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I have been taught that a perpendicular force acting on an object will only change the direction of its velocity, not its magnitude. The explanation that was provided to me is that because there no force in the direction of its velocity, there must be no work done and thus change in the object's kinetic energy. However, I have also studied the vectorial form of the equations of motion

$$\vec v_f = \vec v_i + \vec a\cdot t$$

So I assume that the velocity after a constant force has acted on it for a period of time is simply the vector sum of $\vec v_i $ and $\vec a \cdot t$, which I just assume is the acceleration vector (determined by the force) scaled up by a factor of the time the force acts.Diagrammatically, I took the sum as

Vector Sum of initial and change in velocity

It is obviously clear from the diagram that the new speed (which is the length of the $\vec v_f$ vector) is bigger than the previous speed. This can be confirmed by Pythagoras's theorem, the magnitude of the $\vec v_f$ vector is

$$ |\vec v_f|=\sqrt{|\vec v_i|^2 + |\vec a|^2 \cdot t^2} $$

which is very clearly not equal to $|\vec v_i| $

So, I conclude that the speed of the object has changed even though the force acting on it was perpendicular to its velocity. So my question is, what have I missed?

Is my assumption that the final velocity is the vector sum of the initial velocity and the product of acceleration and time incorrect? Or have I made a mistake somewhere else?

Thank you in advance for your help.

Qmechanic
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PhysicsPotato
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2 Answers2

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In drawing your triangle, you assume that the object has already moved in the direction of your force (maybe just a little bit). In doing so, you forget that the force then stops being perpendicular.

What you have (re)discovered is simply that a force which is constant throughout space cannot remain perpendicular to the motion of a free particle if it lasts a finite amount of time: the particle will simply start to move in the direction of the force, picking up velocity as it starts moving in that direction.

However, if $\vec{F}$ is not of the same magnitude and direction everywhere, one can produce a force that remains perpendicular. The canonical example is that of a body exerting gravity. The direction of the force is always towards the particle (i.e. not the same everywhere). It also decreases with distance, but that's really not an essential point here. Now, if a test particle at a distance $d$ moves past the other body with the right velocity, this force will always remain perpendicular. Can you guess what kind of shape the trajectory of the test particle will trace out? Hint: think about the symmetry of the problem.

Danu
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  • I recognise my mistake, but then that would lead to another question, How do I deal with forces that act at an angle to the velocity? Would I have to integrate the variable force over time? My guess to your question is a circular path (considering the symmetry). You mentioned however, that the test particle has to move with the "right" velocity. What would the right velocity be and how would I find it? – PhysicsPotato May 17 '14 at 10:59
  • @PhysicsPotato If you are familiar with the laws of circular motion, you might know that it is possible to express the acceleration experienced by the particle (which, of course, only serves to change it direction ;) ) can be expressed in terms of the velocity. Because we're dealing with a gravitational problem, it can also be expressed in terms of the gravitational force. Matching these two to eliminate the acceleration will give you the answer to your question. – Danu May 17 '14 at 11:09
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Everything in the diagram is perfect even I had this doubt for long. When the force is applied perpendicular to the velocity, it is applied for a veryyyyy short period of time( t approaches 0). Hence the a*t in the equation you got becomes 0 and the magnitude of initial and final velocity become the same.

SatvatOfficial
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