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While analyzing the case of a force and consequently an acceleration acting perpendicular to the velocity of a given body, I do understand that force's component along the velocity will be 0 causing no change in speed, i.e. magnitude of the velocity (along initial velocity), but will it not add a component in the direction along the force and perpendicular to the velocity.

Even if we assume that at the given instant, the change caused by the acceleration along it does not contribute significantly and magnitude effectively remains the same, how can we state that over a larger span, it wont contribute anything.

Also, if we're neglecting the instantaneous change in magnitude, why then are we accounting the change in direction?

I've tried to search about it but could'nt find a satisfactory answer. Pls correct if I'm thinking anything incorrect

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    $\begingroup$ It's a good thing it doesn't, or else the earth would be orbiting wayyy too fast $\endgroup$
    – Señor O
    Commented Nov 16, 2023 at 21:40
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    $\begingroup$ In my mind at least there is an intuitive leap that is wrong... there is the idea that velocity vectors add, and force vectors add, so adding a force vector perpendicular to a velocity vector must somehow create a perpendicular velocity that adds to the original... $\endgroup$
    – Michael
    Commented Nov 17, 2023 at 12:26

9 Answers 9

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It is true that a force perpendicular to the velocity will change only the direction of the velocity. It's key to note that as the direction of the velocity changes, so too must the direction of the applied force in order to remain perpendicular to it.

At a particular instant, applying a perpendicular force causes the direction to change. If you keep applying that same force for any longer than an instant, you're now applying a force with component in the direction of velocity, resulting in a change in speed. We require that the perpendicular force change direction along with the velocity's direction - swinging a tennis ball on a string, for example, the direction of the tension force changes continuously along with the direction of the ball's velocity, and the ball moves in a circle at a constant speed. As a contrasting case, imagine a non-rotating spaceship that is moving forward and fires a lateral thruster along its center of mass - in this case, the spaceship does pick up speed, because after the first instant, the thrust vector is no longer perpendicular to velocity (the ship keeps its forward speed and adds lateral speed).

You agree that at any one instant, a perpendicular force has no component in the direction of velocity, and therefore can't cause a change in speed. It's critical to recognize that this is true at every time point, because we have explicitly defined the force to be perpendicular to velocity. If there is any force component in the direction of velocity that results in a change in speed, we have violated the premise that the force was perpendicular. It doesn't matter how long of a timespan we consider, as we have defined the applied force in a way that cannot affect speed - its effect on speed is zero at all time points, so it cannot have any effect on speed no matter the length of time.

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    $\begingroup$ It's key to note that as the direction of the velocity changes, so too must the direction of the applied force in order to remain perpendicular to it. I feel like this is the simple line that the other answer is missing, and honestly all you need to answer this question $\endgroup$
    – Señor O
    Commented Nov 16, 2023 at 21:40
  • $\begingroup$ As a non-physics person, I read this question and thought: what? of course acceleration increases! It was only when I considered the "key to note" bit here that the understanding occurred. Now I have to wonder why my physics instructor didn't mention that with more emphasis... $\endgroup$
    – CGCampbell
    Commented Nov 17, 2023 at 11:01
  • $\begingroup$ @CGCampbell When I was in high school I was confused about OP's exact problem. Similar to you it turned out I just didn't understand that the force was always perpendicular to velocity, and if my teacher just emphasized that then I wouldn't be confused, so why didn't he? Well now years later it just seems obvious that force is perpendicular to velocity at all times, and if I didn't have that prior experience I wouldn't think it would be a point of confusion when explaining to others. So I think its just not always clear what needs emphasis when teaching. $\endgroup$
    – Er Jio
    Commented Nov 17, 2023 at 15:19
  • $\begingroup$ I'm guessing that the OP's confusion comes from less-than-clear communication of what the problem is, not from errors in getting a solution. Here we want a force that stays perpendicular to the velocity - and once that's clear there is no confusion. Physics is great and important and stuff... and communicating well with other humans about it is where most of our actual work happens :-) $\endgroup$
    – stochastic
    Commented Nov 18, 2023 at 4:23
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We can use a work/energy argument to deduce that the object's speed (the magnitude of its velocity) will remain constant, as follows.

A force $\vec F$ that acts on an object does work $\int \vec F . d\vec {s}$ on the object where $d\vec {s}$ is the displacement of the object. However, if $\vec F$ is always perpendicular to the object's velocity $\vec v = \frac {d\vec {s}} {dt}$ then $\vec F . d\vec {s}$ is always $0$, and so the force $F$ does no work on the object. Since no work is done on the object (assuming there are no other forces acting on the object), the kinetic energy of the object $\frac 1 2 m |\vec v|^2$ must remain constant and so its speed $|\vec v|$ must remain constant.

