# Homework Help: Circular + parabolic motion

1. Jul 9, 2009

### songoku

1. The problem statement, all variables and given/known data
A point object of mass m is connected to an inertialess string of length L. The other end of which is fixed to a point O. At time t = 0, the object is assumed to begin to move horizontally in a vertical plane from the bottom point A (OA = L) in the clockwise direction with an initial speed Vo. If $$\sqrt{2gL}$$ < Vo < $$\sqrt{5gL}$$ (g=acceleration due to gravity), then at a point B the magnitude of the force acting on the object from the string becomes zero, where OB = L and the velocity of the object is perpendicular to OB. We restrict ourselves to the case 0 < $$\theta$$ < $$\pi$$/2. From the point B, for a while, the object takes a parabolic orbit till a point C, where OC = L. In the case $$\theta$$ = $$\pi$$/3, find the angle $$\varphi$$

2. Relevant equations

3. The attempt at a solution
I've found the speed V = $$\sqrt{gL sin\theta}$$ and the initial speed Vo = $$\sqrt{(2+3sin\theta)gL}$$. I also got the maximum elevation (with respect to the location B) = $$\frac{V^2 cos^2\theta}{2g}$$

What should i do next?

thx

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2. Jul 17, 2009

### songoku

thx

3. Jul 17, 2009

### queenofbabes

Why do you think the mass goes from circular to parabolic to circular orbit?

4. Jul 17, 2009

### songoku

from circular to parabolic, i guess it's because the tension becomes zero.
but i don't think the mass will move on circular orbit again after parabolic.
or it will because the tension is not zero anymore?

thx

5. Jul 18, 2009

### songoku

the answer is pi but i still can't get to it....

Last edited: Jul 18, 2009
6. Jul 18, 2009

### queenofbabes

Yes, the trajectory will change to parabolic when the tension is zero. At some point the parabolic trajectory will move out of the area bounded by the circle of radius L about the origin. But of course, since the mass is bounded by the string it cannot do so and it will resume its circular trajectory, its centripetal acceleration provided by the tension.

So your main problem is to find when this parabolic trajectory intersects the boundary of the circle.

7. Jul 19, 2009

### songoku

i've tried to solve this but....
i don't know the idea to find the intersection.

i even draw a triangle BOC and try to find angle BOC but fail.

8. Jul 19, 2009

### queenofbabes

You can work out the coordinates where the mass starts the parabolic trajectory, right? Set up parametric equations (using the normal kinematic equations) and work out the equation of the parabola, and solve simultaneous equations with the equation of the circle. IMO that's the most straightforward way.

9. Jul 19, 2009

### songoku

this is my work

the parabolic trajectory starts at (-L cos $$\theta$$, L sin $$\theta$$)
the speed when it starts to move parabolic = V = $$\sqrt{gL sin \theta}$$
the angle with horizontal when the parabolic motion starts = 90o-$$\theta$$

using kinematic equations :
x = Vx t

t = $$\frac{x}{V_{x}}$$

= $$\frac{L cos\theta}{V sin \theta}$$

y = Vy t - $$\frac{1}{2}$$g t2

= V cos $$\theta$$ $$\frac{L cos\theta}{V sin \theta}$$ - $$\frac{1}{2}$$g [$$\frac{L cos\theta}{V sin \theta}$$]2

= $$\frac{L cos^{2}\theta}{sin \theta}$$ - $$\frac{1}{2}$$g $$\frac{L^{2}cos^{2}\theta}{g L sin \theta sin^{2}\theta}$$

= $$\frac{\sqrt{3}}{6}$$ L - $$\frac{\sqrt{3}}{9}$$ L

= $$\frac{\sqrt{3}}{18}$$ L

equation of circle : x2 + y2 = L2

x2 + [$$\frac{\sqrt{3}}{18}$$ L]2 = L2

x $$\approx$$ L

then y $$\approx$$ 0, so $$\varphi$$ $$\approx$$ $$\pi$$

i'm still not sure of my answer...

Last edited: Jul 19, 2009
10. Jul 19, 2009

### queenofbabes

I retract my words: I tried it myself and the algebra gets into a mess quickly. I'll try to come up with something, then reply again.

Edit: So far I could only cheat and sub the answer of pi back into my work and verify that it is a solution, I don't think that's sufficient. I can't come up with a neat way to solve this, perhaps someone else can help out?

Last edited: Jul 20, 2009
11. Jul 20, 2009

### songoku

and that means that my work is wrong...T_T

12. Jul 20, 2009

### queenofbabes

When setting up the parabola, you have to remember the parabolic trajectory doesn't start at the origin. But I'm no longer convinced that this is a good way of approaching this problem...

13. Jul 20, 2009

### songoku

14. Jul 20, 2009

### songoku

i've been dealing with this one for more than a week..T-T

15. Jul 20, 2009

### rl.bhat

You have mentioned that the string is inertia less. Is it inextensible?
If it is yes, how can it take the parabolic orbit, where distance between O and vertex of the parabola is greater than L?
In the vertical circular motion,the forces acting on the particle are weight of the particle and the centripetal force. How is that at a point B the magnitude of the force acting on the object from the string becomes zero?

16. Jul 20, 2009

### queenofbabes

According to my calculations the distance between O and vertex < L....hmm have I made a mistake?

