Why does a change of direction imply an acceleration?

We know that it takes no energy to change the direction of a vector, we know also that it takes no energy to displace a body in motion if a push is applied exactly at a right angle on its center of mass. Considering that, if that is true, can you explain why a simple change of direction without any increase of speed is considered an acceleration?

asked Jun 29, 2015 at 10:20 user83553 user83553 $\begingroup$ Your two premises seem false. Can you provide references for them? $\endgroup$ Commented Jun 29, 2015 at 10:25

$\begingroup$ @RedGrittyBrick, do you know the formula to calculate the energy required to change the direction of a vector? What force/energy do we need to change the direction of a body m = 1, v = 10 m/s by 1°, or 1rad? $\endgroup$

Commented Jun 29, 2015 at 10:32

$\begingroup$ As @RedGrittyBrick said, your premises aren't correct. If you apply a "push" you are exerting a force on the object, thus doing work, thus expending energy. It requires a force to change the direction of the trajectory of an object (in non-relativistic mechanics). For example, a constant inward radial force is needed to keep a body in orbit with constant speed. The magnitude of the body's velocity is constant, while the direction is changing. But the presence of that force means it must be constantly accelerating. $\endgroup$

Commented Jun 29, 2015 at 17:13

$\begingroup$ @KyleArean-Raines, if my premise is not correct and you need force/work/energy to divert a body B (m=1,v=10m/s) by 1 degree, how come nobody knows the formula to calculate it? Do you know how much work/joules is needed to change B's direction 1°? $\endgroup$

Commented Jun 30, 2015 at 4:04

$\begingroup$ I think people want you to work it out for yourself. Use Newton's second law. $\endgroup$

Commented Jun 30, 2015 at 10:55

4 Answers 4

$\begingroup$

Answer to the question in the title?

Two vectors are only equal to each other if they are the same (this is a general rule: equality means the things compared are identical). That means having the same direction as well as the same magnitude.

So how could changing the direction of motion not be acceleration?

Don't get hung up on fact that in 1 dimension acceleration always involved changing the magnitude, just ask if the final and initial velocities are the same or not.

Comment on the body text

In the body you ask about kinetic energy. There are two issues that come up with that:

  1. Kinetic energy (or any energy) is a scalar, and so it can remain the same while the velocity changes (as in uniform circular motion).
  2. Changes in (or differences of) velocity are the same for all inertial observer, but changes in kinetic energy are not. If you are thinking of the work-energy theorem you'll notice that this makes the work done frame-dependent as well. But this is not a surprise because the length and direction of the path is frame-dependent.
answered Jun 29, 2015 at 13:50 dmckee --- ex-moderator kitten dmckee --- ex-moderator kitten 86.5k 9 9 gold badges 174 174 silver badges 340 340 bronze badges

$\begingroup$ Thanks, could you clarify if my premise was wrong? What force/impulse/energy is needed to change the direction of motion of a body (m=1, v= 10m/s) by one degree or radian? $\endgroup$

Commented Jun 29, 2015 at 14:12 $\begingroup$ Impulse is the right quantity to ask about and it comes directly from the impulse-momentum theorem. $\vec = \Delta \vec

= m \,\Delta \vec$, so that $|J| \approx m |v| \sin\theta$ (where the approximation is good for small angles). But frankly I think that your problem is that you are treating works like "force" and "energy" as if your colloquial understanding of them is good enough rather than treating them as having precise meanings. You shouldn't be guessing at which concept applies, you should be able to deduce which one to use. $\endgroup$

Commented Jun 29, 2015 at 14:18

$\begingroup$ If I want to change the speed of B (M=1, v=10m/s), that is to increase it by 1m/s, I can use various means, "words" or "concepts", but it always boils down to giving 10.5 J, and an extra 1m/s will cost an additional 11.5. Now, if I want to change its direction by 1. 2 degree[s], can I know in advance how much energy I have to spend ? $\endgroup$

Commented Jun 29, 2015 at 15:08

$\begingroup$ Notice that to an observer passing in a train moving at $10 \,\mathrm$ you add $0.5\,\mathrm$ to the mass, but still change it's velocity by $1\,\mathrm$. (That is what I've written as (2) above.) The work done is not a frame-independent thing. The importance of invariance may take a while to get used to, but it deeply fundamental. There are frames in which any particular instantaneous change of direction of an orbiting satellite takes the form of a pure change in instantaneous speed: but that could be from stopped or to stopped so the KE change may have either sign. $\endgroup$