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Note the following. Let $v$ be the modulus of the velocity and $\vec{F}$ a force perpendicular to it. Now, $$\frac{d}{dt}v^2=2\vec{v}\cdot \vec{a}=2\vec{v}\cdot \frac{\vec{F}}{m}$$ So you see clearly that algebraic considerations lead you to conclude that if the force is perpendicular to the velocity, the modulus of the velocity is constant. But why? Imagine the body making a slow turn under such a force. The force indeed changes the velocity along its direction, but the force also changes to accomodate the turning of the velocity in order to keep being perpendicular to it, in such a way that the length of the resulting vector is unchanged, and just its direction is altered

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  • $\begingroup$ This seems to show that the magnitude of the force is not significant. Whether it would moon, earth, sun or the most extreme black hole - each turned on for an instant - they would only change the direction of the moving object. I can't wrap my head around that. $\endgroup$
    – TAR86
    Commented Nov 17, 2023 at 9:10
  • $\begingroup$ @TAR86 the magnitude of the force is by no means insignificant; it will determine how the body will turn. $\endgroup$ Commented Nov 21, 2023 at 22:45
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I want to highlight something you said:

Even if we assume that at the given instant , the change caused by the acceleration along it does not contribute significantly and magnitude effectively remains the same , how can we state that over a larger span , it wont contribute anything.

This is a fundamental principle in physics -- if something has no effect for an instant at every instant then it will have no effect even when you add up all those instants. This isn't obvious to everyone right away, and it actually makes certain assumptions about how math and the universe works, but it does seem to be true.

That is, if a force doesn't change the speed of an object in instant 1, and doesn't change it in instant 2, and doesn't change it in instant 3, then the speed won't be different between instant 1 and instant 3.

What sometimes causes confusion is that the force at instant 1 is not the same as the force at instant 2 or at instant 3. This can be hard to visualize, but it's necessary. The only way for a force to be perpendicular to a velocity at every instant is for the force to also change direction at every instant.

If you tried to start with a velocity to the North and a force to the East and then kept the force pointing East the speed wouldn't change in the first instant but every instant afterward the Eastward force would make the speed increase. It's only by constantly changing the direction of the force that you can keep the speed constant.

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The magnitude of a vector $\vec v$ can be expressed as the dot product $v = \sqrt{\vec v\cdot \vec v}$ so its derivative can be expanded as

$$\frac{\mathrm dv}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt}\sqrt{\vec v\cdot \vec v} = \frac1{2\sqrt{\vec v\cdot\vec v}}\left(\frac{\mathrm d\vec v}{\mathrm dt}\cdot\vec v + \vec v\cdot\frac{\mathrm d\vec v}{\mathrm dt}\right) = \frac1v\left(\vec v\cdot\frac{\mathrm d\vec v}{\mathrm dt}\right)$$

If acceleration (which is by definition $\vec a = \frac{\mathrm d\vec v}{\mathrm dt}$) is perpendicular to velocity then $\frac{\mathrm d \vec v}{\mathrm dt}\cdot \vec v = 0$, and therefore there is no change in magnitude: $$\frac{\mathrm dv}{\mathrm dt} = 0$$ We can see what does change by expanding the derivative of $\vec v$: $$\frac{\mathrm d\vec v}{\mathrm dt} = \frac{\mathrm d(v\hat v)}{\mathrm dt} = \frac{\mathrm dv}{\mathrm dt}\hat v + v\frac{\mathrm d\hat v}{\mathrm dt}$$ We know $\frac{\mathrm dv}{\mathrm dt} = 0$ so the total change in the velocity is due to a change in its unit vector $$\frac{\mathrm d\vec v}{\mathrm dt} = v\frac{\mathrm d\hat v}{\mathrm dt}$$ which represents the direction of $\vec v$. Hence a perpendicular force/acceleration acts purely by changing the direction of velocity.

How can we state that over a larger span , it wont contribute anything.

It might be insightful to answer this by tracking the speed over time.

The velocity after a small time $\mathrm dt$ is $\vec v_{\mathrm dt} = \vec v_0 + \vec a\mathrm dt$. Since $\vec v_0$ and $\vec a$ are always perpendicular, the new magnitude is just the Pythagorean theorem: $$(v_{\mathrm dt})^2 = (v_0)^2 + (a\mathrm dt)^2$$ After another instant $\mathrm dt$ the speed is $\vec v_{2\mathrm dt} = \vec v_{\mathrm dt} + \vec a\mathrm dt$ with a corresponding magnitude: $$(v_{2\mathrm dt})^2 = (v_{\mathrm dt})^2 + (a\mathrm dt)^2 = (v_0)^2 + 2(a\mathrm dt)^2$$ This pattern will continue forever so after $n$ instants of time the magnitude is $$(v_{n\mathrm dt})^2 = (v_0)^2 + n(a\mathrm dt)^2$$ In terms of a finite duration of acceleration $t$ (corresponding to $n=\frac{t}{\mathrm dt}$), the resulting magnitude is $$(v_t)^2 = (v_0)^2 + ta^2\mathrm dt$$ So we can see that the change in magnitude after a finite time is of order $\mathrm dt$ which is infinitesimal. In conclusion the change in speed does accumulate over time, but it's infinitely small and so it doesn't contribute when dealing with a finite speed.

