Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It might sound stupid but seriously I couldn't find why?

We have a mass with a constant speed, which is acted on by a unit force which is always at right angles to its direction of motion.

Why do the mass travels a circular path!?

share|cite|improve this question
Or it could be helical trajectory... – QGR Feb 17 '11 at 9:41
No its circular, but i couldn't find why. – omeid Feb 17 '11 at 9:42
No, it really could be a helical trajectory too. It's not accelerating in the direction perpendicular to the plane of the circle, but that doesn't mean it's not moving in that direction at a constant rate. Therefore if nothing's actively keeping it still in that direction, the path will really be a circular helix almost all the time. (Also, if the path is a circle in one reference frame there are infinitely many other frames in which it's a helix.) – Keenan Pepper Feb 17 '11 at 13:32
It's not circular, and it's not constant speed. If you see circular path - it's just a rare coincidence. – BarsMonster Feb 17 '11 at 14:58
@BarsMonster Yes, it is constant speed. If the force is at a right angle to the motion, the work done is zero. Also, the projection of the motion into the appropriate plane would be circular, because the speed and acceleration are both constant. – Mark Eichenlaub Feb 17 '11 at 17:27
up vote 5 down vote accepted

As Newton did in his Principia Mathematica with an example similar to yours, imagine what happens over a short time $dt$ to the velocity of the mass. Using $F = m\,dv/dt$ gives

$dv = F\,dt/m$

The force is at right angles to $v$ and so $dv$ is at right angles to, rather than along $v$. Do the same for the next $dt$ and so on, adding up all the contributions of $dv$ for one complete revolution and you will get $\sum dv = 0$. The additional wobble to the circular motion becomes zero as $dv \rightarrow 0$

share|cite|improve this answer
Awesome! I tried to vote up twice but it dosn't works :D – omeid Feb 17 '11 at 11:26
That why I did that for you @Omeid ;-) – Ivo Flipse Feb 17 '11 at 11:48
@Ivo, Cheers man!!! – omeid Feb 17 '11 at 11:52
@Omeid, glad I could help out ;) – John McVirgo Feb 17 '11 at 12:05
thanks alot man I could never though of looking at acceleration as dv/dt that really helped man !!! – omeid Feb 17 '11 at 12:14

Assume the motion is confined to a plane. If the force is always at a right angle to the motion, the speed is necessarily constant. The circular shape follows due to symmetry.

To find the change in speed of an object, we can project the force vector acting on it onto the velocity vector. For example, the force below will create a large change in speed, because it is pushing mostly the same way an object is going.

This next force is just as big, but it will only create a small change in speed because it is pushing only partially in the direction of motion.

enter image description here

This final force will create no change in speed because it is perpendicular to the motion. Its only effect is to change the direction of motion.

enter image description here

Because the problem specifies that the force is always perpendicular to the motion, the object's speed is constant.

Once we know the speed is constant, it follows from symmetry that the motion is a circle. If the motion were, say, a parabola, then different parts of the trajectory would "feel" different to the particle. For example, the very bottom of a parabola is highly curved, but if you go up the sides a way, it is not.

enter image description here

The green part of the parabola, at the bottom, has a very short radius of curvature. The red part has a very long one. A ball traveling at constant speed feeling a constant force should not have such different types of motion at different times. Instead, since the force on it is always the same relative to itself, it should always be doing the same thing. There should not be any part of the curve that is different than any other part.

Other shapes, like an ellipse or a spiral, are also different in some parts than in others. Only the circle has the same curvature everywhere, and so the motion must be a circle.

share|cite|improve this answer
If there's something about this answer you don't understand or disagree with, please leave a comment rather than simply downvoting. – Mark Eichenlaub Feb 17 '11 at 19:46
if your asking me I haven't down voted. :D – omeid Feb 18 '11 at 0:59
@Omeid I was asking whoever downvoted - it's a pretty standard practice. – Mark Eichenlaub Feb 18 '11 at 1:55
I didn't down vote, but it does seem an ott combination of unnecessary diagrams and unnecesary symmetry arguments to describe the fundamental reasons which can be done in a few sentences. The force is defined to act at right angles to the current velocity and so the change in velocity must be at right angles to its current velocity. This point could have benefited from one diagram to show that only the direction changes, rather than its speed over dv. – John McVirgo Feb 20 '11 at 1:11
@John I disagree. I don't think that saying, "There is a differential equation and the solution is a circle," or saying, "You can numerically integrate it and it comes out to be a circle," is insightful. I wanted the answer to be comprehensible to someone with very little physics background, since it is clear that the original poster is a beginning student. Further, I wanted to provide some intuition on what's so special about a circle. The reason for a circle is that the problem has a special symmetry, so the answer has one. That's an important point that's left out of other answers. – Mark Eichenlaub Feb 20 '11 at 21:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.