uniform circular motion: tension and velocity relationship

The question is:

A stone (m =0.35[kg]) is whirled in a vertical circle attached to a string of length r =95[cm].
At the exact top of the path the tension in the string is 3 times the stone's weight.
What is the stone's speed at this point?

I know the follow equations help me:
v = f( 2(pi)(r) / t ) where f is a frequency (usually given, but here not)
a[c] = (v)^2 / r which gives me the centripetal acceleration.

My issue is I don't know any other way using what's given to find a time which (I think) is needed to solve this problem. Any guidance is much appreciated.
ZenotureAsked:
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ozoCommented:
Can you find a relationship between a and v?
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ZenotureAuthor Commented:
well I know a is the derivative of v, and if I were to take the 2nd equation listed above and solve for v I would have:

v = sqrt(a*r)

But without 'v' or 'a' I still can't solve for it.... unless you mean by a system of equations?
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ZenotureAuthor Commented:
but wait, it's the same equation, just solved for a different variable, so that won't work...
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ozoCommented:
You know a and r, and want v
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ZenotureAuthor Commented:
How do I know a?
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ozoCommented:
At the exact top of the path the tension in the string is 3 times the stone's weight.
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ZenotureAuthor Commented:
Are you suggesting that 3(0.35) = 1.05 is the acceleration?
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ZenotureAuthor Commented:
Ok, so for future reference, am I to know "the path of tension" is the acceleration? Maybe I'm just reading it wrong...
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ozoCommented:
What is weight?
What are the units of acceleration?
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ZenotureAuthor Commented:
weight = the mass = 0.35

units of acceleration is distance/time^2

But there is no time given here, from what I can tell, that's my issue.
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ozoCommented:
weight = the mass
no
what are the units of mass?
what are the units of tension?
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ZenotureAuthor Commented:
Ok, I just figured out what you meant.

sum of forces must equal 0, therefore we know Fg = Ft = m*a = 3(0.35), where 3 = a

which gives us the speed of 1.69[m/s]... right?
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ozoCommented:
What is the tension in a string that is not being whirled in a circle?
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ZenotureAuthor Commented:
the force of gravity, since it would be pulling on the mass of the stone, which would then pull the string with it as well
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ZenotureAuthor Commented:
>>what are the units of mass?
>>what are the units of tension?

mass is kilograms, and tension is a force, so newtons, I presume, which is kg*m/s^2 which still does not solve my time dilemma.

Would it be possible to just tell me what I'm missing here? We've been at this for about 2 hours now and I don't feel like we've gotten anywhere with it.
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BillDLCommented:
I think ozo may be outside getting arthritis in his elbow swinging a heavy stone on a string while trying to capture it with a strobe flash ;-)
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ozoCommented:
So what is the tension in a string at the exact bottom of the path when the velocity is 0?
What is 3 times that tension?
What is the relationship between acceleration and tension?
What is the relationship between tension at the top of the path and tension at the bottom of the path?
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ZenotureAuthor Commented:
Not a bad idea, providing the string is tied tight enough, the stone not heavy enough to break the string's tension, and the velocity of the stone (if it breaks midair) slower than his reaction time to move the camera and not shatter it, or any part of himself in the process =)
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ZenotureAuthor Commented:
1.05 [kg*m/s^2] = 0.35 [kg] * ( 2(pi)0.95 [m] ) / t^2 [sec]

simplified, 0.7089 = t [sec],

v = m/s = 2(pi)0.95 / 0.7089 = 8.42 m/s

if that isn't it, I'm completely lost...
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ozoCommented:
Would the answer change if you swung the stone on the moon?
Would the answer change if the stone was bigger or smaller?
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ZenotureAuthor Commented:
Would the answer change if you swung the stone on the moon?
Yes, no gravity is existent

Would the answer change if the stone was bigger or smaller?
No, only if it's mass were different.
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ozoCommented:
How does gravity affect the answer?
How would mass affect the answer?
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TommySzalapskiCommented:
If you freeze the frame (which is legal) then you'll see gravity giving the stone 1 unit of pull down. Since it actually has 3 units of pull up, then it must have a centripital acceleration of 4*g (where g = 9.8 m/s^2)

So a = 4*9.8 and r is given, so v is easy to get.
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TommySzalapskiCommented:
The reason Ozo was asking so many questions was because this looks like a homework problem so we can't just give answers. It looked like you were getting frustrated and confused, so I thought it was appropriate to give a little extra help. Everything I said can be found online anyway in sample problems.
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ZenotureAuthor Commented:
_alias99,

It is homework, and I was not expecting a full solution, all I simply asked for was a way to find the time, from there I could figure out the rest and solve for what was needed, as TommySzalapski has done, as well as countless others who have helped me for the past year I've been here using this system.

