# Ball rotating in a circle without rope/string - why doesn't the ball fall after the peak? [duplicate]

I am currently learning about the problem of a ball rotating in a vertical circle - where we are interested in determining the minimal height $$h$$ required in order for a ball sliding from rest down a frictionless track (which consists of of a slope followed by a vertical circle) not to fall before completing a full round on the circular part. As far as I understand, the way to do it is to determine the minimal velocity required for the ball to reach the peak of the circle (marked by $$v=?$$ in the figure below) without falling off.

My question is -

What guarantees that once the ball reaches the peak it will not fall in any other point on the circle that is more distant than the peak?

Looking forward to your help. Thanks!

• It's a good answer but it approaches the issue less quantitively. When I first learned this, I was never explained why checking only the top is sufficient. I figured I would put a detailed post about it somewhere :) Dec 4 '20 at 1:06
• This is similar to calculating the velocity required for a satellite to remain in orbit. Dec 4 '20 at 1:08
• Well yes, except there (assuming circular orbit), the centripetal force remains the same, so it's not as interesting. I thought this question was quite nice, as it allows for justifying why we only check the top point when trying to find $v_{\rm min}$. When I studied this, every textbook just presented the $mv^2/r=mg$ statement and did not bother to specify why it holds along other points in the trajectory. Dec 4 '20 at 1:09
• The track may be frictionless but air isn't. The car confuses the issue (is it powered, or just rolled from h?). The requirement for the points of the (1/4 circle) more distant from the peak is v needs to offset g until the bottom 1/2 of the track is reached. Anyone in the car would be wise to carry some excess v. Dec 4 '20 at 2:21
• Hi guys, thanks for your comments! I do understand the general reasoning behind the fact that the water bucket (or car) doesn't fall off, but can't understand why if it arrives to the top of the circle, it must complete a full circle. Dec 4 '20 at 16:03

For the ball to travel in a circle, the following must be true: $$F_c = \dfrac{mv^2}{r}.\tag1$$ $$F_c$$ is the centripetal force. It is the component of the net force that is directed towards the inside of the circle. There will be circular motion if Eq.(1) holds. Now, the bottom hemisphere of the circle is not a problem, since the only path the ball can take is along the surface.

In the top hemisphere, consider the trigonometric angle as shown:

Therefore, it is obvious that:

$$F_c = mg\sin\theta + N\tag2$$

side note: many explanations will simply give you the condition that $$mv^2/r=mg$$ and call it a day. But there is more to it.

Notice that when the angle $$\theta$$ increases from $$0^\circ$$ to $$90^\circ$$, the velocity of the particle will go down (due to conservation of energy -- as we gain potential energy, we lose kinetic energy).

However, notice that $$\sin\theta$$ will increase as $$\theta$$ increases from $$0^\circ$$ to $$90^\circ$$. This means that at the very top, the speed is at a minimum, but the gravitational force (centripetal component) will be at a maximum. This creates a potential problem, because of the possibility that $$mg \sin\theta + N > mv^2/r$$. If that is the case, the motion will no longer be circular.

To ensure that the path is indeed always circular, recall the following 'must:' $$F_c = \dfrac{mv^2}{r}$$ must be true. So, ask yourself, where would that relationship not be true, so that way we know what condition to check. Obviously, that relationship may not hold when the particle is at the very top of the circle. The reason being: the speed is minimal, and the force of gravity pulling the object towards the center is maximal.

And so at the very top, for there to be circular motion, you must have:

$$\dfrac{mv^2}{r} = mg + N\tag3$$

Now if the normal force is $$0$$ (after all, normal force comes up when the ball goes fast enough to press against the surface), then the minimum value $$F_c$$ can have (such that the path is circular) is:

$$\dfrac{mv^2}{r} = mg \tag4$$

the reason that it's the minimum value is: if $$\tfrac{mv^2}{r}>mg$$, then you have the normal force that will act on the object, hence you will have: $$\tfrac{mv^2}{r}=mg+N$$, and the path is circular. But, since the normal force cannot be less than zero, the smallest value $$F_c$$ can have at the top is $$mg$$

Anyway, by Eq. (4), you get that the minimal speed at the top is: $$v=\sqrt{gr}$$. If the speed is any less than that, then the gravity will pull the object in towards the center too strongly and the path will no longer be circular.

Now, a very important part:

The reason that this TOP CONDITION is SUFFICIENT (i.e. why that's the only thing we must check for):

When the angle $$\theta$$ is is exactly $$90^\circ$$, then $$mg\sin\theta$$ will have its maximum value of $$mg$$, whereas $$mv^2/r$$ will have its minimum value (because speed is at its minimum).

It follows that when $$\sin\theta$$ gets smaller (i.e. the angle is not $$90^\circ$$), then $$mg\sin\theta$$ will no longer be at its maximum. On the other hand, because you gain KE, the speed $$v$$ will increase! This implies that $$mv^2/r > mg\sin\theta$$. This should make total sense; if we assume that at the top, $$mv^2/r = mg$$, then when we are no no longer at the top, the $$mv^2/r$$ term increases due to gain in KE, whereas the $$mg\sin\theta$$ term will get smaller (due to sine term). Therefore, if we know that $$mv^2/r=mg$$ at the top, then we also know that when $$\theta\neq90^\circ$$, that $$mv^2/r > mg\sin\theta$$.