If at point B, the component of the weight in the direction of the string = centripetal force, won't the tension be zero then?

17. Jul 21, 2009

### rl.bhat

In that case the diagram given in the post is wrong.
Yes. It is possible. In that case what about the other component of the weight? What is its role in the projectile motion of the particle?

18. Jul 21, 2009

### queenofbabes

Yeah I agree the diagram is abit misleading.

As the for the other component of the weight, I thought it wouldn't matter once it deviates from its circular orbit and enters a parabolic trajectory, then we can just treat it as normal projectile motion...

19. Jul 21, 2009

### songoku

about the diagram, i just copy it from the original question
I also agree it is misleading....

EDIT :
my last interpretation was wrong. I think this is what the problem is:
the other component of the weight that is perpendicular to the tension at B will resist the projectile motion, so the dash line should be inside the circle. Then, at C the string becomes taut again so there is tension.

Am i right?

Last edited: Jul 21, 2009
20. Jul 21, 2009

### rl.bhat

Yes. you are right.
Now, At B velocity V = ( 2gLsinθ)^1/2 = Vo
Its vertical component = Vv = Vocosθ
Its horizontal component = Vh = Vosinθ
Now the particle is falling freely with a horizontal velocity Vh which remains constant.
The vertical component starts decreasing, becomes zero then increases until it reaches C. At this point the magnitude of this velocity is same but the direction is reversed.
The time taken for this change is given by Vv = - Vv + gt
Or t = 2Vv/g
Now BC = Vh*t
Now you know side OA, OB and BC. By knowing θ, can you find angle ABC? And from that φ

21. Jul 21, 2009

### songoku

i get V at B = (gL sinθ) ^1/2
My work :
At B, I think weight is the only force acting on the object, so
$$centripetal force = \frac{mv^{2}}{r}$$

$$mg sin\theta = \frac{mv^{2}}{L}$$

V = (gL sin θ)^1/2

Do you mean angle OBC?
Using kinematic equation, I'll get BC and after that using cosine rule to triangle OBC, I'll get angle OBC.
Angle BCO = angle OBC, so angle BOC = 180$$^{o}$$- 2 x angle OBC
Finally, $$\varphi$$ = BOC + $$\pi$$ \ 3

I've tried it but i don't get pi as the final answer...

I also confused that when using V = (gL sin θ)^1/2, I get BC = 3/4 L. I think because the answer should be pi, angle BOC should be 120$$^{o}$$ and then BC must be greater than L.

Using V = (2gL sin θ)^1/2, I get BC = 3/2 L. It's make more sense but the answer is still not pi and i don't know how to get V = (2gL sin θ)^1/2...

Aaaaah, it's so frustrating...T-T

22. Jul 21, 2009

### rl.bhat

You are right. It is typo. V = (gLsinθ)^1/2
If θ is 60 degree, symmetry does not allow φ to be 180 degree. I have my own doubts about the final answer.

23. Jul 21, 2009

### queenofbabes

I have verified that pi is correct. I'll type a complete answer when I get back home

24. Jul 21, 2009

### rl.bhat

Use the projectile equation
-y = x*tanθ - gx^2/2v^2cos^2θ.
Ηere y = Lsin60, θ = 30 degree and v^2 = gLsin60.
Solve for x. If it is L + L/2 then x is along the reference line and φ = 180degrees because on the reference line Lcos 60 = L/2.

25. Jul 21, 2009

### queenofbabes

Ok, it IS possible to get a neat(-ish) solution to this after all.

First consider point B. For there to be no tension in the string, component of weight in that direction = centripetal acceleration. From that we get, given theta = pi/3,
g cos(pi/6) = (v^2)/L, thus $$v^2 = \frac{\sqrt{3}gL}{2}$$

Considering components, we can get $$v_x^2 = (v cos(\frac{\pi}{6}))^2 = \frac{3\sqrt{3}}{8}gL$$, and $$\frac{v_y}{v_x} = \frac{1}{\sqrt{3}}$$.
These will be useful later.
Coordinates of point B is $$(-\frac{L}{2}$$,$$\frac{\sqrt{3}}{2}L)$$

Thus setting up equation of parabola, bearing in mind origin is at the centre of the circle.
$$x = -\frac{L}{2} + v_xt$$
$$y = \frac{\sqrt{3}}{2}L + v_yt - \frac{1}{2}gt^2$$
sub for t from x eqn, eventually $$y = (\frac{1}{3\sqrt{3}})(-\frac{4}{L}x^2 - x + 5L)$$

sub y into equation for circle, $$x^2 + y^2 = L^2$$
expanding, you eventually end up with x^4 polynomial : $$x^4 + \frac{L}{2}x^3 - \frac{3}{4}L^2x^2 - \frac{5}{8}L^3x - \frac{1}{8}L^4 = 0$$
Recognise that x = -L/2 is a solution (that was point B), so factorizing we get $$(x + \frac{L}{2})(x^3 - \frac{3}{4}L^2x - \frac{1}{4}L^3) = 0$$
After factorizing the cubic bit (after some trial and error, or using quickmath.com etc) you get $$(x-L)(2x+L)^3 = 0$$

x = L is a solution, corresponding to φ = pi.

Whew!

Last edited: Jul 21, 2009