Regardless of the method chosen, I signed up and pay for this system so I wouldn't have to sit here for 5+ hours doing one problem and feeling like I'm getting no where with it. I understand that it was late and not many were on available to help, but that is the only time I have available to work out those kinds of problems, and if this is the type of "help" I'll be getting from now on I don't see the point of paying monthly for this service when I could go on other forums and post for free with the same turn-around time.

TommySzalapski,

I kind of get what you were doing there, but what I don't understand is why it works that, simply because within that 3-hour run around I did google searches and didn't find any examples or solutions which explained it in any way close to yours.

You're saying that the upward force is 3 times whatever the downward force is, but Newton's 2nd Law states that the sum of all forces must equal 0, so how is it that you get 4*9.8 = a?
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SamGHCommented:
dont forget that f=mv^2/r
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hdhondtCommented:
On a side issue, to ozo's question "Would the answer change if you swung the stone on the moon?" Zenoture said:

Yes, no gravity is existent

Without gravity, how did the astronauts manage to walk and drive a rover on the moon?
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TommySzalapskiCommented:
The rock has three forces acting upon it. Gravity and tension are pulling it down. Gravity is pulling at a force of mg and tension is 3mg. This means the upward force due to the spinning must be mg + 3mg. Since we need the acceleration, we divide out the mass and get 4g or 4*9.8m/s^2 (or 9.81 or however your book likes to round it).

Of course the moon has gravity. He knows that. It's a bit lighter though. As a side note the almost complete lack of atmosphere on the moon makes it so all these first year physics problems where we neglect air resistance actually give you accurate results on the moon.
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TommySzalapskiCommented:
A lot of people accidentally make problems much more complicated than they need to be. I tend to see them in the simplest ways and so skip steps a lot. I hope the above is easier to understand now that I put the mass back in.
As a sanity check, apply the problem the same way to some obvious versions.
If the string was dangling limp, gravity would be pulling down at 1mg and tension would be pulling up at 1mg so the force due to the spinning would be 0 as expected.
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phoffric\Commented:
When I took this physics class a long time ago, I could easily spend 3 hours on a problem of this nature on the first encounter. Sometimes I'd fall asleep on the book, wake up 8 hours later and know the answer. On the second encounter of a problem of this nature, the solution came much faster to me.

Physics is very frustrating. That's what makes it interesting to some students.

I looked at ozo's posts, and saw absolutely nothing wrong with them. I think a methodology to the solution was at hand. If you try to solve physics problems without understanding the dimensional analysis that needs to be done, then the student will be working with a bunch of meaningful, but meaningless formulas.

When I saw what ozo was trying to accomplish here, I respectfully declined from contributing to this post. But I was monitoring it to see (hopefully) the enlightment that the author would experience when solving the problem.
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TommySzalapskiCommented:
Physics is very frustrating. That's what makes it interesting to some students.
That must be why I liked it so much. It's almost like a riddle sometimes.

then the student will be working with a bunch of meaningful, but meaningless formulas.
I agree completely. I hate formulae unless I understand why they work. If I know where the formula came from, I don't need to remember it, I can just rederive it (usually).

I also was only monitoring the post and leaving it to Ozo, when I saw the frustration (and the delete request) I thought it had reached a point.

