GREPhysics.NET: GR9277 #93

GREPhysics.NET

GR9277 #93

Problem

GREPhysics.NET Official Solution

Alternate Solutions

\prob{93}

9277_93

THe figure above shows a small mass connected to a string, which is attached to a vertical post. If the mass is released when the string is horizontal as shown, the magnitude of the total acceleration of the mass as a function of the angle $\theta$ is

$g\sin\theta$
$2g\cos\theta$
$2g\cos\theta$
$g\sqrt{3\cos^2\theta+1}$
$g\sqrt{3\sin^2\theta+1}$

Mechanics $\Rightarrow$ }Boundary Condition

Getting low on time, one should begin scoring points based more of testing strategy than sound rigorous physics. At the initial release point, the acceleration is due to gravity and the tension is 0 (no centripetal acceleration). The only choice that gives $a(\theta=0) = g$ is choice (E).

See below for user comments and alternate solutions!

Alternate Solutions

his dudeness 2010-09-05 15:55:11	Yosun's method is clearly by far the best for this particular problem. However, if anyone is curious about how the relation in (E) is actually derived, here you go: The acceleration can be split into radial and transverse components. Since gravity is the only force contributing to transverse acceleration, We have $a_t=gcos(\theta)$ . To calculate $a_r$ , we note that $a_r=\frac{v^2}{r}$ . From conservation of energy, we have $\frac{1}{2}mv^2=mgsin(\theta)$ , so we get $a_r=2gsin(\theta)$ . All together now: $a_{tot} = \sqrt{a_r^2+a_t^2} = \sqrt{gcos(\theta)^2+4gsin(\theta)^2}$ . A simple trig identity gets us the (E). Reply to this comment
jburkart 2007-11-02 00:23:15	A real way to do the problem is as follows. Kinetic energy is $T=mR^2\dot{\theta}^2/2$ , where $R$ is the length of the string; potential is $U=-mgR\sin\theta$ . Setting the Lagrangian to $L=T-U$ , Lagrange's equation is $\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial\dot{q}}$ . In our case $q=\theta$ and we find $\ddot{\theta}=g\cos\theta/R$ , which then yields $a_\parallel=\ddot{\theta}/R=g\cos\theta$ . But this is only the acceleration along the direction of motion; there is also a perpendicular centripetal acceleration given by $v^2/R$ . To find $v$ we can invoke conservation of energy: $mv^2/2=mgR\sin\theta$ , so $v^2=2gR\sin\theta$ , so the centripetal acceleration is $a_{\perp}=v^2/R=2g\sin\theta$ . Taking the norm of the the two acceleration components, we get $a=g\sqrt{\cos^2\theta+4\sin^2\theta}=g\sqrt{1+3\sin^2\theta}$ , answer (E). A common mistake would be to omit the calculation of the centripetal acceleration, which would yield an incorrect answer of $g\cos\theta$ , which ETS kindly does not give as an option. Reply to this comment

Comments

his dudeness
2010-09-05 15:55:11

Yosun's method is clearly by far the best for this particular problem. However, if anyone is curious about how the relation in (E) is actually derived, here you go:

The acceleration can be split into radial and transverse components. Since gravity is the only force contributing to transverse acceleration, We have $a_t=gcos(\theta)$ . To calculate $a_r$ , we note that $a_r=\frac{v^2}{r}$ . From conservation of energy, we have $\frac{1}{2}mv^2=mgsin(\theta)$ , so we get $a_r=2gsin(\theta)$ .

All together now: $a_{tot} = \sqrt{a_r^2+a_t^2} = \sqrt{gcos(\theta)^2+4gsin(\theta)^2}$ . A simple trig identity gets us the (E).

physicsres.com
2014-11-15 12:55:37

good answer just a typo alert.
$\frac{1}{2}mv^2= mgr sin\theta$

unurbat
2016-06-28 23:24:16

Where did the \"r\" disappear when you try to find radial acceleration?

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Alternate Solution - Unverified

elzoido238
2008-11-06 11:02:47

Answer choice C should be 2g sin $\theta$ .

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NEC

Poop Loops
2008-10-25 20:22:52

Oh Jesus, people, get all this nonsense away. Calculations on the GRE? That's suicide!

