# Forces acting on the water in a bucket when the bucket is vertically spun on a string?

From my understanding, the only force acting on the water is its own weight. The tension in the string is pulling on the bucket but not the water, right?

However, mathematically that does not seem to be the case. If we assume (this comes from a question and this was the information given) that:

• The circular motion is uniform
• The angular speed ($\omega$) is 5 rad s$^{-1}$ (for both the water and the bucket, since both are turning through the same angle per second)
• The radius of the circle ($r$) is 1m

Then:

Let the mass of the water be $m_w$ and the centripetal force on the water be $F_w$.

$$F_w = m_w \ \omega^2 \ r$$ $$F_w = 25 \ m_w$$

This means that there must be one or more forces acting on the water who's net downward component, when the bucket is vertically above the centre of rotation, is $m_w (25 - 9.81)$. What force am I missing?

• Note: This is not a homework question, so I have not used the homework tag. It is a question on scientific understanding using a question as the framework with which to ask it. – Pancake_Senpai Jul 11 '17 at 17:54
• Have you considered the fact that the water is in a container? There then has to be some force on the water to keep it in that shape, and not falling to the earth. – mphy Jul 11 '17 at 18:07
• Would this force increase as the angular speed of the water increases? – Pancake_Senpai Jul 11 '17 at 18:49
• Have you tried swinging anything in a vertical circle on the end of a string at a constant speed? Can't be done. I wish people would stop asking this question. – M. Enns Jul 11 '17 at 19:06
• Hi Pancake_Senpai. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. – Qmechanic Jul 11 '17 at 19:33

I'm going to start with a calculus-heavy explanation and then transition into a lighter one, if you don't mind.

# What does centripetal mean? (feat. Calculus)

So it's not 100% right to talk about "centripetal force" as if it were some sort of special force that can be added to other forces. The right way to think about it is to break it up into its constituent words: centripetal is not a title, but it just means "pointing towards the center of motion", and any force which can cause circular motion can hypothetically be centripetal.

In other words, in order to travel in a circle of radius $R$ with some angle $\theta(t)$ with time derivatives $\dot \theta$ and $\ddot \theta$ your position vector has to be $$\vec r(t) = [x(t), y(t)]= R~[\cos\theta(t),\sin\theta(t)]$$ and therefore your velocity has to be (one derivative wrt. time) $$\vec v = \frac{\partial \vec r}{\partial t} = R~\dot\theta~[-\sin\theta, \cos\theta]$$and therefore by the product rule your acceleration has to be $$\vec a = \frac{\partial^2\vec r}{\partial t^2} = R~ \ddot \theta~ [-\sin\theta, \cos\theta] - R ~\dot\theta^2 ~[\cos\theta, \sin\theta],$$ and finally therefore the sum of all of the forces incident on you must be:$$\sum_i \vec F_i = m \vec a = m~R~ \ddot \theta~ [-\sin\theta, \cos\theta] - m~R ~\dot\theta^2 ~[\cos\theta, \sin\theta].$$ This has a centripetal or radial component, given by the second term, and a tangential or angular component, given by the first term. For a sum-of-forces to keep you moving in a uniform circle, $\theta = \omega~t+\theta_0$ and then $\ddot \theta = 0$ and the first term goes away; if your circle is vertical rather than horizontal then generally your bucket will be going faster near the bottom than the top and so this first term also matters -- but it's not what you're asking about.

# How does this apply to what you're talking about?

Now at the very top of the loop, the centripetal component of the force comes from both the tension on the rope and the gravity of the water itself. You can imagine getting rid of the first part by spinning it less and less fast until the bucket just barely becomes weightless and the rope just-so-barely slackens at the top of the loop. The equation tells us that this will be when $-m R \dot \theta^2 = -m g,$ so the instantaneous angular velocity $\dot\theta$ will be $\sqrt{g/R}$ measured in radians per unit time. If you try to do the loop faster, you will notice that you're pulling on the bucket harder and harder, because then $m~g + T = m R \dot\theta^2$ where $T$ is your tension force.

If you tried to do it slower, then you would run into a problem with ropes: they are very good under tension, lousy under compression, because they're not rigid. So they can't support a negative tension, and the bucket will fall away from that circular trajectory.

# Less calculus: think about free-fall parabolas.

The above is all about a lot of calculus; how can I explain this with only a minimum of calculus?