Zenoture, the more you figure out yourself, the more you learn. Always start a physics problem by drawing a diagram with all the forces. This is how I was taught to do it, and it works very well (especially if you are a visual learner, which I am not).
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phoffric\Commented:
>> The Experts are NOT allowed to directly assist you in solving homework problems.
   I did learn however that experts are allowed to directly assist you in solving homework provided that the offering is only a small part of a larger problem (and if the small part, which may be still unknown to the author, is considered not too difficult by the ZA and moderators, even if other experts disagree).
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phoffric\Commented:
>> A "small part" is considered as hint, not solution.
That is what I used to think until the EE academic policy was recently clarified to me. The "reasonable test" that I had been using up till recently was whether a professor would be concerned if a solution or a partial solution (however small) in a post would be a concern to the professor. But now my understanding is that we can provide partial solutions providing that we explain step by step why the solution works. (I am not really happy with this EE policy because of the "reasonable test" that I just noted.

For academic questions asked by non-students who are self studying, my previous understanding was that we should just provide guidance and not offer full solutions. (One ZA wrote that the Academic policy still applies even if the academic question is not homework.) But now that I understand the policy better, I no longer RA posts having solutions if there is an explanation of why it works. Take for example, the question where a new EE member (who says he is self-studying) and has not provided a single line of code to a non-trivial question. We see experts providing code solutinos with explanations:
    http://www.experts-exchange.com/Programming/Languages/C/Q_26939029.html#a35339559
    http://www.experts-exchange.com/Programming/Languages/C/Q_26939029.html#a35346711
    http://www.experts-exchange.com/Programming/Languages/C/Q_26939029.html#a35346734

I only bring this up, because it does appear that it is OK to give answers in EE as long as the expert provides an explanation as to why the solution works. Personally, I prefer helping the student discover the solution path by pointing them in the right direction and answering their specific questions. I think that as long as they are not missing their deadlines, many students get a good deal of satisfaction, and a longer term understanding of the material.

My misunderstanding of the EE Academic policy came to light in this question where the author was having difficulty synchronizing communications between the client and server:
    http://www.experts-exchange.com/Programming/Languages/Q_26913837.html#a35224171

The author's erroneous code was here:
    http://www.experts-exchange.com/Programming/Languages/Q_26913837.html#a35223410
        n = read( sock, pch, size_to_read);  
        sizedata = n;  
        printf("n is %d and size_to_read is %d\n", n, size_to_read);  
        if(n < size_to_read)  
        {         
                error("Error in receiving data buffer");  
        }

Open in new window

The solution which I consider non-trivial was offered here:
    http://www.experts-exchange.com/Programming/Languages/Q_26913837.html#a35224171
int nread = 0;  
   while (nread < (int)size_to_read)  
   {  
        int result = recv(sock, &buffer[nread], size_to_read - nread);  
        if (result <= 0)  
        {  
             error();  
             break;  
        }  
        nread += result;  
   }

Open in new window

I was told that since this code was just a small part of a larger project, and was just showing a standard way of the server handling receive requests, then it was Ok to write the code for the student (afterall, this code can be found online in many client/server links).

I RA this solution thinking that it violated EE Academic Policy. But I learned: "why the rule of academic integrity exist. It is NOT to prevent askers getting the assistance they need, it IS to stop unscrupulous experts from abusing the asker’s desire for a solution to earn quick points." I wrote in the RA that "I now understand that providing short code can really help an author do his homework faster and with greater understanding."

While I think the EE Academic Policy is not in conformance with my "reasonable test", at least now I understand what it is, and no longer RA when a post provides a solution (as long as there is a tutorial explanation).

I think the approach ozo took in this problem was just fine. But problems do arise with authors when they need a solution fast due to an impending deadline to a complicated assignment. Otherwise, if under less pressure, then I think they appreciate the discovery process.
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ZenotureAuthor Commented:
Actually, this "off-topic" discussion was more informative than the solutions provided to me. Regardless, I asked the question be closed because none of the solutions provided helped me understand what was going. Additionally, requiring 3+ hours to solve a problem that could have easily been solved in one if all of these nonsense rules were not questioned, then I would have gotten a solution and understood it better, instead of wasting my time and passing my deadline, there was a reason why I wanted only to know a part of how to find the velocity, because I could do it from there.

The deadline has passed, and still no solution has been given, nor do I plan to play these games with the mods and experts for another week trying to understand something will lead to questions within questions of questions, so please close this question, or it will continue in the way it is going now.
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ZenotureAuthor Commented:
Modalot,

Look at the question's history, it has been attempted to be closed, but was objected by mods and experts, so please close this question. Thank you.
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phoffric\Commented:
Did you ever figure out how to do this problem?
Do you know the final answer? Post it please before this question is deleted.

