# B Centripetal force and spinning discs

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1. Jul 13, 2016

### Frank Castle

Apologies for a possibly very basic question. I was recently asked by someone to explain to them the answer to the following classical mechanics (uniform circular motion) problem:

Consider two discs of different radii, attached to one another such that they are concentric, and each containing a rim around its outer edge (such that a ball can fit within the width of the rim). The discs are attached to a spin wheel such that one can freely spin them around, however, since they are attached to one another, they do not move independently.

Suppose now that one has two balls (of identical size and mass) and one places one in the rim of the outer disc (with largest radius) and the other one in the rim of the inner disc (with smallest radius), then one starts the discs spinning. We suppose that the rotation of the discs is fast enough such that the rim of each disc provides a strong enough centripetal force to keep each of the balls following a circular path (each remaining pushed against the rim of of their respective disc).

Since the discs are fixed to one another, such that neither of their motion is independent of one another, they spin with the same angular velocity.

The question is then, as the rotation of the discs starts to slow down, which ball will fall towards the bottom of the disc (and onto the floor) first and why?

(** If it helps, I've attached an image at the bottom of this post to detail the problem pictorially (apologies for the drawing skills! **)

My answer to this was that the ball in the rim of the inner disc (with smaller radius) will fall towards the floor first, my reasoning being that the smaller the radius, the greater the tangential velocity of the ball will need to be to maintain a circular path and overcome gravity (analogising to the case of satellite orbit around a planet). Hence, since both discs have the same angular velocity, and the inner disc has a smaller radius, the tangential velocity of the ball in the rim of the inner disc will be smaller and so as the angular velocity decreases (equivalently, the rotation of the disc slows down), the ball in the rim of the inner disc will fall towards the ground first.

I'm fairly sure that this is the correct answer, but the person I was questioned by wasn't satisfied, and furthermore, wanted to see the maths behind the argument.

I have to admit, I'm unsure whether I've done this part correctly. I assumed that it would be easier to work in the rotating reference frame of the disc for the analysis. This being the case, with respect to this reference frame, the ball is at rest and the only force acting on it is gravity, and the (fictitious) centrifugal force, hence we have, $$\mathbf{F}_{gravity} - m_{ball}\mathbf{a}_{centripetal}=\mathbf{F}_{gravity}+\mathbf{F}_{centrifugal}=\mathbf{0}$$ (where $\mathbf{a}_{centripetal}=\frac{v_{ball}^{2}}{r_{disc}}\hat{\mathbf{r}}$). Thus we require that the magnitude of these two forces be equal $$\frac{GM_{Earth}m_{ball}}{R^{2}}=\frac{m_{ball}v_{ball}^{2}}{R}\Rightarrow v_{ball}^{2}=\frac{GM_{Earth}}{R}$$ This agrees with my answer above. However, the bit I'm sure isn't quite correct, is that actually the radius in the centrifugal force equation should be the radius of the disc, whereas the radius in the gravitational force equation should be that of the Earth.

Last edited: Jul 13, 2016
2. Jul 13, 2016

### CWatters

You are correct about your mistake. You seem to have turned rdisc into R and then cancelled an R each side.

If r is the radius of the disc and R the radius of the earth it should be

vball2 = GMearthr/R2

or you could use..

g = GMearth / R2

to simplify.

3. Jul 13, 2016

### Aniruddha@94

Wouldn't it be easier to use the condition that at the highest point in the rotation, the centrifugal force must be greater than ( or equal, in the limiting case) to the force of gravity. This is in the lab frame of reference. $mr\omega^2\geq mg$
And since $mg$ and $\omega$ are equal in both the cases, as we reduce $\omega$, the ball in the smaller disc falls first.

4. Jul 13, 2016

### Frank Castle

Yes, good point (so obvious, I feel really stupid!)

Is the rest of the explanation that I wrote correct at all? (What confuses me is that the maths seems to imply that the ball at the smaller radius will have a smaller tangential velocity than the ball at the larger radius?!

That is true, but in the lab frame of reference there is no centrifugal force, right (only the true centripetal force). Is the point that, in order to get circular motion in the first place, the centripetal force has to be greater than or equal to the force of gravity acting on the balls?

5. Jul 13, 2016

### CWatters

Yes that's correct. If they share a common axle they take the same time to make one revolution. A point on the larger disc has further to go in that time than the small one because the circumference is greater.

Will have to answer your last q later - got to take the kids to rugby training sorry.

6. Jul 13, 2016

### Frank Castle

Ok cool, that's what I thought.

Sure, no worries. Appreciate you taking the look! It's been bugging me for a while, so it would be great to get it properly sorted in my head.

7. Jul 13, 2016

### CWatters

The problem with that description is that at the top gravity provides some or all of the centripetal force. In this case I think it's better to say..

