Why is speed constant when centripetal acceleration is constant?

[gionole]

TL;DR Summary

centripetal acceleration

It's actually getting little boring and makes me angry why all the videos/articles show centripetal acceleration formula and presume that speed is constant.

I want to prove backwards, i.e we know the constant perpendicular force acts on an object from the center and why object starts to move in circular motion with constant speed. I'm not looking for intuitive answer with (work, energy and so on), but the proof.

v2 = v1 + v⊥ (vectors)

I think the logic I'm making a mistake is I presume that 𝑣⊥=𝑎𝑑𝑡 This presumes that during 𝑑𝑡 time interval, 𝑎 is constant, hence this gives us > 0 value which means 𝑣⃗ 2>𝑣⃗ 1
After seeing another question, I realize 𝑎 is not constant during 𝑑𝑡 but even if it's not and changing, then we could split 𝑑𝑡 again in even smaller intervals(𝑑𝑡1,𝑑𝑡2,𝑑𝑡3,...) and say that 𝑣⊥=𝑎1𝑑𝑡1+𝑎2𝑑𝑡2+𝑎3𝑑𝑡3+...... This way, 𝑎 is not constant over 𝑑𝑡. but now this should give us 0 and the only way this is possible is if over small 𝑑𝑡 interval, 𝑎 increases, then decreases and overally results in 0. What's the non-intuitive, solid proof that this is what happens?

proofs like this(https://physics.stackexchange.com/a/770948/366606) are horrible to be honest. Just making dt->0, doesn't say anything solid.

Could there be some good proof ? maybe could you draw a picture ? I understand how centripetal acceleration formula is derived, but they assume that speed is constant and thats how they derive with 2 similar triangles, but i'm tired of this. I need to prove speed is constant, not vice versa.

 

[vanhees71]

If you have a particle in uniform circular motion, by definition
$$\vec{x}(t)=a (\cos(\omega t),\sin(\omega t))$$
with $\omega=\text{const}$. Then the velocity is
$$\vec{v}=\dot{\vec{x}}=a \omega (-\sin(\omega t),\cos(\omega t))$$
and the acceleration
$$\vec{a}=\dot{\vec{v}}=-a \omega^2 (\cos(\omega t),\sin(\omega t))=-\omega^2 \vec{x}.$$
As you see, the acceleration is not constant, because it's changing its direction. Only its magnitude is constant.

 

[gionole]

If you have a particle in uniform circular motion, by definition
$$\vec{x}(t)=a (\cos(\omega t),\sin(\omega t))$$
with $\omega=\text{const}$. Then the velocity is
$$\vec{v}=\dot{\vec{x}}=a \omega (-\sin(\omega t),\cos(\omega t))$$
and the acceleration
$$\vec{a}=\dot{\vec{v}}=-a \omega^2 (\cos(\omega t),\sin(\omega t))=-\omega^2 \vec{x}.$$
As you see, the acceleration is not constant, because it's changing its direction. Only its magnitude is constant.

Thanks for the answer. Though, you already start your logic by assuming uniform circular motion. but what I'm asking is what causes uniform motion(such as w = const). I need the proof that what causes uniform speed is actually true.

 

[jbriggs444]

I think the logic I'm making a mistake is I presume that 𝑣⊥=𝑎𝑑𝑡 This presumes that during 𝑑𝑡 time interval, 𝑎 is constant, hence this gives us > 0 value which means 𝑣⃗ 2>𝑣⃗ 1

Yes, if $dt$ were a finite interval during which $a$ is constant in both magnitude and direction, this would mean that the ending velocity would have a larger magnitude than the starting velocity.

But we are talking about a limiting process. If we subdivide the actual curved arc more and more finely so that each individual $dt$ gets smaller and smaller, the cumulative $\Delta v$ computed by summing all of the $dt$'s over a fixed arc length gets smaller and smaller as well.