One way to think of this is to think of the parabolas that things make as they fall, versus the graph of the circle. A normal object in free-fall describes the path $$[x, y] = [v_x~t+x_0, -\frac12 g~t^2+v_y~t+y_0].$$ It achieves its maximum height at the time $t_\text{max} = v_y/g.$ It describes therefore a parabola with maximum height $y_\text{max} = y(t_\text{max}) = -\frac12 g~(v_y/g)^2 + v_y~(v_y/g) + y_0 = y_0 + \frac12 v_y^2/g,$ which should both be familiar if you think about its energy $\frac12 m v_x^2 + \frac12 m (v_y - g t)^2 + m g y$ and what must happen when all of that second term of energy gets absorbed into the third term, the first being quite constant.

Let's choose our coordinates so that $y_0 = 0$ and $x_0 = -v_x~t_\text{max}$ so that at its maximum it is at $x=0.$ This equation simplifies to: \begin{align}x(t) =&~ v_x~(t - t_\text{max})\\ y(t) =& -\frac12 g~(t - t_\text{max})^2 + y_\text{max}\end{align}and this you can see by substitution describes the parabola $$y = y_\text{max} - \frac12 \frac{g}{v_x^2}~x^2.$$

Now, a circle whose center is at height $h$ and which has radius $R$ obeys a different equation, $$x^2 + (y - h)^2 = R^2,$$ but it is not too different, because we can write $$y = h \pm R~\sqrt{1 - \left(\frac x R\right)^2}.$$ The $+$ is for the top half of the trajectory; the $-$ is for the bottom half.

Now for very small $x$, at the very tippy-top of the trajectory, we can make an approximation. This approximation is that for very small $u$ the graph of $\sqrt{1 - u}$ is approximated by the line $1 - \frac12~u.$ (For proof, just look at the graph.) Of course I got this approximation by calculus, but you don't need to know how I know this in order to also know it.

So actually circular motion, at the very tippy-top, looks like the parabola $$y=h + R\left(1 - \frac12 \left(\frac xR\right)^2\right) = h + R - \frac{x^2}{2R}.$$So the first "no surprise there" thing should be that clearly these two equations are similar in the sense that $y_\text{max} = h + R.$ Sure, if the center is at height $h$ and the radius of the circle is $R$ then the highest it goes is $h + R.$ So let's stipulate that we choose $h$ such that this is true.

## Comparing parabolas

Now overlap these three graphs on top of each other. In this graph I am showing you a circle ($h=2, R=1$) and a parabola which matches the circle up near the tippy-top, and also a parabola inside that one, which let's imagine describes something that's falling naturally -- that is its trajectory.

Now if I ask you, "can a rope of that radius from that center turn that falling trajectory into a circular one?" you might be able to see that the answer is heck no. What's the problem? The problem is, the thing is not moving fast enough at the top! It falls inside the circle, falling way too fast compared to how it is going around the circle, so your rope will just slacken immediately. To make that thing go in that circle, you would have to "push it out" towards the circle--but ropes don't bear compression loads, only tensions. (In fact this is basically what prevents you from doing a full circle around a swingset, is that you start to describe one of these freefall parabolas off of the circle, and the chain holding you on the swing goes slack. As the chain snaps back once you're done falling, it loses some of the energy, more as you go higher and higher, until either you go over the top successfully or the energy it steals is equals the amount of energy you're putting in on each swing.)

Now suppose the parabola fits outside the circle: that's shown in this graph. If that parabola sits above the circle's approximating parabola, then yes, the rope can certainly pull that particle in and complete the circular trajectory. And if they perfectly overlap? Then that's the case where the tension is $0$ right at the very top of the circle. I said above that this happens when $g= R\dot\theta^2,$ but remember that I also said the velocity is $v = R\dot\theta$ and at the top of the trajectory that is purely in the x direction, so we could say $g = v_x^2/R$ at the top. And what do you get when you equate those two parabolas that we got without calculus? You get the same equation: $g = v_x^2/R.$

## Wait, that's only talking about the bucket, right?

Well, no. So the key here is that we're talking about anything which is freely falling, so if your bucket is full of sand then we can imagine each sand particle freely falling. If you were to use a stiff rod of radius $R$ rather than a rope to swing the bucket-full-of-sand, and you follow that first graph where the bucket is not moving fast enough at the top of the trajectory, then the bucket will follow the circle but the sand will follow the inner parabola and dump on your head. It's the same for water, with only minor changes because the water sticks to itself and the bucket a little.

But if the water/sand is going fast enough at the top that it wants to occupy a parabola outside of the circle that you're trying to swing it in, then it will crash into the bucket that's holding it (or other grains of sand which are crashing into the bucket -- same difference) as it tries to go around that circle, and the bucket will therefore hold it in. And that force has to come from somewhere and of course it comes from the tension you put on the rope/rod.

• Thank you very much for that detailed answer. It's 11pm here, so I'll read through it properly tomorrow morning, and mark it as accepted once I've done so. – Pancake_Senpai Jul 11 '17 at 22:00