"To get circular motion gravity must provide less than or equal to the required centripetal force."

If a ball was fired from a canon horizontally it would (initially) follow a curved path due to gravity. In other words gravity provides the centripetal force needed for the ball to move with an initial radius Rgravity. If the ball was constrained by a wheel or rope to follow a tighter/smaller radius eg Rsmall < Rgravity then gravity doesn't provide enough centripetal force. The wheel or rope must provide the short fall. If the wheel has a larger radius Rlarge then gravity provides too much centripetal force and the two paths Rlarge and Rgravity diverge - the ball falls off. When the radius is "just right" (so the ball doesn't quite fall off) gravity is providing just enough centripetal force and the wheel or rope is providing none.

8. Jul 13, 2016

### Aniruddha@94

Not lab frame.. It should've been in the frame of the ball. Sorry for the mistake

9. Jul 14, 2016

### Frank Castle

Good point.

I've thought about the problem a bit further and I think I may have figured out a solution which is hopefully correct.

(In the following we work in the inertial lab reference frame).
If we take the extreme case in which the ball is at the top of the disc (analogous to the case of a rollercoaster going around a loop and being at the top of the loop). In this case, assuming that the tangential velocity of the ball is just enough to maintain circular motion, the normal force (provided by the rim on the disc) will be zero (since the ball will only just be touching the rim of the disc) and so the only force (neglecting air resistance, etc.) acting on the ball is gravity, and since we have stated that the ball (only just) maintains circular motion, the gravitational force must act as the centripetal force causing the ball to continue along its circular path around the rim of the disc. In doing so we can imply what the minimum tangential velocity of the ball must be (or equivalently a minimum angular velocity) in order for it to maintain a circular path. Since the gravitational force is providing the centripetal force at the top of the disc, we must have that $$\frac{GM_{Earth}m_{ball}}{R^{2}}=\frac{m_{ball}v_{ball}^{2}}{r}\Rightarrow v_{ball}^{2}=\frac{GM_{Earth}}{R^{2}}r=gr$$ (where $g=\frac{GM_{Earth}}{R^{2}}$)
Thus, in order for the ball to maintain circular motion around the disc, it's tangential velocity must satisfy $$v_{ball}\geq \sqrt{gr}$$ which is the result we arrived at before.
The problem I still have, is that I know intuitively that the ball in the inner disc requires a greater tangential velocity to maintain circular motion (instead of falling to the floor under the influence of gravity), but this is not what the formula above is implying?!

My thoughts on this are that it is more helpful to analyse the angular velocity. Indeed, for uniform circular motion, it is always true that $v=\omega r$, where $\omega$ is the angular velocity (or speed, to be technically precise). Hence, in our case, $$\sqrt{gr}\leq\omega r\Rightarrow\omega\geq\sqrt{\frac{g}{r}}$$ and so we see that the inner (smaller) disc requires a greater angular velocity than the outer (larger) disc in order for the ball to maintain circular motion (and not fall to the ground). Since, at each instant in time, both inner and outer disc have the same angular velocity, we see that as the angular velocity decreases, the ball in the rim of the inner disc will fall to the ground first (since the ball in the rim of the outer disc requires a smaller angular velocity to maintain circular motion).

This explanation seems correct to me, but I maybe wrong?!

Last edited: Jul 14, 2016
10. Jul 14, 2016

### CWatters

Centripetal force acts towards the centre of rotation. The equation mv2/r gives you a figure for the required centripetal force. It isn't telling you how much centripetal force is actually available.

In order for the ball to fall to the floor it's radius of travel must become even smaller. That normally means the centripetal force must increase which it cannot do in this situation because the actual centripetal force is provided by gravity which is constant.

To make the ball fall the required centripetal force must be reduced so that gravity is sufficient. That's why it falls when the ball is going slower.

11. Jul 14, 2016

### Frank Castle

By this do you mean that $v^{2}=gr$ must be satisfied in order for there to be a sufficient centripetal force to prevent the ball from falling to the ground as it moves round the top of the disc? If so, this makes sense, because then as the ball slows down (due to the decrease in angular velocity of the disc) this condition will no longer be satisfied (since $gr$ is a constant) and so the ball will fall to the ground. Given this, is the reason why the ball in the disc with the larger radius stays in circular motion for longer because the tangential velocity of the ball increases with radius (since $v=\omega r$) and so, as the angular velocity of the disc is decreasing, the ball at the larger radius will have a greater velocity and so will satisfy the condition $v^{2}=gr$ for longer before eventually (sometime after the ball at the smaller radius has) fall to the ground?!

