moment of Inertia cancels with the denominator. So the ratio of the two pendulums is $\sqrt{ r_X / r_Y }$ = $\sqrt{ 4 }$ = 2

The answer is B.

A little bit extra work, but since these are practice tests, it's good to know several ways to solve these problems so that you can solve them more generally.

gravity
2010-11-04 23:42:02

Also note that the moment of inertia in that equation is measured from the support.

The ratio is still the same factor.

http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

Reply to this comment

apr2010
2010-04-08 08:23:48

One could also use:

1) Know that $\omega = \sqrt{mgR/I}$ with m mass, g gravitational constant, R radius and I moment of inertia from a few problems before with the two pendulums.

2) Get Moment of Inertia $I_1,I_2$ with $I = mR^2 + md^2$

3) Get ratio $\omega_2 / \omega_1$ .

Reply to this comment

venus19000
2009-10-30 10:55:25

I don't know if this is a correct approach, but I used $w^2 = \sqrt{\frac{k}{m}}$ .

This gave me $T_X = \sqrt{\frac{m}{k}}$
and $T_Y = \sqrt{\frac{m/4}{k}}$ .

$T_Y = \frac{1}{2}\sqrt{\frac{m}{k}} = \frac{1}{2}T_X$

Therefore the period of Y was $T_X/2$ .

flyboy621
2010-10-22 16:27:04

This is not correct. You got lucky.

The period depends on the size, but not the mass.

flyboy621
2010-10-22 16:48:52

At least, that is what Yosun was saying. I think Marten down below has it right.

From wikipedia, the angular frequency of small-angle oscillations for a physical pendulum is

$\omega\approx\sqrt{\frac{mgL}{I}}$

where L is the distance from the suspension point to the CM, in this case the radius for each hoop. So the period DOES depend on the mass no matter how you do it. Again, Marten got it right.

Reply to this comment

99percent
2008-11-05 06:52:16

apr2010
2010-04-08 08:24:35

Awesome

ali8
2011-07-12 13:28:32

That's how I get at this problem.

Reply to this comment

99percent
2008-11-05 06:50:43

Reply to this comment

Tommy Koulax
2007-10-26 19:41:22

I ment T/ $\sqrt{2}$ not $\sqrt{z}$

Tommy Koulax
2007-10-28 11:53:43

Wow, I just had a brain fart there. The factor is 4, not 2 like I was saying in my stuper.

Reply to this comment

Tommy Koulax
2007-10-26 19:40:02

Rx=2Ry not 4Ry. Solving the problem as done above leads to T/ $\sqrt{z}$ . Something is not right.

Reply to this comment

marten
2007-10-24 13:32:54

If we have $\omega$ = $\sqrt{\frac{mgr}{I}}$

And I figure the mass moment of inertia for a hoop rotating about a point on the rim to be:

$I = mr^2 + mr^2$ by the parallel axis theorem, I don't see how the angular frequency is the same as a pendulum.

ewhite2
2007-10-27 13:19:12

marten, I think you are right. However, since it is just the ratio of periods we are interested in, this extra factor of 1/2 inside the radical will cancel giving the same solution.

kyros
2007-10-31 16:30:33

Feelfree to work it out using your formula - I just did, and it comesout to T/2.

marten
2007-11-13 09:29:30

Thanks for the comments. I see what you're saying about the factors canceling out. I'll have to work it out to see for myself, but since the Nov 3rd test is finally done, and I've been studying since June, I'm not very inclined to more problems just quite yet!

Reply to this comment

Andresito
2006-03-29 10:33:23

There is a MimeTeX failure after the ratio of periods. The ratio is quite confusing written in that way. Thanks for the answers worked out Yosun.

Reply to this comment

Post A Comment!

You are replying to:

You can do it this way - and this is the way I did it first time around - but you might not get the same physical pendulum on the test and I'm noticing that I'm using this equation a lot of the time on practice tests, so I suggest knowing how to do it this way too: The period for a physical pendulum is (memorize this): T = 2 $\pi$ $\sqrt{ I / m g r_{cm}}$ Where $r_{cm}$ is the center of mass. The moment of Inertia is in the form of mr $^2$ (don't care about the constant in front). The mass and one r in the moment of Inertia cancels with the denominator. So the ratio of the two pendulums is $\sqrt{ r_X / r_Y }$ = $\sqrt{ 4 }$ = 2 The answer is B. A little bit extra work, but since these are practice tests, it's good to know several ways to solve these problems so that you can solve them more generally.

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone