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 04-24-2012, 10:45 AM #1 Senior Member     Status: Pre-Medical Join Date: Jun 2011 Posts: 239 Centripetal Force Q SDN Members don't see this ad. (About Ads) Two identical objects are rotating in a uniform circle about the earth. One object is twice the distance from the center of the earth as the other. What is the ratio of the centripetal forces acting on the objects? a) 1/2 b) 1/4 c) 1/8 d) 1/16 Help please. __________________ "If people let government decide what foods they eat and what medicines they take, their bodies will soon be in as sorry a state as are the souls of those who live under tyranny.” -Thomas Jefferson
 04-24-2012, 10:54 AM #2 1K Member     Status: Pre-Medical Join Date: Dec 2009 Location: Where the rain grows Posts: 1,896 F=ma. For uniform circular motion, a=v^2/r and F=mv^2/r. Twice the radius means half the force -> A.
04-24-2012, 10:58 AM   #3
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 Originally Posted by milski F=ma. For uniform circular motion, a=v^2/r and F=mv^2/r. Twice the radius means half the force -> A.
That's what I thought too. This is a Q from a TBR FL, and they say the answer is 1/4. Because the centripetal force is supplied by gravity, they equate the centripetal force with the gravitational force. Therefore the force is proportional to 1/r^2, not 1/r.

04-24-2012, 11:07 AM   #4
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 Originally Posted by brandonh4 That's what I thought too. This is a Q from a TBR FL, and they say the answer is 1/4. Because the centripetal force is supplied by gravity, they equate the centripetal force with the gravitational force. Therefore the force is proportional to 1/r^2, not 1/r.
They have a very good point. What I wrote would be true if the objects were rotating around Earth at the same angular velocity which is certainly not the case here. In other words - ops, I was wrong.

 04-24-2012, 11:09 AM #5 2K Member   Join Date: Apr 2010 Posts: 2,406 the v in mv^2/r is angular velocity? i thought it was linear
04-24-2012, 11:14 AM   #6
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 Originally Posted by chiddler the v in mv^2/r is angular velocity? i thought it was linear
It is. They are proportional, so saying that v1=v2 is the same as saying ω1=ω2.

And "same angular velocity" is a bit easier than saying that they have the "same magnitude of velocity."

04-24-2012, 11:16 AM   #7
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 Originally Posted by milski They have a very good point. What I wrote would be true if the objects were rotating around Earth at the same angular velocity which is certainly not the case here. In other words - ops, I was wrong.
Hmm... so if it were to state that the velocities were the same, the centripetal force would now be proportional to r^-1 rather than r^-2?

I'm just confused because they don't offer different velocities as one of their reasons to use the gravitational force equation rather than centripetal force equation.

 04-24-2012, 11:23 AM #8 2K Member   Status: Pre-Medical Join Date: Mar 2010 Location: NY Posts: 2,184 Wait, what? How do you set gravitational and this equal so you get r^2??? __________________ sector9, mauberley, flodhi1, flatearth22, MedPR, Neuronix, Catalystic, LizzyM, PharMed2016, Fencer, DrMidLife, nadaba, Gnomes, thlaxer, [04/28/12 MCAT]: Without them, I could not be where I am now. The most f'ed up, psychotic thing I've ever read on SDN.
04-24-2012, 11:25 AM   #9
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 Originally Posted by pfaction Wait, what? How do you set gravitational and this equal so you get r^2???
My brother and I have been arguing this for the past half hour. The words "Newton should die" and "f&ck you" have been spoken numerous times.

 04-24-2012, 11:26 AM #10 2K Member   Status: Pre-Medical Join Date: Mar 2010 Location: NY Posts: 2,184 Oh my ****ing god. Did they do this? FGrav = Fac GMm/rsq = Fac mass is the same since they're identical GMm/rsq = Fac What douchebags. Now it makes sense. GMm/rsq = Fac of object 1 GMm/rsq = Fac of object 2 They're saying the gravitational causes the centrip acceleration, which is true, and not asking you to solve any further. I would have also done Fac = mv^2/r to get 1/2. ****.
04-24-2012, 11:27 AM   #11
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 Originally Posted by brandonh4 Hmm... so if it were to state that the velocities were the same, the centripetal force would now be proportional to r^-1 rather than r^-2? I'm just confused because they don't offer different velocities as one of their reasons to use the gravitational force equation rather than centripetal force equation.
if you state that angular velocity is the same, then it is r^-1. since linear velocities are the same in this case, angular velocity drops.

alpha = v/r

as r increases while v stays constant, alpha decreases.

 04-24-2012, 11:29 AM #12 2K Member   Join Date: Apr 2010 Posts: 2,406 what i don't understand is how to come to the conclusion that angular velocities differ in each. my thinking was hold everything else constant - distance changes. therefore, it is half. what prompts one to look at the gravitational force equation?
04-24-2012, 11:30 AM   #13
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 Originally Posted by chiddler what i don't understand is how to come to the conclusion that angular velocities differ in each. my thinking was hold everything else constant - distance changes. therefore, it is half. what prompts one to look at the gravitational force equation?
TPR says you should be asking "what causes centripedal acceleration", so in this case, it's FGrav... I would have never thought of this on my exam.

04-24-2012, 11:30 AM   #14
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 Originally Posted by pfaction Oh my ****ing god. Did they do this?
Yes, but now you have 2 conflict equations, one that says it's proportional to 1/r and one that says it's proportional to 1/r^2. The best reasoning for using the gravitational equation was brought up by saying that, since the velocities were not stated to be equal, you can't use the centripetal force equation, whereas everything is constant except r in the gravitational force equation.

If I'm understanding things right...

 04-24-2012, 11:32 AM #15 2K Member   Status: Pre-Medical Join Date: Mar 2010 Location: NY Posts: 2,184 Bro I got it wrong for the same reason you did, only when you stated they used FGrav did I connect it together. I mean look at it this way: Fac = mv^2/r; Fac = w^2*r. TBR rapes my mind continually with this ****.
04-24-2012, 11:33 AM   #16
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 Originally Posted by milski F=ma. For uniform circular motion, a=v^2/r and F=mv^2/r. Twice the radius means half the force -> A.
Using that equation assumes that their tangential velocities are equal, which is not the case. The tangential velocity is also dependent on the radius from earth. The easiest way to solve this is set centripetal force and then using the gravitational force equation, exactly how TBR explains.

04-24-2012, 11:35 AM   #17
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 Originally Posted by pfaction TPR says you should be asking "what causes centripedal acceleration", so in this case, it's FGrav... I would have never thought of this on my exam.
what do you mean?

04-24-2012, 11:35 AM   #18
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 Originally Posted by Tatertots Using that equation assumes that their tangential velocities are equal, which is not the case. The tangential velocity is also dependent on the radius from earth. The easiest way to solve this is set centripetal force and then using the gravitational force equation, exactly how TBR explains.
Couldn't you achieve the same tangential velocity just by launching the farther satellite it into orbit... I don't know, with more velocity?

04-24-2012, 11:36 AM   #19
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 Originally Posted by Tatertots Using that equation assumes that their tangential velocities are equal, which is not the case. The tangential velocity is also dependent on the radius from earth. The easiest way to solve this is set centripetal force and then using the gravitational force equation, exactly how TBR explains.
he corrects himself two posts down.

 04-24-2012, 11:38 AM #20 1K Member     Status: Pre-Medical Join Date: Dec 2009 Location: Where the rain grows Posts: 1,896 There's nothing conflicting about the two equations. F=mv^2/r is correct but both v and r change, so you cannot say anything directly from here unless you calculate what the magnitude of the velocity will be. It's much easier to consider that the only force acting is gravity and use that it is proportional to 1/r^2, like Tatertots said.
04-24-2012, 11:48 AM   #21
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 Originally Posted by brandonh4 Couldn't you achieve the same tangential velocity just by launching the farther satellite it into orbit... I don't know, with more velocity?
For a given tangential velocity there will be one acceleration (and one force) which keeps the body in uniform circular motion. If you vary the force, you can vary the velocity at which the body rotates around earth at a fixed distance. But since there are no additional forces described in the problem, the only force is due to gravity. That means that for a certain distance there will only one velocity at which UCM continues. That velocity will be different for different r.

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04-24-2012, 12:01 PM   #22
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 Originally Posted by brandonh4 Couldn't you achieve the same tangential velocity just by launching the farther satellite it into orbit... I don't know, with more velocity?
Nope. The tangential velocity is a result of the gravitational force and that's it in tho question. See milaki below for a good explanation.
Quote:
 Originally Posted by chiddler he corrects himself two posts down.
Yeah I saw- I was slow. Took me 10 minutes to answer because my PI interrupted me.
Quote:
 Originally Posted by milski For a given tangential velocity there will be one acceleration (and one force) which keeps the body in uniform circular motion. If you vary the force, you can vary the velocity at which the body rotates around earth at a fixed distance. But since there are no additional forces described in the problem, the only force is due to gravity. That means that for a certain distance there will only one velocity at which UCM continues. That velocity will be different for different r. --- I am here: http://tapatalk.com/map.php?ph0bqo
This times 100. Remember Keplers laws - velocity is dependent on the radius of orbit.

04-24-2012, 12:05 PM   #23
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 Originally Posted by milski There's nothing conflicting about the two equations. F=mv^2/r is correct but both v and r change, so you cannot say anything directly from here unless you calculate what the magnitude of the velocity will be. It's much easier to consider that the only force acting is gravity and use that it is proportional to 1/r^2, like Tatertots said.
a) Is this likely to appear on the test?
b) How would you be able to differentiate it?!

04-24-2012, 12:09 PM   #24
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 Originally Posted by pfaction a) Is this likely to appear on the test? b) How would you be able to differentiate it?!
a. possible.

b. we have to ask ourselves some questions. first of all, what is changing in this question? we see that as radius increases, velocity changes too due to a changing force.

looking at F=mv^2/r, does it make sense to use this equation when there are three variables changing? no. what is an alternative so we can compare only two?

gravitational force works, since masses of the objects are constant and we can compare radius and force directly.

the important thing to learn, having read this thread, is that just because the question does not explicitly say velocity is changing doesn't mean you count out the possibility.