The limit of the cumulative $\Delta v$ as $dt \to 0$ is zero.There is a different and more idiosyncratic way that I like to think about this conundrum. In deciding that velocity is increasing under the effect of a "radial" acceleration over a bunch of constant $dt$'s, you are running time forward. You are basing your acceleration estimate on the starting velocity for each the interval. Naturally, you would get a result with velocity increasing into the future.

If you ran your simulation backward then you be basing your acceleration estimate on the ending velocity for each interval instead. You would get a result with velocity increasing into the past.

In my mind's eye, I see this as observer bias. An analysist who prefers the future to the past will see an increasing velocity.

An unbiased analysist will see that the two distinct results cannot both match reality. If one reduces the step size in the simulation, one sees that the discrepancy is reduced. Never eliminated, of course, just reduced. The correct physical result is obtained in the mathematical limit of the reduction process.

 

[gionole]

The limit of the cumulative is zero

I don't think this is the proof to be honest. I can say that in v = v0 + at. I will make dt -> 0 here, and even in any motion, i prove that v = v0. Do you like this ? i don't think so.

 

[PeroK]

TL;DR Summary: centripetal acceleration

It's actually getting little boring and makes me angry why all the videos/articles show centripetal acceleration formula and presume that speed is constant.

Could there be some good proof ? maybe could you draw a picture ? I understand how centripetal acceleration formula is derived, but they assume that speed is constant and thats how they derive with 2 similar triangles, but i'm tired of this. I need to prove speed is constant, not vice versa.

If you apply a force to an object in motion, then the component of the force in the direction of the velocity changes the object's speed. And the component of the force perpendicular to the velocity changes it's direction.

In general, therefore, a force of constant magnitude does not result in uniform circular motion. You only get uniform circular motion if the force is always perpendicular to the instantaneous velocity.

 

[PeroK]

PS planetary orbits are elliptical in general with near circular orbits as a special case.

 

[Nugatory]

but what I'm asking is what causes uniform motion(such as w = const). I need the proof that what causes uniform speed is actually true.

To do that, we start with $\vec{F}=m\vec{a}$ where $\vec{F}$ has constant magnitude $F$ and always points towards the center. This is a differential equation that we solve for $\vec{x}(t)$, the trajectory of the moving object given its initial position and velocity. For any given distance from the center there is one particular combination of initial position and velocity we will find that the trajectory is the one given by @vanhees71 above, leading to the constant speed and acceleration conclusion.

 

[sophiecentaur]

Thanks for the answer. Though, you already start your logic by assuming uniform circular motion. but what I'm asking is what causes uniform motion(such as w = const). I need the proof that what causes uniform speed is actually true.

This is only true for uniform circular motion; that sets the conditions for this particular discussion; there is no 'proof' about initial conditions. Only when you have a 'piece of string' to restrain a particle into a circular path will the centripetal force / acceleration have constant magnitude. There are many other systems with non-uniform circular motion; we only learn how to treat them when we have established the easiest situation.

 

[vanhees71]

To do that, we start with $\vec{F}=m\vec{a}$ where $\vec{F}$ has constant magnitude $F$ and always points towards the center. This is a differential equation that we solve for $\vec{x}(t)$, the trajectory of the moving object given its initial position and velocity. For any given distance from the center there is one particular combination of initial position and velocity we will find that the trajectory is the one given by @vanhees71 above, leading to the constant speed and acceleration conclusion.