===

BTW - I notice that when you attempt to solve a physics problem, you plug in numbers a little too early, IMO.

Suppose you delayed entering numbers, and instead, just worked with symbols. Suppose in some problem you derived a relationship of 5 variables. For example, a*b = c*d + e.
   (This is a highly contrived example, just to make a point - no intended relationship to any physics or math problem that I know of.)

           a*b = c*d + e

If you were given a, c, d, e, then you could solve for b. If the problem were worded differently and you were given a,b,c,d, then you could solve for e.

By deferring the number insertion to the end, you end up with a relationship that you can start thinking about in terms of physical phenomena, and even using your intuition about motion or statics to see if the relationship is correct. Another nice thing about deferring to the end is that some of the terms cancel out, so there is less arithmetic to do, and less chance for error.
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ZenotureAuthor Commented:
the answer was 6.1, and it was a guess based on another question, which gave the formula 2*sqrt(g*r), so I got lucky. Still, the formula that gave me the right answer was not explained to me, nor do I know how to derive it simply from the question, and not "cheating" else where.

As for plugging in early, unfortunately, that was the way i was taught. We were given formulas, and taught how to make FBD's (and use them to find formulas) but I might have messed up on my equations, or even set the wrong variables equal to one another, simply because I was taught a = 9.8, but that is only on earth, and the a = g, so whenever I see "a" I think 9.8 when it is not always so (i.e m*g != m*a unless free falling)
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phoffric\Commented:
>> a = 9.8, but that is only on earth
And I think you know by now that this only near the surface of the earth. Satellites orbiting around the earth experience a lower gravitational acceleration.

In FBD's you can use F1, F2, etc. Sometimes, actually quite often, you see components cancelling out due to symmetry, thereby saving some arithmetic steps (as well as increased understanding of the model you are drawing). For example, you have a horizontal bar hanging from a ceiling by two equal length wires attached to each end of the bar and meeting at the ceiling at a single point. Here there is some trigonometry involved, but thanks to symmetry, you will be able to simplify the formulas provided that you defer plugging in numbers until you have the relationships between the symbols representing your model. Here on EE, it is also easier IMO for us to evaluate your progress if you use symbols rather than numbers, since now we have to do more arithmetic ourselves instead of spending quality time on the model's relationships.
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TommySzalapskiCommented:
This is where my post was headed. I'll give a more detailed explanation here. By the way, I wholly agree that keeping everything in symbols until the last second is by far the best way to do it. They kind of forced me to do it that way in my physics class (they asked for symbolic answers on most questions)

The stone 'feels' 3g tension and gravity applies 1g acceleration. This means the total acceleration on the stone is 4g toward the center of the circle about which it is rotating. Since there are no other forces in play, this must also equal the centripital acceleration.
You know a = v^2/r so v = sqrt(a*r)
a = 4g and r = .95 and g = 9.8 so v = sqrt(4*9.8*.95) = sqrt(37.24) = about 6.1
If you had asked for more detail as to why I'd have given it. I see this is getting deleted today, so if you don't mind, I'll hit object so you can see this explanation. I hope it helps your understanding somewhat.
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ozoCommented:
you know a relationship between a and v for a given r in circular motion,
so all you need is to determine the appropriate a
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phoffric\Commented:
Tommy,
Since you are explaining your solution, I think you should clarify that you cannot add a 3g tension (which is a force) to a 1g acceleration. You should explain how you get the 4g without appearing to add different model elements.
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TommySzalapskiCommented:
Ah, yes. Sorry. I'm skipping steps again.
Technically, the tension is a force with a magnitude of 3gm and the force due to gravity (weight) is gm. So the sum of forces is 4gm. Since F=ma, then a = F/m or 4g.
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phoffric\Commented:
>> the stone not heavy enough to break the string's tension
LOL.
    A student once asked about a pendulum hanging from a ceiling about whether the ceiling could break (and this was actually stopping him from solving a HS problem). He eventually got a full college scholarship after going from an F to an A. His problem was that he was thinking to realistically about real-world situations and couldn't handle the simpler idealistic introductory problems. His capability of thinking deeply about a problem was hindering his progress at the level in which he had to perform.
    So, if you were serious, then you fall into the deep thinker category. In fact, I expect that you will be asked to solve problems where the string breaks (and worse, you have two objects and two strings, with one connected to a ceiling - then you pull on the bottom object with a force, and are asked which string breaks - and the answer is that it depends on the force and the two masses). I advise that you work with symbols!

