# Motion in a vertical circle problem

1. Mar 11, 2015

Suppose a mass $m$ is whirled at instantaneous speed $v$ on the end of an inextensible string of length $R$ in a vertical circle. Let $\theta$ be the angle the string makes with the horizontal, and let $W$ be the weight of the particle.
According to the equation:
$$T = \frac{mv^2}{R} - W \sin{\theta}$$
$\frac{mv^2}{R}$ must be greater than or equal to $W \sin{\theta}$ since tension can never be negative.
But if both terms are equal, then the tension of the string is zero. The particle no longer follows a circular path if $T = 0$, since this is equivalent to the absence of a string holding on to the particle, right?
So the condition for pure circular motion (no coriolis or linear radial acceleration terms) should really be:
$$\frac{mv^2}{R} > W \sin{\theta}$$
without an equal sign, is this correct?

2. Mar 11, 2015

### PhanthomJay

Be sure of correct plus and minus signs in your equation when considering theta above or below the horizontal.
When above, and if T is 0 at the very top of the circle, what does your free body diagram show for forces, and what does Newton laws tell you at that point? Can T be zero here and still have centripetal motion?

3. Mar 12, 2015

If $T$ is zero, then the only force acting on the particle is its weight. The particle would follow a parabolic path (since the component of weight $W \sin{\theta}$ won't always point towards the centre of the circle). Is my analysis correct?

Last edited: Mar 12, 2015
4. Mar 12, 2015

This is a somewhat unrelated question: does the component $W \sin{θ}$ point towards the origin in the absence of $T$ (i.e. the absence of a string)?

5. Mar 12, 2015

### PeroK

You're over-thinking this. If T = 0, then that is the instant that the particle begins to depart from its circular orbit. So, if $sin(\theta)$ increases, it will lose its circular orbit. But, if that happens when $sin(\theta) = 1$, then it remain in a circular orbit, as $sin(\theta)$ will immediately reduce and non-zero tension will be regained.

6. Mar 12, 2015

Excerpt from K&K Mechanics:
"The maximum value of $W \sin{\theta}$ occurs when the mass is vertically up; in this case $\frac{mv^2}{R} > W$. If this condition is not satisfied, the mass does not follow a circular path but starts to fall; $\ddot{r}$ is no longer zero."

7. Mar 12, 2015

### PeroK

Yes, I'm the one who's over thinking this! Although, it might be an intertesting question!

Suppose the string is cut when the mass is at the top.

$v_x^2 = Rg$

$x = \sqrt{Rg}t$
$t = \frac{x}{\sqrt{Rg}}$
$y = R-\frac{1}{2}gt^2 = R-\frac{x^2}{2R}$
$y^2 = R^2 - x^2 + \frac{x^4}{4R^2}$
Compare this with a circular orbit:

$x^2 + y^2 = R^2$
$y^2 = R^2 - x^2$

So, the parabolic orbit is higher than the circular one, so T = 0 at the top leaves the mass in a circular orbit. The tension will be regained.

8. Mar 12, 2015

So $T = 0$ is only allowed if it occurs at the top. If $T = 0$ anywhere else, the particle would no longer be in circular motion.
In other words $W \sin{\theta}$ is allowed to be equal to $\frac{mv^2}{R}$ only at the top of the circle. At all other points it must be less than $\frac{mv^2}{R}$.
According to your equation(s), the parabolic trajectory coincides with the circular path at $(0,R)$, at all other points, the parabolic trajectory is higher, right?
I have a question that might not be relevant to this topic (it's related to the difference between circular and parabolic trajectories). If we cut the string at any point in the upper portion of the trajectory, won't $W \sin{\theta}$ still point towards the origin at all times?

Last edited: Mar 12, 2015
9. Mar 12, 2015

### PeroK

If the mass has enough energy to get to the top with $v^2 = Rg$ then it stays in it's circular orbit. If it has less energy, then it won't make it to the top. The point it leaves its circular orbit must coincide with T = 0.

10. Mar 12, 2015

Thanks PeroK!
What about the second part of my question? How do I explain the fact that although there is a component of weight that is always pointing towards the origin, the particle does not undergo circular motion?

Last edited: Mar 12, 2015
11. Mar 12, 2015

### PeroK

If the string is cut, W will be vertical. It's the string that creates the circular orbit, not gravity.

12. Mar 12, 2015

I understand. My question is really about the components of a force in polar coordinates. Any force can be decomposed into a radial component and a tangential component, but not all trajectories are circles. How do I explain that?

13. Mar 12, 2015

### PhanthomJay

14. Mar 13, 2015

I understand that radial and tangential components of the acceleration of a particle in parabolic motion add up to $g$. But I still don't understand why this doesn't result in circular motion (in terms of polar coordinates). What makes parabolic motion parabolic in polar coordinates?
Also, in the case of motion in a vertical circle, while $W \sin{θ}$ doesn't point radially inward at all points below the horizontal, it does for $0 < θ < \pi$.
But when we cut the string, the particle follows a parabolic path.
What's going on here?

Last edited: Mar 13, 2015
15. Mar 13, 2015

### PhanthomJay

I am not sure I understand the question. A particle moving on a curved path causes centripetal forces in the direction of the center of curvature of that path at that point. The path does not have to be circular, it just has to be curved. Like a parabola, for instance, where the curved parabolic path is dictated by the combined horizontal and vertical motions.
And why should it not? For the circular motion of an object being whirled by a string in a vertical plane, the weight component and the non-zero string tension provide the centripetal force, and the fixed string length keeps that curved motion circular. When the string is cut, the tension force no longer exists, so the only force acting is the weight force. That's free fall. And since you have a horizontal component of speed when the string is cut, you get your typical parabolic trajectory. Unless the string is cut at 0 degrees or 180 degrees, in which case you just get vertical motion.

16. Mar 13, 2015

If it's the fixed length of the string that causes the motion of the particle to remain circular, doesn't this mean that the motion is circular because of the geometric constraint involved, rather than the forces? My question is: what is the actual cause of circular motion (in terms of forces, that is)? Because if the only thing required for circular motion is:
(a) a force directed towards the origin
(b) the force is perpendicular to the velocity of the particle
Then it's not very clear to me (in terms of polar coordinates) why projectiles are parabolic and not circular. I can explain why they are parabolic in terms of cartesian coordinates, but I can't in terms of polar coordinates.
There's something I'm missing here.

17. Mar 13, 2015

### PhanthomJay

Let the origin (the free end of the rope) for circular motion be at (0,0). For circular motion, you not only need a centripetal force that is directed radially toward the origin at (0,0), but also that force must always be directed toward the origin at every point along the path. For parabolic motion, the centripetal force at points along the path is directed not to a particular origin, but to a continually changing center of curvature . Maybe that's why the use of polar coordinates are letting you down.

18. Mar 13, 2015

I think I get it now. How do I mathematically prove that the center of curvature is changing, which implies parabolic motion (not the other way around)?

19. Mar 14, 2015

### PhanthomJay

Sorry , the proof is beyond my expertise. Try googling on 'changing center of curvature'. Or maybe others can help.

20. Mar 16, 2015