# How can a particle in circular motion about a fixed point accelerate, if the point doesn't too?

When a particle is performing uniform circular motion attached to a string about a fixed centre, at any instant of time its acceleration is directed towards the centre but the centre has no acceleration. But I was taught in school this is not possible because of the string constraint:

The accelerations of the ends of a string are the same if the string is not slack.

Where am I wrong?

• What do you mean by string constraint? – ayc Nov 2 '18 at 10:18
• I was taught in school that accelerations of both the ends along the string is same if the string is not slacked a.k.a the string constraint. – Harsh Somani Nov 2 '18 at 10:30
• I think your problem may be the assumption that the center has no acceleration. Tie a weight to a string and twirl it around your hand. You will observe that your hand accelerates. – jamesqf Nov 2 '18 at 17:15
• Acceleration is the wrong word here. Force is conserved; acceleration is not (except in the specific circumstance that both objects have equal mass and have no other forces acting on them). – Brilliand Nov 2 '18 at 19:20
• @HarshSomani it would be very useful to mention that in the question. – David Z Nov 3 '18 at 9:48

I suspect what you were actually taught is either that the velocity component along the string is the same on both ends, or that the force / tension is the same. One might think that both of these imply also equal acceleration component, by way of simple $$f(x) = g(x) \Rightarrow f'(x) = g'(x)$$ consideration or via Newton's second law, respectively. But actually neither is the case!

To take the derivative of these velocities, you need to take into account that the direction changes. \begin{align} \mathbf{v}_0(t) =& 0 \\ \mathbf{v}_1(t) =& r\cdot\omega\cdot\begin{pmatrix}-\sin(\omega\cdot t)\\\cos(\omega\cdot t)\end{pmatrix} \end{align} The radial component of $$\mathbf{v}_1$$ is indeed always zero $$\mathbf{v}_1(t)\cdot\mathbf{e}_{\mathrm{r}} = r\cdot\omega\cdot\Bigl(-\cos(\omega\cdot t)\cdot\sin(\omega\cdot t) + \sin(\omega\cdot t)\cdot\cos(\omega\cdot t)\Bigr) = 0$$

but the radial component of the acceleration is nonzero. You need to first take the derivative of the velocity vector, then project it into the radial direction, not take the derivative of the radial velocity only. $$\frac{\mathrm{d}\mathbf{v}_1}{\mathrm{d}t} = r\cdot\omega^2\cdot\begin{pmatrix}-\cos(\omega\cdot t)\\-\sin(\omega\cdot t)\end{pmatrix} = -r\cdot\omega^2\cdot \mathbf{e}_{\mathrm{r}}.$$ With Newton's law you can't argue because this assumes masses. For a true circular motion around a center, you need an infinitely massive central anchor and a finite-mass satellite. The center has then virtually zero acceleration despite handling a significant force. Not so for the satellite.

I'm guessing you mean the string constraint that Tension must be equal in both directions at all points in the string except the endpoints, where the tension at the endpoints must be equal and opposite?

So for an object moving in circular motion around a fixed point attached to a string, you're right that the object is moving in a circle because of the tension from the rope giving centripetal force. I think your confusion is coming from, shouldn't the center point also feel a tension and thus accelerate?

So the answer comes from the definition of a "fixed point"! In real life this means nailing something to the ground, or gluing it down, or placing it between a rock and a hard place, etc. This means that the center will indeed feel tension, but it will also feel some resistive force (usually normal or frictional forces) that will keep it from accelerating.

If the center point was not "fixed", then the circular motion would immediately stop, the string would go immediately slack, and the problem would become much more complex.

• actually I meant the accelerations of both the points along the string should be same by spring constraint – Harsh Somani Nov 2 '18 at 10:32
• I think I'm a little confused by your comment. "Strings" in the conventional sense don't obey Hookean physics, i.e. the tension in a rope is not given by $F = -k(x - x_0)$. Indeed, this makes sense because Tension is not a conservative force, and hookean forces are the product of a derivative of a potential, which Tension is not. – CuriousHegemon Nov 2 '18 at 10:33
• The string constraint is the condition that the length of an ideal string is constant. For solving problems, the string constraint translates to Sum of acceleration for all ends of a system of strings is zero. – Harsh Somani Nov 2 '18 at 10:36
• @HarshSomani I think I see what you mean. So the spring constraint just means that the forces on the ends are the same. But that doesn't mean that the net force at the center is equal to the tension! There is always some normal or frictional force canceling out the pull of the tension at the center when we are looking at central motion. – CuriousHegemon Nov 2 '18 at 11:04
• @NeilShah Be aware that the OP keeps switching between "string constraint" and "spring constraint". I think they mean the former. – BioPhysicist Nov 2 '18 at 11:21

To distill (previous answers are correct, but perhaps unnecessarily long if I understand your point of confusion):

For a rigid system moving without rotation, all points in the system move with the same acceleration.