Here's how you think of it:

When $\theta$ is $0$ , the mass wants to go straight down, so a = g.

That's it. Only E survives that.

But, if you had a different choice of answers, you can also think that acceleration is a maximum at $\theta = \frac{\pi}{2}$ , because it also has centripetal acceleration, which adds to the gravitational force at that point. This obviously corresponds to sin, so there you go.

ajkp2557
2009-11-03 13:57:32

"Calculations on the GRE? That's suicide!"

Best. Advice. Ever. The PGRE is entirely about test taking strategy. Yes, once you are in grad school it will be very important to know how to correctly work every problem, but spending four minutes on a problem on a test that only gives you 1.7 minutes per problem isn't a smart approach. You get the same credit for the answer, regardless of how you got it.

Work smarter, not harder. Best of luck!

Reply to this comment

NEC

ayabepaula
2008-10-22 00:36:22

time= zero

${a_{total}}=g$

$U=mgh$
$F=mg=m{a_{total}}$

time

$t={t_{bottom}}$

${a_{total}}=2g$

$mgh=m\frac{v^{2}}{2}$

${v^{2}}=2gh$

${a_{total}}=2g=\frac{v^{2}}{h}$

$F=m.{a_{total}}={F_{cp}}=m{a_{cp}}=m(2g)$

runcibleshaw
2015-08-16 06:54:34

I don\'t understand why the acceleration is 2g at the bottom. The total acceleration is the combination of g, which points down and the centripetal acceleration which points up, so it must be less than g total, right?

Reply to this comment

NEC

wangjj0120
2008-10-08 12:42:19

I am confused, why can't I use
$mgsin(\theta)=T=$ force on string, so $a_{perp}=0$ , and $mgcos(\theta)=m(l\ddot{\theta})=ma_{para}$ , where $l$ =length of string. From this, $a_{para}=gcos(\theta)$ is correct, but $a_{perp}$ is wrong. Obviously the concept is incorrect here, but why?

iplayterran
2010-03-13 17:03:00

The tension force $F_T$ on the string is the sum of $mg\sin(\theta)$ and the centripetal force $\frac{mv^2}{r}$ . If you use Newton's second law,

$\sum F = F_T + F_g$

and decompose the forces into radial and angular components, you have

$\sum F = (mg\sin\theta + \frac{mv^2}{r})\hat{r} + (-mg\sin\theta)\hat{r} + (mg\cos\theta)\hat{\theta}$

$\sum F = \frac{mv^2}{r}\hat{r} + mg\cos\theta\hat{\theta}$

find velocity using conservation of energy, then find the magnitude of the above vector gives the correct answer E.

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Answered Question!

jburkart
2007-11-02 00:23:15

A real way to do the problem is as follows. Kinetic energy is $T=mR^2\dot{\theta}^2/2$ , where $R$ is the length of the string; potential is $U=-mgR\sin\theta$ . Setting the Lagrangian to $L=T-U$ , Lagrange's equation is $\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial\dot{q}}$ . In our case $q=\theta$ and we find $\ddot{\theta}=g\cos\theta/R$ , which then yields $a_\parallel=\ddot{\theta}/R=g\cos\theta$ . But this is only the acceleration along the direction of motion; there is also a perpendicular centripetal acceleration given by $v^2/R$ . To find $v$ we can invoke conservation of energy: $mv^2/2=mgR\sin\theta$ , so $v^2=2gR\sin\theta$ , so the centripetal acceleration is $a_{\perp}=v^2/R=2g\sin\theta$ .

Taking the norm of the the two acceleration components, we get $a=g\sqrt{\cos^2\theta+4\sin^2\theta}=g\sqrt{1+3\sin^2\theta}$ , answer (E).

A common mistake would be to omit the calculation of the centripetal acceleration, which would yield an incorrect answer of $g\cos\theta$ , which ETS kindly does not give as an option.

kroner
2009-10-06 22:29:16

You don't need to do all that work to find the tangential acceleration. It's just the tangential component of gravity, which is $gcos\theta$ .