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If we represent the movement in a 2D Cartesian plan, the velocity vector at any instant is $\mathbf v = (v_x,v_y)$. The velocity modulus is $\sqrt{v_x^2 + v_y^2}$. Now suppose that the derivative of the velocity is perpendicular to the velocity: $$\mathbf a = \frac{\mathbf{dv}}{dt} = (\frac{dv_x}{dt},\frac{dv_y}{dt})$$ If they are perpendicular: $$\mathbf {v\cdot a} = 0 \implies v_x\frac{dv_x}{dt} + v_y\frac{dv_y}{dt} = 0 \implies \frac{1}{2}\frac{d(v_x^2 + v_y^2)}{dt} = 0$$ That is, we come to the conclusion that the modulus is constant, because it's derivative is zero. We can also depart from the last part, (the hypothesis that the modulus is constant) and come to the conclusion that the vectors are perpendicular (or the acceleration is zero).

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I want to come at this from a slightly different perspective.

If you were writing a computer program to simulate a ball on a string but you wanted to simulate it with forces in Cartesian coordinates, the most basic simulation would have the exact same issue that your intuition is claiming there should be:

say you wanted your $1kg$ ball to go in a circle with a $1m$ radius at $1\frac{m}{s}$. That would mean you'd need: $$F=m\,a$$ $$F=m\,\frac{V^2}{r}$$ $$F=1\,\frac{m\,kg}{s^2}=1N$$

So lets pick the simplest simulation where we advance one time step while assuming the forces don't change during that time step:

A ball on a arc with a velocity arrow tangent to the arc and acceleration arrow perpendicular to the arc

Here I've scaled the acceleration according to my time step of about 0.2s so that it's easier to add the impulse due to the acceleration.

If we advance time only updating the acceleration and velocity at the end of the step we'd end up with something that looks like:

Same arc, this time the ball is off the arc at the end point of the former velocity arrow. There are faded acceleration and velocity arrows coming from the ball representing the old acceleration and velocity. A new velocity vector is coming from the ball as the sum of the old velocity and old acceleration.

Now this is obviously not right as the ball is no longer on the circle. The reason for that is that we used the original velocity for the full timestep while in reality the velocity will smoothly vary from the initial to the final velocity. The final velocity is pointing significantly inward, so you can see that the ball would get closer to the line if it used some of the final velocity within the time step.

The final velocity shown is similarly wrong. It's just the original acceleration applied over the entire timestep. You can see that the final velocity would definitely have a larger magnitude if this were how it actually worked. However, you can also see that the acceleration at the end is facing slightly backwards. If we used the acceleration that was smoothly varying it would be exactly the right amount to keep the magnitude of the velocity the same, just like if we smoothly varied the velocity then the path of the ball would stay on the line.

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It's really very simple, because both force and Velocity are vectors, not scalars. That is, they have a direction and not just a value. When you exert a force on an object, the resultant acceleration (F=ma) is also a vector and must have the same direction as the force that created the acceleration.

Thinking about the force as the rate of change of the momentum, since momentum is also a vector, the same logic applies. Momentum is mv (mass times the velocity - also a vector - with a direction). So again, the change in the momentum vector must lie in the same direction of the vector representing the force.

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Most answers rightly point out that a force, in order to achieve what you are stipulating (it changes the direction of the object, not its speed), must be constantly (at each instant) perpendicular. This entails that the object is, due to the applied force, constantly changing its direction and the force is at the same also re-adjusting its own direction, so as to always be pointing in one which has no component in the object's ever-changing direction.

But I would add that you will not see this unless the force in question has a certain modulus, so that the magnitude of the acceleration that it causes is always that of uniform circular motion, i.e. $a = v^2/r$.

In practice you can achieve this by either constantly adjusting the magnitude of the applied force or initially fixing the velocity of the object.

The first thing is what happens when the object is on the end of a string attached to a pivot. In this case, it is held in circular motion by tension, which is (like normal force) a constraint force, meaning that it self-adjusts so as to achieve its aim: normal force has the appropriate modulus to avoid penetration of the object into the ground, while tension exerted by a string also self-adjusts its modulus so as to avoid that the object flies away from circular trajectory and self-adjusts its direction thanks to the pivot around which the string rotates.

The second thing is what happens with gravity. The modulus of the force of gravity is what it is, it does not self-adjust. But then if you fire a projectile from the top of a mountain and you want that it adopts a circular trajectory, following the curvature of the Earth, without ever falling closer to it or escaping away, you must imprint on it the adequate velocity, so that $v^2/r$ matches acceleration due to gravity.

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