In any other case (i.e. if there is any change in orientation of the system) this is no longer true.

accelerations of both the ends along the string is same if the string is not slacked

If you have a string placed in the shape of s in vacuum and if you start pulling it from one end it finally becomes l i.e straight.Here you can say that string isn't slackened because acceleration of both ends in same.

In case of circular motion i.e particle rotating about a fixed centre the string provides the necessary force to keep the particle moving around the centre and this force is called as tension.

Now,since string isn't slackened, is the acceleration of both ends same?

I want to explain what's happening in terms of forces rather than acceleration:

Newton's third law of action and reaction states that if the string exerts an inward centripetal force on the particle, the particle will exert an equal but outward reaction upon the string,the reactive centrifugal force.

The string transmits the reactive centrifugal force from the particle to the centre, pulling upon the centre. Again according to Newton's third law, the centre exerts a reaction upon the string, pulling upon the string. The two forces upon the string are equal and opposite, exerting no net force upon the string (assuming that the string is massless), but placing the string under tension i.e no slackening of the string.

The reason the centre appears to be "immovable"(not accelerating) is because it is fixed. If the rotating ball was tethered to the mast of a boat, for example, the boat mast and ball would both experience rotation about a central point.

I was taught in school that accelerations of both the ends along the string is same if the string is not slacked a.k.a the string constraint.

A lot of good answers already, but I will start by discussing what context you heard this in.

This was probably told to you in the scenario of two objects connected by a string, and then you pull horizontally on one object to pull the entire system. In this case you are right. If the string is not slacked, then the accelerations of each end of the string must be the same. At this point the string acts like a massless rigid body. Since the whole system will have a single acceleration, it must be that all points (not just the ends) of the string have the same acceleration. If this were not the case, then the string would either be stretching at some points, or folding in on itself at some points.

Now, in the case of the rotating object attached to a string, you actually have a bigger "issue" than you realize. Technically all points along the string have a different acceleration! This is because, assuming a constant angular velocity $$\omega=v/r$$, $$a=\frac{v^2}{r}=\omega^2r$$

So if the heart of your question is asking why the "string constraint" doesn't apply here, you should be looking at all points along the string. Not just the end in the center.

The reason this occurs is because as you move closer to the center of the circle, the linear velocity becomes smaller and smaller. Since acceleration is change in velocity, this means that the acceleration will also become smaller. At the center of the circle, the velocity is constant ($$0$$), so the acceleration must be $$0$$ as well.

So the resolution to your question really is just that the "string constraint" is not a true constraint in this system (unless you find a way to reword it to be more general). What you learned in school was probably just said in the context of a particular problem, and was not meant to be a generalization to how strings behave in all contexts.

I think what you're asking is "How can a particle accelerate towards a point without ever getting closer?"

Acceleration is "change in Velocity," and Velocity is the combination of speed and direction. So acceleration can mean "a change in speed", "a change in direction", or a combination of the two. In order for any object to travel in a circle, it must change direction; it must accelerate.

The particle on the string is in a kind of equilibrium caused by the string. If it didn't accelerate enough toward the point, its current velocity would move it away from the point; if it accelerated too much, it would get closer. But the string causes the particle to remain at a constant distance, causing just enough acceleration to maintain the orbit.

The force at both ends of the string is the same, not the acceleration. Your rotating object is obviously accelerated, making it move in a circle. So what's with the force on the other end of the string? It's "fixed", which means that in physical reality it is tied in one way or another to planet Earth, and the force acts on planet Earth, partly making its rotation wobble, partly making its orbit wobble. It's just that Earth has some non-trivial mass and some non-trivial rotational inertia, so you are not likely to see much of a measurable effect unless we are talking about some comparatively big particle like the moon (which is not tied by a piece of string but rather by gravity, for which strings are pretty glad).