Using Centripetal Forces to find the radius of a circle

AI Thread Summary
The discussion revolves around calculating the radius of a circular path for a rock attached to a rope, using centripetal force equations. The user initially struggles with finding the radius without knowing the angle θ but combines equations involving tension and centripetal force to derive a solution. After some algebraic manipulation, they arrive at a radius of approximately 0.97 meters. Other participants confirm the calculations and clarify that gravitational acceleration should be used directly, not multiplied by mass. The final consensus is that the calculated radius of 0.97 meters is correct.
QuickSkope
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Homework Statement



A 100g (0.1kg) rock is attatched to a 1.0m rope and spun around in a circle with a period of rotation of 1.0s. What is the Radius of the circle that it forms?

Homework Equations




Fc = (mV^2) / r
V= (2∏r/T)
LCosθ = r


The Attempt at a Solution



Im quite stick here, as I am not sure how I could ever find r without knowing the θ of the circle.

First thing I tried was making a triangle like so:

Illustration in 2D

Im going to start with the simple equation TCosθ = Fc. We also know that (mV^2)/r = Fc. If you put those 2 equations together, you get TCosθ = (mV^2)/r. This leaves us with the equation:

TCosθ = (mV^2)/ r

Furthurmore, we know that V = 2πr/P (Where P is period, should normally be T but there's Tension in this equation, so ill just do P)

TCosθ = m (2πr/P)^2 / r

The period of rotation is 1, so the equation simplifies again to:

TCosθ = m (2πr)^2 / r

We can also sub r into the formula we just moved in.

TCosθ = m (2π (Cosθ)^2) / Cosθ

From here, we can make the equation:

T = (m (2π) ^2 * (Cosθ)^2) / (Cosθ)^2 (Correct? I am not sure if that's logical)

Now the (Cosθ)^2 can cancel out, making:

T = m (2π)^2

Sub in the values, we have T = (0.1)(2π)^2
= 0.4π^2

We now know Tension, Which we can use in the triangle.

Triangle illustration

If I use Sin, I can find that the angle is 14.37321*.

We can use this θ in the equation LCosθ = r
(1)(Cos14.37321) = r
r = 0.97 m.

Is that correct?
 
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Your method and result look fine.

Note that you might consider the problem in terms of similar triangles (I know, geometry was a looong time ago :smile:).

attachment.php?attachmentid=57981&stc=1&d=1366247405.gif


Consider the ratio ##g/a_c##...

You can use the rotational motion form for centripetal acceleration, ##a_c = ω^2 r##, where ##ω = 2\pi/T##. A little algebra and you can find an expression for r that does not require sines or cosines.
 

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gneill said:
Your method and result look fine.

Note that you might consider the problem in terms of similar triangles (I know, geometry was a looong time ago :smile:).

attachment.php?attachmentid=57981&stc=1&d=1366247405.gif


Consider the ratio ##g/a_c##...

You can use the rotational motion form for centripetal acceleration, ##a_c = ω^2 r##, where ##ω = 2\pi/T##. A little algebra and you can find an expression for r that does not require sines or cosines.

Do you mind explaining that a little? I'd like to check my answer, as I think 0.97m is a little large for the radius concidering you only make a full turn per minute. Or if you could do the math for this perticular situation and make sure it comes out correct, that would be awesome. Its for marks, so I just don't want to get it wrong.

Thanks :D
 
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QuickSkope said:
Do you mind explaining that a little? I'd like to check my answer, as I think 0.97m is a little large for the radius concidering you only make a full turn per minute. Or if you could do the math for this perticular situation and make sure it comes out correct, that would be awesome. Its for marks, so I just don't want to get it wrong.

Thanks :D

The problem says that the period of rotation is 1 second, not 1 minute. So ω is ##2\pi/1sec##.

L is the length of the rope. r is the resulting radius of the circle that the mass follows at one revolution per second. ##a_c## is the centripetal acceleration acting on the rotating mass. g is the gravitational acceleration. You should be able to pick out similar triangles in the diagram. Equate the side-length ratios.
 
Yea, sorry you're right. 1 second.

Ohh, so the Theta in both Triangle Ac/g and Lr are the same, correct? So you can equate the two, like this?

Tanθ = g/ac
Tanθ = √1^2 - r^2

g/ac = √ 1-r^2

g/ (2π)^2 = √ 1-r^2

(9.8 / (2π^2)) ^ 2 = 1-r^2

r^2 = 1 - (9.8 / (2π^2)) ^ 2

r^2 = .938

r= .969

Thats what I get when I do similar triangles, which I am 99% sure isn't right. Is the g (.1)(9.8) or mg, or is it just 9.8?
 
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QuickSkope said:
is the g (.1)(9.8) or mg, or is it just 9.8?
It's g, not mg. The mass canceled out.
 
Alright, so that makes both of them the same answer :). I end up with 0.97. Can someone do it and verify that that is right?

Thanks :)
 
QuickSkope said:
Alright, so that makes both of them the same answer :). I end up with 0.97. Can someone do it and verify that that is right?

Thanks :)

I agree with .97
 
Great, thanks :D
 
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