04-24-2012, 12:10 PM   #25
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 the important thing to learn, having read this thread, is that just because the question does not explicitly say velocity is changing doesn't mean you count out the possibility.
That's exactly what I didn't know/remember, that as radius increases the velocity decreases, so two variables no good - have to focus on one that works for both.

04-24-2012, 12:24 PM   #26
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 Originally Posted by pfaction That's exactly what I didn't know/remember, that as radius increases the velocity decreases, so two variables no good - have to focus on one that works for both.
As the radius increases velocity decreases when something is in orbit(because angular velocity is not the same) with gravity but don't let this confuse your general circular motion questions. If two objects traveling in circular motion have the same w then the one with the largest radius is going to have the highest tangential velocity because v=rw.

04-24-2012, 12:27 PM   #27
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 Originally Posted by pfaction a) Is this likely to appear on the test? b) How would you be able to differentiate it?!
On the phone, so tl;dr only:

A) It's a good question, I would use it if I was writing the test.
B) There is no distinction to be made - UCM is UCM and the formulas hold. What went wrong for me and OP is that we assumed the velocities are the same. There is nothing in the problem indicating that. You can start with that equation, derive what the velocities will be, plug them in and get the same result.

 04-24-2012, 12:55 PM #28 Senior Member   Status: Medical Student Join Date: Jul 2011 Posts: 315 So wait. As v1 approaches v2 The centripetal force proportionality r^-2 approaches r^-1 Meaning there is a whole continuity of variable proportions in between... I am quite confused because this doesn't seem to make sense/
04-24-2012, 03:19 PM   #29
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 Originally Posted by Tatertots As the radius increases velocity decreases when something is in orbit(because angular velocity is not the same) with gravity but don't let this confuse your general circular motion questions. If two objects traveling in circular motion have the same w then the one with the largest radius is going to have the highest tangential velocity because v=rw.
So in our problem, if they both had the same w, would the one traveling 2r away have a higher velocity? And thus higher centrip accel?

04-24-2012, 05:08 PM   #30
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 Originally Posted by pfaction So in our problem, if they both had the same w, would the one traveling 2r away have a higher velocity? And thus higher centrip accel?
Yes. It would need some sort of additional force to make that happen, typically a rocket engine.

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 04-24-2012, 10:21 PM #31 Banned   Status: Pre-Medical Join Date: Apr 2012 Location: Winter of our discontent Posts: 5,669 d^2χ/dt^2=-k^2(1+m) χ/r^3+Σi k^2mj [(χj-χ/ρj^3) – (χj/rj^3)] This is the general equation of motion of a planet or comet (m) with the sun as origin of coordinates. m subscript j is the action of disturbing planets. This same equation can be used for the motion of a satellite around its primary. Newton only describes the forces acting between two point masses that are inertial, not rotating or subject to acceleration. If the distance between m subscript a and m subscript b is r, then thf force acting between them is F=k^2mamb/r^2 K depends on units of mass, time and length. If the sun, for example is taken as unity and mean solar day and astronomical unit as length then k= Gaussian constant (0.01720209895) this is exact. Last edited by Tatiana3325; 04-24-2012 at 10:33 PM.
04-24-2012, 10:32 PM   #32
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 Originally Posted by Tatiana3325 d^2χ/dt^2=-k^2(1+m) χ/r^3+Σi k^2mj [(χj-χ/ρj^3) – (χj/rj^3)] This is the general equation of motion of a planet or comet (m) with the sun as origin of coordinates. m subscript j is the action of disturbing planets. This same equation can be used for the motion of a satellite around its primary.
well duh. so obvious! how could we forget this crucial equation!

04-24-2012, 10:35 PM   #33
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 Originally Posted by chiddler well duh. so obvious! how could we forget this crucial equation!
Well if I had said that solar system dynamics were not really Newtonian you would have dismissed my comment.

That mcat question in the initial post is vague by the way. But I'm sure you guys can figure it out. You're over thinking it.

Last edited by Tatiana3325; 04-24-2012 at 10:59 PM.

04-24-2012, 11:01 PM   #34
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 Originally Posted by Tatiana3325 Well if I had said that solar system dynamics were not really Newtonian you would have dismissed my comment. That mcat question in the initial post is dumb by the way. Its really vague but I'm sure you guys can figure it out. You're over thinking it.
all i meant was that the equation you posted wouldn't really help here because the vast majority of premeds are not well versed with such physics equations.

but that was rude of me anyway, sorry.

04-25-2012, 01:42 AM   #35
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 Originally Posted by chiddler all i meant was that the equation you posted wouldn't really help here because the vast majority of premeds are not well versed with such physics equations. but that was rude of me anyway, sorry.
.

Last edited by Tatiana3325; 04-25-2012 at 06:57 AM.

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