# Short-duration forces

In circular motion, it is said that the centripetal force acts only for a very very short period of time, hence is able to only change the direction but not magnitude of the velocity.

Similarly in a collision, the impulsive-collision forces act for a very short period of time, but they are able to change the direction as well as magnitude of velocity, considering an inelastic collision. The collision lasts for a very short period of time, almost instantaneous (with reference to Newtonian mechanics, not considering Einstein's special relativity).

*Why is that Why is collision force able to change magnitude when it acts for a short period of time as well?

The reasoning that collision forces have a very high magnitude won't work here, as say they have a magnitude $100,000\text{ N}$, then imagine the circular motion to be of a mass $1\text{ kg}$ and speed $100,000\text{ ms}^{-1}$ and radius $1\text{ m}$, now we have both centripetal and collision forces equal but still centripetal force is able to only change direction.

Similarly we can always adjust both the forces and circle's radii accordingly, so that impulse provided by both the forces, centripetal and collision forces are always equal.

we calculate impulse provided by the centripetal force in circular motion to be such that for the differential time $dt$ , we evaluate impulse for ,the force remains constant in terms of both
magnitude and direction and according to continuity of time , there are infinite of such instances giving us a non-zero length span of such a time .

The reasoning that centripetal force acts perpendicular to direction of motion also won't work since then problem will be asking that momentum doesn't change even in the direction of the force which is perpendicular to path, even in the perpendicular direction there is no impulse and if there's impulse, it may lead to violation of Energy conservation, which can't happen.

These two mode of reasoning, I have tried and they fail here.

All conditions are sort of idealistic here, following all of Newton's and Euler's laws.

All masses are to be considered to be point masses.

Impulse is $\int^{t_2} _{t_1} Fdt$ where $F$ is the force and $t_2-t_1$ is the span of time of finite length we considered.

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This "In circular motion , it is said that the centripetal force acts only for a very very short period of time" appears to be a misunderstanding. The analysis presented considers a short interval of time so that only a small error is made be neglecting second order effects, then calculus is applied to get the solution in the limit of vanishing time steps making that approximation rigorous. All applications of calculus to physics work this way. – dmckee May 24 '13 at 14:58
Note, I've removed many incorrect (in standard English usage) spaces before punctuation marks. I left the ones in the quote because I have not seen the source, but unless you cut-n-pasted that from an authoritative source those are almost certainly wrong as well. – dmckee May 24 '13 at 15:02
Who said that? References? – ja72 May 24 '13 at 15:48

Actually, the reasoning that the force in circular motion is perpendicular to the direction of motion does work, and is in fact the correct answer. You appear to be saying that this violates conservation of energy by changing the object's momentum. However, the centripetal force only changes the direction of the momentum, not the magnitude, and thus conservation of energy is conserved. (Of course, conservation of momentum is broken; this is because conservation of momentum only applies when there's no external force, and circular motion requires an external force.)

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You have simply stated again what was referred in the question . You haven't responded how a short-duration force's effect during collision is able to change magnitude but in case of centripetal force it is able to change only direction . How and why this happens , is the question . – user23503 May 25 '13 at 15:43

Duration of force application has nothing to do with magnitude/direction change. In fact, we can see the effect of force on differential scale.

Centripetal force changes only direction because its perpendicular to the direction of velocity. This can only accelerate particle radially. As motion in two perpendicular direction is independent of each other, tangential speed is unaffected.

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Try approaching this from a different perspective.

Consider a particle of mass $m$ moving along a circular trajectory in the x-y plane. The x(t) and y(t) coordinates are:

$x(t) = \cos(\omega t)$

$y(t) = \sin(\omega t)$

The particle's momentum components are thus:

$p_x(t) = m\dot x(t) = -\omega m \sin(\omega t)$

$p_y(t) = m\dot y(t) = \omega m \cos(\omega t)$

The components of force on the particle are thus:

$f_x(t) = \dot p_x(t) = -\omega^2 m \cos(\omega t)$

$f_y(t) = \dot p_y(t) = -\omega^2 m \sin(\omega t)$

So, the time varying force components acting on the particle do change the momentum components and more, they do not act for a short period of time.

There's nothing mysterious here. This is just application of Newton's laws and it not unlike other applications of Newton's laws.

What is, evidently, causing your consternation is this:

$p^2 = p^2_x + p^2_y = (\omega m)^2$

or maybe:

$\vec p \cdot \vec f = 0$

But, honestly, I don't see why this should lead you to the following:

In circular motion, it is said that the centripetal force acts only for a very very short period of time, hence is able to only change the direction but not magnitude of the velocity.

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In circular motion, it is said that the centripetal force acts only for a very very short period of time, hence is able to only change the direction but not magnitude of the velocity.

Perhaps this is the right place to address specifically such "pictures":

No -- in circular motion (especially in "uniform circular motion", or at least in "circular motion without stopping") there is no "period of time" (or: "duration"), however short, during which the acceleration (especially "the centripetal component of acceleration") could be called constant. Instead, the acceleration is constantly changing;
the magnitude of jerk is always non-zero. (For uniform circular motion this magnitude is "$\omega^3$".)

Of course, calculating "acceleration at some particular time/instant" involves considering "average acceleration during finite duration" and determining the "limit, as the duration vanishes". But thus calculated, the acceleration still changes "from instant to instant".

Similarly, calculating "velocity at some particular time/instant" involves considering "average velocity during finite duration" and determining the "limit, as the duration vanishes". The velocity, thus calculated, also changes "from instant to instant"; and for uniform circular motion, at each instant the correspondingly calculated acceleration and velocity are perpendicular to each other.

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