# Homework Help: Maximum angle in a vertical circular motion

1. Oct 7, 2014

1. The problem statement, all variables and given/known data

A person with the mass m swinging in a pendulum motion (vertical circular motion) hangs by a rope (with negligible mass) that can stand a tension of two times the person's mass. I am trying to calculate the maximum angles of the circular motion without the rope breaking.

2. Relevant equations

I know that somehow I'll have to manage how to solve for which angles equals T>2m.
I know that it's a circular motion, which can be described as ac = = [PLAIN]http://theory.uwinnipeg.ca/physics/circ/img24.gif[I]r[/I]. [Broken]
I just don't get how to get going with this one! Can someone help me going in the right direction?

3. The attempt at a solution

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Last edited by a moderator: May 7, 2017
2. Oct 7, 2014

### TSny

Some hints:

1. At what point of the swing will the tension be greatest?

2. If $v_a$ is the speed of the mass at the point of question 1, can you find an expression for the tension at that point in terms of $v_a$, m, r, and g?

3. How can you relate the maximum angle to the speed $v_a$?

(Note: your expression T > 2m doesn't make sense. On the left you have force while on the right you have mass.)

3. Oct 7, 2014

Thank you, I'll try to work those hints and see if that gets me anywhere!
Oh, I meant to write T > 2mg (or this perhaps isn't even relevant?)

4. Oct 7, 2014

Initially I'm just trying to get my head around what the assignment even mean. I guess the Tension varies with the speed, thus the highest amount of tension should be where the maximum speed takes place which should mean the point on the circle that has the same horizontal coordinate is 0.

So figuring out which angle that causes some speed, v that doesn't exceed the Tension equal to 2 mg is a proper way to view this? (Or the maximum angle that equals the maximum tension that is).

Last edited: Oct 7, 2014
5. Oct 7, 2014

### TSny

Try to use Newton's second law as applied to circular motion to find an expression for the tension at the point where the mass has maximum speed. This should yield a relation between the speed and the tension at that point.

6. Oct 9, 2014

I'm having a hard time getting my head around this. But at the equilibrium point at the bottom of the arc the speed is the highest. And this point is good for operating on since there's no motion in the horizontal direction at this instant, right? Centripetal acceleration can be described as F_c = v^2/r.

Perhaps there's a way to solve for when the speed is zero? How do I get the angle into all of this? It's fairly abstract to operate on this without any numbers (but I guess it's the best way to learn stuff proper).

7. Oct 9, 2014

Can the tension perhaps be written as F_c + mg at the eq. point? m((v^2/r)+g)?

8. Oct 9, 2014

And I'm guessing the conservation of energy has to be included somehow! That mgh = mv^2/2 in the eq. point is of major importance. Which leads to that the speed the eq. point speed v = sqrt(2gh).

Am I onto something here?

Last edited: Oct 9, 2014
9. Oct 9, 2014

### TSny

Yes, the tension at the bottom of the swing is $T = mv^2/r + mg$. Hopefully, you got that result using Newton's 2nd law.

10. Oct 9, 2014

### TSny

Yes, use conservation of energy. Think about the meaning of h and how you can relate h to the maximum angle.

11. Oct 9, 2014

I'm clueless when it comes to relating the angle to h. Help me please!

12. Oct 9, 2014

### TSny

Draw a figure showing the person at the bottom of the swing and also at the point of maximum angle. Can you indicate the quantity $h$ on your diagram? Try to construct a right triangle so that you ran relate $h$ to the maximum angle, $\theta_0$, and the length, $L$, of the rope.

13. Oct 9, 2014

OK, so I guess I can write the height h as h = L-L*cos(θ). But I don't have any values to insert and it's the maximum angle I'm looking for. I'm very confused here.

14. Oct 9, 2014

And I noticed that the assignment says that the string holds for 3*m, not 2*m as I obviously happened to write in the first post! Sorry about that!

15. Oct 9, 2014

### TSny

Good.

Don't worry about numbers yet. Keep writing things in symbols.

Try using your expression for h in your energy equation. Also, in the energy equation you can substitute for v^2 at the bottom of the swing in terms of the tension T using your earlier equation that relates T to v^2.

16. Oct 10, 2014

As of now I'm just stuck in algebraic manipulations without knowing what I'm really doing. I'm really stuck here. T = mv^2/2+mg = mgh.
With the masses cancelled out it leaves me v^2/2+g = gh. That leads me to v^2/2+1 = h.

2h = 4 <---> h = 2?

17. Oct 10, 2014

### TSny

Oops. Consider what you've written here. Some of your terms have dimensions of force [T and mg], whereas the other terms have dimensions of energy.

Maybe it will help if you go back to your equation for the tension at the bottom of the swing: $T = mv^2/r + mg$. What does this equation tell you about the speed $v$ when the tension is at its maximum allowed value?

18. Oct 10, 2014

Well if I enter $T = 3mg$ (the maximum tension before the string breaks) and insert that in $T=mv^2/r+mg$ and substitute $T$ with $3mg$ and solve for $v$ I get $v=sqrt(2gr)$ Perhaps that algebraic expression's irrelevant? I do not know $r$ and have no idea how to get it either..

19. Oct 10, 2014

### TSny

Good. Now you have an expression for the maximum speed that the person can have at the bottom of the swing without the rope breaking.

Don't worry about not knowing the value of r. Keep writing things in symbols.

Try working some more with your energy equation now that you have an expression for v at the bottom.

20. Oct 10, 2014

Well..

I on one hand have $mgh$ which can be written as $mgL(1-cos(theta)$.
And on the other I have $mv^2/2$ which can be written as $v^2=2gL(1-cos(theta))$

If use the relationship between $mgh$ and $mv^2/2$ and the conservation of energy with the algebraic manipulation as seen above that gives me $mv^2/2=mg(L-Lcos(theta)) -> 2gL=Lcos(theta)v^2 -> 2g/v^2=cos(theta)$

And I'm sitting here, hour after hour and just trying to work the algebra. I'm trying every single combination of everything that I won't even recognize anything that is going to help me when I stumble upon it! :(

21. Oct 10, 2014

### Orodruin

Staff Emeritus
Remember that you have also related the velocity to the radius by the requirement on the tension.

22. Oct 10, 2014

You mean $v^2=2gr$? I'm sorry if I'm a bit slow, but I've been spending days on this assignment now and it's frustrating to try to solve for everything algebraically as it most often seems like I'm stuck in a endless loop of manipulation without knowing exactly what the end result looks like or should look like. I have so many unknowns!

23. Oct 10, 2014

### TSny

OK, good. You're almost home.

Don't forget that you have also showed that $v = \sqrt{2gr}$.
Try using this in the equation $v^2=2gL(1-\cos(\theta))$

You'll have to think about how $r$ is related to $L$.

24. Oct 13, 2014

OK, so $L$ must be equal to $r$ since it's the magnitude of the string in the bottom of the arc, and since it's turning around a fix point in a circular motion it's also the radius of the circle.

$v^2=2gL(1-cos(theta))$ where $v^2=sqrt(2gr)$ leads to
$2gr=2gL(1-cos(theta))$ and using the fact that $r=L$ gives me
$cos(theta)=0$ which is the same as 90 degrees or $pi/2$ radians. This would mean that the relative angle to the eq. point would be from a total horizontal angle and fall down a negative $pi/2$ radians until the eq. point is reached.

Does this make any sense? That an angle bigger than $pi/2$ would break the string?

25. Oct 13, 2014

### TSny

Yes, that looks good. If the maximum angle that the rope makes to the vertical is 90 degrees, then the rope would be just at the breaking point when the person reaches the bottom of the swing.