# Homework Help: Vertical Circle Motion

1. Sep 10, 2014

### rulo1992

1. The problem statement, all variables and given/known data Suppose a non-uniform circular motion where a particle of mass "m" is attached to a string, which rotates on a vertical plane. Once an initial velocity is provided to the particle at the lowest point of the trajectory, no further forces act on the particle. (Air drag is negligible)
1. Which is the minimum velocity that the particle requires to reach the highest point of the trajectory?
2. If initial velocity is two-times the one we calculated in question 1, what would be the velocity in the highest point? Calculate the acceleration too:
3. Calculate the maximum height the particle can reach if the velocity is half the one calculated in 1.
4. Calculate the initial velocity if the particle rotates only 2$\Pi$/3 radians:

2. Relevant equations
$1/2$mv^2bottom=$1/2$mv^2top+mg2r
Centripetal acceleration=(v^2)/(r)
v=ωr

3. The attempt at a solution
For no.1 , given that KE is fully transformed into potential energy at the highest point of the circle,KE is 0 in the right side of the equation so: v=2√(gr)

For number 2, if velocity is 4√(gr) then $1/2$m(4√(gr))^2=$1/2$mvtop^2+mg2r →vtop=2√(3gr)

So I tried to calculate the centripetal acceleration and I got an=(2√(3gr)/r
But know I don't know how to calculate the tangential acceleration with only the values I know and now I'm stuck on this problem. Pleasse help me!

Last edited: Sep 10, 2014
2. Sep 10, 2014

### haruspex

If all KE converted to PE, how is it still moving in a circle?

3. Sep 10, 2014

### ehild

The speed is minimum at the top of the circle. How is tangential acceleration related to speed?

ehild

4. Sep 10, 2014

### rulo1992

Well question no. 1 asks for the minimum velocity required to get to the top of the circle, so the particle does not necessarily move in a cirlcle after that point.

5. Sep 10, 2014

### ehild

It will not even reach that point along the circle as the string becomes slack before.

ehild

Last edited: Sep 10, 2014
6. Sep 10, 2014

### rulo1992

If the velocity vector(tangent to the trajectory) is antiparallel to the tangential acceleration vector, speed decreases.
But that doesn't answer my question :(
Sorry but this is my first post and I don't know how to use latex

7. Sep 10, 2014

### ehild

Well, an other hint: What is the direction of the resultant force at the top of the circle? And how are acceleration and force related?

ehild

8. Sep 10, 2014

### rulo1992

Ohh I get it the resultant force points towards the center of the circle, therefore there's no tangencial acceleration right?

9. Sep 10, 2014

### ehild

It is right!

ehild

10. Sep 11, 2014

### rulo1992

Hi, I have another question, in another section of my homework it asks me to calculate the initial velocity if the object traveled to 2pi/3 radians, but how can I do it if I don't know any angular velocity nor acceleration?

11. Sep 11, 2014

### ehild

Apply conservation of energy again.

ehild

12. Sep 11, 2014

### rulo1992

Can I use polar coordinates? so y=rsen120 when the angle is 2pi/3 radians, and so the height at that point would be r+rsen120?

13. Sep 11, 2014

### ehild

Yes, the height is R(1+sin(120°)).

Edit: Oops. It is R(1-cos(120°))

ehild

Last edited: Sep 11, 2014
14. Sep 11, 2014

### rulo1992

And considering the circle rotates counterclockwise and at 2∏/3 radians tangential acceleration exists again(parallel to velocity, so it accelerates) , is it really possible for the object to only rotate to 2∏/3?

15. Sep 11, 2014

### ehild

The particle moves along the circle. See picture. The height from the lowest point is h=R(1-cos(θ)) .
It does not matter if theta is measured clockwise or anti-clockwise as cosine is even function.

ehild

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16. Sep 11, 2014

### rulo1992

Oh yes sorry my mistake we have to use cos indeed. Another question if you don't mind, will this initial velocity depend on the tension of the string?

17. Sep 11, 2014

### ehild

The initial velocity is given. The tension of the string depends on the speed. It has to provide the necessary centripetal force together with the radial component of the weight.

ehild

18. Sep 11, 2014

### rulo1992

What I don't get is how is it possible to calculate the initial velocity using conservation of energy.
Because if height=(R(1-costheta)) , is it correct to use the formula:1/2m(velocityatbottom)^2=mg(R(1-costheta))? or am I doing something wrong?
I don't get how can kinetic energy be 0, if the particle keeps moving along the cirlce after that point

19. Sep 11, 2014

### ehild

The problem text said that

So the particle does not go around its circular path after covering 2pi/3 radians.

If he particle is fixed to a rigid rotating stick that means the particle stops at that angle, then starts to move backwards.

If it is connected to a string, it still can have speed after reaching that angle, but the radial component of the weight is not enough for the centripetal force, the string gets slack, the motion will not be circular any more.
I think the problem maker expect the solution according to the first case.

ehild

20. Sep 11, 2014

### rulo1992

How are newton's 1st and 3rd laws applied to the forces acting on the particle?
this is the last question of my homework, and I'm having problem answering it correctly in a few lines. Do I have to consider inertial and non inertial frames separately?

21. Sep 11, 2014

### ehild

Stay in the inertial frame of reference. Consider the forces acting on the particle: the tension in the string and gravity. The radial component of the resultant is the centripetal force, while the tangential component accelerates the particle along the circle.

When discussing Newton's 3rd law, be careful. You have two kinds of interaction. One between the particle and the string. The strings pulls the particle with its tension. The particle pulls the string and that causes the tension in the string.
The other interaction is the gravitational pull between Earth and particle.

ehild

22. Sep 11, 2014

### olivermsun

Well you know that the height at the top of the circle is just $2R$ (you can check that $\theta = 180^\circ$ and $\cos 180^\circ = -1$), so that part is fine.

As long as KE > 0 at the highest point in the circle, the motion will continue, so what you do find the velocity $v_\mathrm{min}$ that gives exactly KE = 0. Then you know that the initial velocity $v_0$ has to satisfy $v_0 \gneq v_\mathrm{min}$.

23. Sep 11, 2014

### haruspex

This is wrong. There is a minimum speed at the top of the circle, and it is greater than zero. It won't reach the top if the string goes slack at any time before that.
What forces act on it at the top? What acceleration does it have there?

24. Sep 11, 2014

### olivermsun

Yup you're right. Thanks for catching that.

v^2/R = g is the limiting condition, where gravity is exactly equal to the centripetal force needed to follow the circle (and hence the string need not provide any tension). Does that sound right to you?

25. Sep 11, 2014

### rulo1992

how can I express the forces that act on the particle with x, y coordinates instead of polar coordinates?