====
Another advantage of putting your answer in symbols is that you can then do a dimensional analysis sanity check before you do the arithmetic.

So, instead of
>>     v = sqrt(4  *  9.8  *  .95)
write:
v = sqrt(4  *  g  *  r) = 2 sqrt( g meters/sec²  *  r meters )
    ~ sqrt ( meters/sec²  *  meters ) = sqrt ( meters²/sec² ) = sqrt ( [meters/sec]² ) = meters/sec (ok, this is speed)

====

>> Since F=ma, then a = F/m or 4g.
>> m =0.35[kg]

Hmm, notice that the m=0.35 kg does not need to be computed at this stage to compute a. In fact, the mass seems to have disappeared from the solution. Can that really be the case?
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ZenotureAuthor Commented:
>>
TommySzalapski:
Ah, yes. Sorry. I'm skipping steps again.
Technically, the tension is a force with a magnitude of 3gm and the force due to gravity (weight) is gm. So the sum of forces is 4gm. Since F=ma, then a = F/m or 4g.
>>
Force of gravity should be negative according to the FBD, and the Force of Tension should be positive, making it -gm + 3gm = 2gm. Thus why I am still lost, and why this should have already been closed.

>>
phoffric:
So, if you were serious, then you fall into the deep thinker category.
>>
It was a joke, but it was a serious thought that crossed my mind. I tend to over analyze everything I work on, that's part of the reason why I can write programs really well, but take forever to analyze and simplify things.

Regardless, this question was due 2 weeks ago, and the solution still isn't clear. I figured by now it'd be dead, but since people are still trying, I'm surprised no one has tried actually drawing out the FBD just to make it clear as day what is going on.

I'm still sticking to my guns and making a point though, this question needed an answer 2 weeks ago, and I already have the answer, I'm not going assign points for something that should have been answered two weeks ago.

On a side note, even if I did go further with the question, and ask you to explain everything the way you did, it still would have taken longer than I needed, so that isn't an excuse either.
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TommySzalapskiCommented:
Force of gravity should be negative according to the FBD, and the Force of Tension should be positive

Why should tension be positive? Draw the FBD, the string is at the top of the loop so tension is pulling the stone downward (the negative direction). Gravity and tension are both pulling the string in the same direction so naturally the forces would add together.

I gave that answer the same day you posted the question then you didn't say anything for a week. How could I clarify and explain if you didn't say what you didn't understand?
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TommySzalapskiCommented:
FBD
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ZenotureAuthor Commented:
I've drawn this FBD 5 times now, so obviously I'm drawing it wrong, therefore asking me to draw it again will not solve anything.

>>
I gave that answer the same day you posted the question
>>
My post:
Posted: 04/09/11 11:04 PM
Time Zone: Pacific Daylight Time (GMT-07:00)

Your post:
04/10/11 07:59 AM

Which is
  1. 12 hours after I posted the question
  2. After I asked to closed the question since it was past due and no longer relevant to what I was doing.
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ZenotureAuthor Commented:
Close enough to 12 hours at least, but still past an acceptable turn around time for me.
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TommySzalapskiCommented:
I don't care if you delete the question, I just want you to understand the problem before the question goes away. I drew the FBD above. Do you see why that is correct? Tension pulls the stone toward the string, it doesn't push it away. (Or were you drawing the stone at the bottom of the circle?)
Anyway, I hope it's finally clear and I'm sorry we weren't able to get you to the solution any faster.
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ZenotureAuthor Commented:
that still looks like the gravity is pulling down on the string, making the string have force of +3mg, therefore, still showing me that it's 2mg, hence the confusion.

I understand what you're trying to say, but I drew my FBD just like that, but with the force of gravity downward, and the force of tension upward due to the stone.
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ZenotureAuthor Commented:
Solution provided, even if it is late.
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ZenotureAuthor Commented:
If only it was sooner drawn, it would have helped, but I appreciate the effort.
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