That's a somewhat more complicated task. Just let's calculate it. It's a central-force problem with a potential,
$$V(\vec{x})=F_0 r \; \Rightarrow \; \vec{F}=-\vec{\nabla} V=-F_0 \vec{x}/r.$$
Then we have energy conservation and conservation of angular momentum:
$$\frac{m}{2} \dot{\vec{x}}^2+F_0 r=E=\text{const}$$
and
$$\vec{x} \times \dot{\vec{x}}=\vec{L}=\text{const}.$$
The latter equation means that the trajectory is a plane curve. Then we introduce polar coordinates in that plane, $x_3=0$:
$$\vec{x}=(R \cos \varphi,R \sin \varphi,0).$$
This implies
$$\dot{\vec{x}}=\dot{R} (\cos \varphi,\sin \varphi,0) + R \dot{\varphi}(-\sin \varphi,\cos \varphi,0)$$
Then
$$E=\frac{m}{2} (\dot{R}^2+R^2 \dot{\varphi}^2)+F_0 R$$
and
$$\vec{L}=m R^2 \dot{\varphi} (0,0,1) \; \Rightarrow \; m R^2 \dot{\varphi}=L=\text{const}.$$
Eliminating $\dot{\varphi}$ from the energy equation gives
$$E=\frac{m}{2} \dot{R}^2 + \frac{L^2}{2m R^2} + F_0 R.$$
This is a separable differential equation for $R(t)$, but the integral looks pretty much like an elliptic integral, i.e., we can't solve this in the general case.

Of course, circles around the center are always solutions for any central force. Then $\dot{R}=0$ and $R=\text{const}$, which implies $\dot{\varphi}=L/(m R^2)=\text{const}$, i.e., this leads to the solution given in the inverse way in #2.

 

[A.T.]

I need to prove speed is constant, not vice versa.

Due to symmetry there is no way decide if it should increase or decrease, so it must be constant.

Or apply mechanical power definition:
P = F dot v
which is 0 when F and v are orthogonal, so no energy input/output, thus KE = const

 

[vanhees71]

As my analysis in #10 shows, a central force with constant magnitude does NOT imply a constant speed and thus not $R=\text{const}$. Of course, there are always special solutions with $R=\text{const}$ in any central-force problem. You can also analyze whether it's a stable solution.

 

[Lnewqban]

I want to prove backwards, i.e we know the constant perpendicular force acts on an object from the center and why object starts to move in circular motion with constant speed.

I believe that the nature and magnitude of the tangential velocity is not as important as the movement of that object.
As per Newton's first law, the natural free movement is always linear.

"A body remains at rest, or in motion at a constant speed in a straight line, unless acted upon by a force."
Therefore, when that natural movement is perturbed by a force, the object resists that deviation from the straight line (centripetal force), which induces that reactive centripetal (towards the instantaneous center of rotation) force.

Whenever we have a perturbing vector force and a mass in movement, an acceleration vector is present.
That acceleration vector can be constant in magnitude (if the angular velocity is constant) or not, but it is always changing direction on the plane of rotation, exactly at the rate of the angular velocity.

I don't know how to do it, but I believe that you can prove that the tangential velocity does not need to be constant for you to have centripetal acceleration.
Whenever you see centripetal acceleration of constant magnitude, you either have constant angular velocity and constant radius of rotation, or one increasing as the other decreases, at such rates that the tangential velocity remains constant.

I understand how centripetal acceleration formula is derived, but they assume that speed is constant and thats how they derive with 2 similar triangles, but i'm tired of this. I need to prove speed is constant, not vice versa.

 

[Mister T]

There is such a thing as non uniform circular motion, you know. Tie a string to a rock and swing it in a vertical circle. The motion of the rock is circular, but it's not uniform. The speed is constantly changing.

When you utter the phrase uniform circular motion you are assuming the speed is constant.

 

[pbuk]

I realize 𝑎 is not constant during 𝑑𝑡 but even if it's not and changing, then we could split 𝑑𝑡 again in even smaller intervals(𝑑𝑡1,𝑑𝑡2,𝑑𝑡3,...) and say that 𝑣⊥=𝑎1𝑑𝑡1+𝑎2𝑑𝑡2+𝑎3𝑑𝑡3+...... This way, 𝑎 is not constant over 𝑑𝑡. but now this should give us 0 and the only way this is possible is if over small 𝑑𝑡 interval, 𝑎 increases, then decreases and overally results in 0. What's the non-intuitive, solid proof that this is what happens?