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Alternate Solution - Unverified

kb
2007-10-28 22:49:32

is E is correct, can someone explain why
a(pi/2) = sqrt(3+1)g = 2g?

kb
2007-10-29 16:52:27

sorry, typo

"if E is correct, can someone explain why a(pi/2) = sqrt(3+1)g = 2g?"

and should go under "help!"

jburkart
2007-11-02 00:25:23

Because at that point the mass is experiencing a lot of g's! Try swinging in a swing sometime--you feel a lot of acceleration at the bottom.

evanb
2008-06-26 19:26:17

At the top, the energy is $mgh$ , and at the bottom it is $.5mv^2$ .

This leaves us with $v^2 = 2 g h$ , and $a = v^2 / r$ , but $r = h$ , so $a = 2g$ .

Reply to this comment

NEC

rmyers
2006-12-01 13:29:01

The first four words should be edited out of this solution.

lambda
2007-10-21 22:13:29

I agree. This test can't be finished in the time allowed if one tries to apply "rigorous physics"!

ndubuisi
2011-11-05 01:52:04

But one must answer a quetn well,rather than apply tricks n fail

Reply to this comment

NEC

Andresito
2006-03-30 12:05:12

Thank you for the advice. :)

ayabepaula
2008-10-22 00:03:57

1)http://dev.physicslab.org/img/db0bf4f1-2f9b-428f-81a6-536588e8a7ce.gif

2)http://www.physics-lab.net/applets/simple-pendulum

$U=mgh=\frac{mv^{2}}{2}$
then
${v}^{2}=2gh$

Gravitational acceleration:
********radial direction***************
${a_{cp}} = \frac{v^2}{Radius=h}$
${a_{cp}}=\frac{v^2}{h}$
then
${a_{cp}}= \frac{v^2}{h}=-2{g_{radial}}$ (negative sign --> ${a_{cp}}$ and ${g_{radial}$ --> opposite direction.
${a_{cp}}= \frac{v^2}{h}=-2{g_{radial}}$ ( r-direction)
${a_{cp}}=-2g(cos(90-\theta))=2g(sin90)(sin\theta)$
${a_{cp}}=-2g(sin\theta)$
${a_{cp}}^{2}=4{g^2}sin{^2}\theta$

Gravitational acceleration
********tangential direction********
${a_{tang}}={g_{tang}}$ (t-direction)

${a_{tang}}={g}cos\theta$

${a_{tang}^{2}}={g^{2}}{{\cos^{2}{\theta}}$

********total acceleration*********
${a_{total}}=\sqrt{{{a_{cp}}^{2}}+ {a_{tang}}^{2}}$
***********************************************
using
$cos^2\theta+sin^{2}\theta=1$

${a_{total}}=\sqrt{{g^{2}}\cos^{2}{\theta}+ 4{g^2}sin{^2}\theta}$

${a_{total}}=\sqrt{{g^{2}}(1-{\sin^{2}{\theta})+ 4{g^2}sin{^2}\theta}$

********
therefore:

The magnitude of the total acceleration of the mass as a function of the angle $\theta$ is:
********
${a_{total}}=g\sqrt{3sin^{2}\theta+1}$

********
Paula Ayabe Pereira

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NEC

Post A Comment!

You are replying to:

A real way to do the problem is as follows. Kinetic energy is $T=mR^2\dot{\theta}^2/2$ , where $R$ is the length of the string; potential is $U=-mgR\sin\theta$ . Setting the Lagrangian to $L=T-U$ , Lagrange's equation is $\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial\dot{q}}$ . In our case $q=\theta$ and we find $\ddot{\theta}=g\cos\theta/R$ , which then yields $a_\parallel=\ddot{\theta}/R=g\cos\theta$ . But this is only the acceleration along the direction of motion; there is also a perpendicular centripetal acceleration given by $v^2/R$ . To find $v$ we can invoke conservation of energy: $mv^2/2=mgR\sin\theta$ , so $v^2=2gR\sin\theta$ , so the centripetal acceleration is $a_{\perp}=v^2/R=2g\sin\theta$ . Taking the norm of the the two acceleration components, we get $a=g\sqrt{\cos^2\theta+4\sin^2\theta}=g\sqrt{1+3\sin^2\theta}$ , answer (E). A common mistake would be to omit the calculation of the centripetal acceleration, which would yield an incorrect answer of $g\cos\theta$ , which ETS kindly does not give as an option.

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