# Perpendicular Forces and Changes in Momentum

1. Oct 8, 2012

### Jacob959

Okay, I feel like I am just missing something that should be very easy to see, but I can't seem to wrap my head around this concept. Can anyone explain to me why a force perpendicular to the momentum only changes the direction of the momentum and not the magnitude? By my logic, if Fnet=Δp/Δt, and thus, Δt*Fnet = mΔv, then shouldn't the perpendicular force change the velocity of object in the direction perpendicular to the current direction of momentum? And because the velocity changes in the new direction, wouldn't the magnitude of the momentum change since the magnitude of the velocity changed?

Hopefully you can follow my logic and see where it is flawed! Thanks in advance!

2. Oct 9, 2012

### andrien

of course it will change if it is always pointing in the same direction or at least for a finite time so as to cause this velocity change.If you are confused with some circular motion type thing then you might be aware that the acceleration is not always or for a finite moment points in same direction.

3. Oct 9, 2012

### haael

The problem goes as this:

You have a particle moving on a plane. You have an acceleration vector of the constant magnitude and of the direction always perpendicular to the velocity of the particle. You can write it down as a differential equation.

Now consider infinitesimal accelerations (changes of velocity). You can split an infinitesimal acceleration into two parts: the part parallel to the direction of motion and the part perpendicular to the direction of motion. I don't know if you can see it, but the parallel part is responsible for changing the magnitude of velocity without changing direction and the perpendicular part is for changing direction without changing magnitude. If you don't see it, then just integrate it in your mind :).

With the problem you gave, the acceleration vector is always perpendicular to the direction of motion, so it will never change the magnitude.

Note that the magnitude is determined by the magnitude of acceleration. It is not a boundary condition. The speed of your particle is a solution of the equation.

4. Oct 9, 2012

### xAxis

Yes, the velocity will change in that direction, so will the momentum. But in the direction perpendicular to it, the velocity will be a little smaller, n't it? When you add them, you will see the speed didn't change, therefore magnitude of momentum stays constant.

Last edited: Oct 9, 2012
5. Oct 9, 2012

### rcgldr

The idea here is that acceleration is always perpendicular to velocity. As velocity changes direction, so does the acceleration, so that it remains perpendicular to velocity. So the path changes, but not the speed. As a real world example of this, imagine a car at constant speed taking a variety of turns on a twisty road.

Last edited: Oct 9, 2012
6. Oct 9, 2012

### sophiecentaur

This is an example of a counter-intuitive situation. It is hard not to think in terms of the effect of gravity, say, on an object that's moving horizontally. After a short while, the (free fall) object's momentum (and, of course, Kinetic Energy) will have increased. But the force is only normal to the path of the object initially - so the OP's statement only applies during the first infinitessimal. Thereafter, the force has a component in the direction of the motion so the magnitude of the momentum can increase.
In the case of circular motion (the only example where the magnitude of the momentum does not increase) the easiest argument against a change is that the 'string' doesn't get any shorter. Hence, there is no 'force times distance' involved so there is no work done - so the KE will be unchanged and, hence, neither will the magnitude of the momentum.

7. Oct 9, 2012

### rcgldr

You can have a constant speed path of just about any shape, as long as there are no sharp "corners". The example of a car driving at constant speed in a variety of patterns (circle, ellipse, spiral, sin wave, parabola, ...) should help explain this concept.

8. Oct 9, 2012

### sophiecentaur

Ah yes - but not with only a normal force acting.
(You had me worried there for a second. )

9. Oct 9, 2012

### jbriggs444

Did you just write the opposite of what you meant?

All of the described trajectories can indeed be obtained with only a normal force acting.

If the force is not normal, the speed will not be constant but all of the described trajectories can still be obtained. Along with including some additional shapes that have sharp corners.

Or have I missed the joke?

10. Oct 9, 2012

### sophiecentaur

I was rather thinking that the speed would need to be constant for the magnitude of the momentum to be the same and was then thinking in terms of elliptical orbits, where the central force is not normal all the time. A 'rounded square' would fit the bill as well as a circle but, I guess, if you imagine a perfectly smooth wire track, there could only ever be a normal force on the object - so it could be any shape you liked and the object would maintain constant speed.