You need to learn calculus.

 

[vanhees71]

Yes, and it's very easy in this case. See my posting #2.

 

[Svein]

$F=m \cdot a$ is the equation for bodies moving in a straight line. The corresponding equation for bodies in a rotating system is $D=I\cdot\dot{\omega}$ (moment of force equals moment of inertia times angular accelration).

 

[jbriggs444]

$F=m \cdot a$ is the equation for bodies moving in a straight line. The corresponding equation for bodies in a rotating system is $D=I\cdot\dot{\omega}$ (moment of force equals moment of inertia times angular accelration).

Use a bit of caution with that formulation. It works for rigid rotations where $I$ is constant. If you write with slightly more generality then you get a term from the product rule: $D = \frac{dL}{dt} = \frac{d(I(t) \times \omega(t))}{dt} = I \frac{d\omega}{dt} + \omega \frac{dI}{dt}$

Admittedly, that more general formulation is still quite restricted. It applies to bodies that are sufficiently rigid to have an identifiable rotation rate. (e.g. a skater pulling in her arms).

 

[Svein]

Of course. Just as the original linear formula reads $F=\frac{d}{dt}(m\cdot v)$
.

 

[PeroK]

Of course. Just as the original linear formula reads $F=\frac{d}{dt}(m\cdot v)$
.

$m$ is a constant (and not generally time dependent) in Newton's second law.

A particle moving in any trajectory obeys$$\vec F = m\vec a$$

 

[Mister T]

TL;DR Summary: centripetal acceleration

I understand how centripetal acceleration formula is derived, but they assume that speed is constant and thats how they derive with 2 similar triangles, but i'm tired of this. I need to prove speed is constant, not vice versa.

You need to look in some other books. All that's needed is an assumption that the particle has the speed $v$ at some instant. It doesn't have to be constant.

 

[Svein]

Of course. Just as the original linear formula reads $F=\frac{d}{dt}(m\cdot v)$.

In rocket science $\frac{dm}{dt}$ is negative (the fuel is part of the mass).

 

[PeroK]

In rocket science $\frac{dm}{dt}$ is negative (the fuel is part of the mass).

The rocket equation is a special case where a time-dependent mass is considered. The rocket equation is, however, not a generalisation of Newton's second law. See, for example:

https://en.m.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

Note that in the derivation there are two velocity terms. One is the instantaneous velocity of the rocket and the other is the exhaust velocity.

These are often conflated and the erroneous generalisation of Newton's second law is remarkably common.

 

[A.T.]

As my analysis in #10 shows, a central force with constant magnitude does NOT imply a constant speed and thus not $R=\text{const}$. Of course, there are always special solutions with $R=\text{const}$ in any central-force problem. You can also analyze whether it's a stable solution.

If the force is always perpendicular to velocity any force magnitude will give a constant speed.

If the force is towards a fixed point, and merely instantaneously perpendicular to velocity, then any force magnitude will give an instantaneous rate of speed change of zero, but only certain magnitudes will give constant speed over time. For other magnitudes the force simply won't remain perpendicular to velocity anymore.

 

[vanhees71]

This is something different. In #10 I analyzed the claim that a radial force with constant magnitude leads to constant speed, which is wrong.

It's of course different with what you say, because then the force is given as
$$\vec{F}=\vec{v} \times \vec{f},$$
where $\vec{f}$ can be an arbitrary function (of $\vec{x}$, $\vec{v}$, or whatever more), because then
$$\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \vec{v}^2 \right) = \vec{v} \cdot \vec{F}=\vec{0} \; \Rightarrow \; \vec{v}^2=\text{const}.$$
An example is a charged particle in an arbitrary magnetic field,
$$\vec{F}=q \vec{v} \times \vec{B}.$$