Is any of what I put at the end of my last post (#9) about angular velocity correct at all? i.e. is any of this correct:

It seems to make sense to me, since $\omega$ is the same for both discs. Let $v_{1}=\omega r_{1}$ be the speed of the ball in the smaller disc of radius $r_{1}$ and $v_{2}=\omega r_{2}$ be the speed of the ball in the larger disc of radius $r_{2}$. Clearly $v_{2}>v_{1}$ since $r_{2}>r_{1}$. The condition for circular motion to be maintained is $v=\omega r\geq\sqrt{gr}\Rightarrow\omega\geq\sqrt{\frac{g}{r}}$ and so we have, for each ball to maintain circular motion, the following inequalities must be satisfied $$\omega\geq\sqrt{\frac{g}{r_{1}}}\quad\text{and}\quad \omega\geq\sqrt{\frac{g}{r_{2}}}$$ for the disc $r_{1}$ and $r_{2}$, respectively. From this we see that the smaller disc has to maintain a higher angular velocity in order for the ball to stay in circular motion. Since $\omega$ is the same for both discs, as it decreases, the condition $\omega\geq\sqrt{\frac{g}{r_{1}}}$ will be the first to be violated (since $g$ and $r_{1}$ are both fixed constants) and hence the ball in the inner disc will fall to the ground before the ball in the outer disc.

Last edited: Jul 14, 2016
12. Jul 14, 2016

### CWatters

Unfortunately not. You still seem to think that it's the velocity that produces the centripetal force. That's not correct. At the top centripetal force acts downwards and is produced by gravity. So it doesn't make sense to say that there is "sufficient centripetal force to prevent the ball from falling to the ground". Centripetal force is the force trying to pull it down. Inertia keeps the ball up.

To stay up..

the available centripetal force (gm) must be less than the required centripetal force (mv2/r)
so
gm < mv2/r
or
g < v2/r

If you want to think about it from a velocity perspective you can rearrange the inequality...

To stay up..

v2/r > g
or
v > √gr

The centripetal force required for a smaller radius is greater than the centripetal force gravity is able to provide.

More on angular velocity in a moment.

13. Jul 14, 2016

### Frank Castle

Apologies, very bad wording on my part. I am fully aware that it's not the velocity that produces the centripetal force - indeed, at the top of the disc, in the extreme case, it is purely gravity that is providing the centripetal force acting on the ball. What I'm confused about is how to argue that the ball at the smaller radius will fall towards the ground first (as the rotation of the disc slows down). I don't see how $v>\sqrt{gr}$ implies that the centripetal force required for the smaller radius is greater than the centripetal force that gravity can provide? If the rotation of the disc is fast enough then surely this inequality will be satisfied, since the tangential velocity of the ball will be great enough to "keep it up" as it passes over round the top of the disc.

14. Jul 14, 2016

### CWatters

I had to think about that but yes I believe that's correct.

15. Jul 14, 2016

### Frank Castle

Thanks for taking a look. For some reason it makes more sense to me looking at it from this perspective. I don't know why, but I can't see from the inequality $v\geq\sqrt{gr}$ that gravity is unable to provide the centripetal force required to keep the ball in circular motion around the disc. Surely this depends on the angular velocity of the disc, since this provides the ball with a tangential velocity and so the greater the angular velocity of the disc the greater the tangential velocity of the ball, so at least at some point the inequality $v\geq\sqrt{gr}$ will be satisfied. It is only when the angular velocity of the disc decreases, and hence the tangential velocity of the ball decreases, that this inequality starts to be violated. I think the point is that the ball in the at the smaller radius will have a smaller velocity to start with and so this inequality will be violated earlier than it will be for the ball at the larger radius, right?
Apologies for my ramblings, but I'd really like to get this sorted in my head!

16. Jul 14, 2016

### CWatters

I think you are correct. Because the two discs are linked and share a common angular velocity it makes sense to take the angular velocity approach.

17. Jul 14, 2016

### Frank Castle

Ok. Would you say that from what I've written about that, that I've understood the concept correctly?

From the perspective of velocities, how should one correctly interpret the inequality $v\geq\sqrt{gr}$?
From the derivation earlier, it seems to me that $v=\sqrt{gr}$ quantitatively describes the minimum velocity that the ball must possess in order to make it round the top of the disc without being "pulled" to the floor by gravity (since gravity is the only force providing the centripetal force at the top of the disc). Since the velocity scales proportionally to the radius, then the ball in the of the larger disc will have a larger velocity (and will experience a larger centripetal force than the ball in the rim of the smaller disc, since $F=mv^{2}/r=mr\omega^{2}$). Therefore, as the angular velocity of the discs decreases, the velocities of the balls will decrease proportionally, however the ball in the smaller disc will violate the condition $v=\sqrt{gr}$ first, since it has a smaller velocity than the ball in the larger disc (due to them sharing a common angular velocity), and so it will fall to the ground first.

Last edited: Jul 14, 2016
18. Jul 14, 2016

### Frank Castle

I have a feeling that I've already understood the situation, but I'm now over thinking it and confusing myself?!