Second, we will justify why the object will continue moving in a circle:

The reason the object will continue in a circle is due to the normal force; if you go really fast, the normal force will increase such that $$mv^2/r=mg\sin\theta+N$$, and if you go slower, then the normal force will conveniently decrease, again, such that $$mv^2/r=mg\sin\theta+N$$. BUT, the normal force cannot be negative in the top hemisphere -- the normal force's smallest value is zero.

Therefore, in the top hemisphere, it must always be the case that $$mv^2/r\geq mg\sin\theta$$. And if that is true when $$mg\sin\theta$$ is at its max and $$v$$ is at its min, then it will definitely be true when $$mg\sin\theta$$ decreases and $$mv^2/r$$ increases.

Therefore, checking the speed at the top is a necessary and sufficient condition for frictionless vertical circular motion.

Note: this explanation applies for the top hemisphere. As for the bottom hemisphere of the circle, the object will not break its circular path due to the normal force which (assuming is strong enough, i.e. material is strong enough) will always keep the object in circular motion. The reason is that the normal force will act towards the center of the circle, so it can counteract gravity such that $$F_c=mv^2/r$$.

In summary, the condition for the ball to continue in circular motion is that the speed at the top, must be greater or equal to $$\sqrt{gr}$$. Thus, the minimum speed needed at the top is $$v_{\rm min} = \sqrt{gr}$$.

This answer was long, but I hope it clears everything up.

Suppose the car reached the top point with a velocity $$v$$. The centripetal acceleration must be $$g \implies g = \frac{v^2}{r}$$

If the track was removed, the problem was to find the trajectory of a particle with an horizontal initial velocity $$v$$, having a constant acceleration $$g$$ downwards. We know that the solution is a parabola in this case.

Considering the origin the center of the circle, the equation of the parabola comes from: $$x = vt$$ and $$y = r - \frac{1}{2}gt^2$$

$$y = r - \frac{1}{2}g\frac{x^2}{v^2}$$

Substituting the value of $$g$$ $$y_{par} = r - \frac{1}{2}\frac{x^2}{r} = r\left(1 - \frac{1}{2}\frac{x^2}{r^2}\right)$$

The equation of the circle is: $$y_{cir} = \sqrt{r^2 - x^2} = r\sqrt{1 - \frac{x^2}{r^2}}$$

It is necessary to proof that the "natural path" of the parabola is always higher than the circle (they are equal to $$r$$ for $$x=0$$).

Expanding the square root in a Taylor series at x = 0:

$$y_{circ} = r\left(1 - \frac{1}{2}\frac{x^2}{r^2} - ...\right)$$

The first 2 terms are the same as the parabola, but follows an infinity of negative terms that makes $$y_{cir} < y_{par}$$

Centripetal acceleration and the normal force.

$$\sum F_r = m a_c$$

with $$F_r$$ being the component of force in the radial direction ($$+$$ sign is for forces towards the center of the circle).

At the peak of the circle, there are two forces affecting the ball: the weight of the ball $$w = mg$$ (gravitational force), the normal force $$N$$ exerted by the track on the ball. The vector sum of these two forces is equal to the centripetal force, $$F_c = m a_c$$.

Plugging in $$a_c = \frac{v^2}{r}$$, you get the following:

$$mg + N = m \frac{v^2}{r} \rightarrow N = m ( \frac{v^2}{r} - g )$$

As you can see, for the ball to stay on the track, the centripetal term must "defeat" gravity in order for the normal force to be positive. Once the normal force turns zero or negative, then the ball will no longer stay on the track and will fall.

If we want to minimise h, then we have to consider that normal reaction at the top of the track is zero.

The reason we can't solve for v=0 at the top, because then normal reaction will become zero even before reaching the top point, and as a result of which, the car will lose the track.

That means there will be some speed at the top of the track too.

As the car approaches the top most point, v will gradually decrease, and therefore Normal reaction will decrease, and the instant it reaches the top, normal reaction will be zero for one instant (so as to minimize h), but car still have speed, which will again push the car to the track, and normal reaction will again be working after that one instant.

Now, as the car again starts to come down, v will increase, and the component of gravity in the direction perpendicular to the motion decreases, but someone has to balance the centrifugal force, as v is increasing and gravity component is decreasing, both implies that normal has to increase for the entire remaining half trip.

This proves that, if in this situation, h is large enough that the car reaches the top of the track, it will always complete the remaining half.

• Just because the car reaches the top of the track does not mean the full circle will be completed. It needs enough kinetic energy at the top such that it stays on the path while going down. Dec 3 '20 at 20:46
• @user256872, I am talking in context of the given situation, if in this situation, h is large enough that the car reaches the top along the vertical half circle, then it will most certainly cover the remaining half too, in ideal situations. Dec 4 '20 at 4:19
• @user256872 "a very important part", your this point is exactly what I have explained in my answer Dec 4 '20 at 4:37
• Saying, "if in this situation, h is large enough that the car reaches the top of the track" is pointless. Because said situation happens assuming $h$ is large enough such that the speed at height $2r$ will be $\sqrt{gr}$, which of course implies that the car reaches the top. By definition of your 'situation,' the car reaches the top of the track. I will however admit that when I first read it, I misunderstood what you meant. I removed the downvote. Dec 4 '20 at 4:53
• @Letsintegreat thank you! your answer helped my intuition. Dec 4 '20 at 15:59