1
$\begingroup$

Can motion of a particle be circular if the radial acceleration is zero, but the tangential acceleration is not $0$?

$\endgroup$
  • 5
    $\begingroup$ What discussion, reading, or thoughts have prompted you to ask that question? Do you understand the meaning of the terms "radial" and "tangential"? $\endgroup$ – Bill N Jun 24 at 12:27
8
$\begingroup$

Let's look at the general acceleration vector in polar coordinates$^*$:

$$\mathbf a=\left(\ddot r-r\dot\theta^2\right)\hat r+\left(r\ddot\theta+2\dot r\dot\theta\right)\hat\theta$$

If we want our object to remain on the same circle, we must have $\dot r=0$ and $\ddot r=0$. This means our acceleration must have the form: $$\mathbf a=-r\dot\theta^2\hat r+r\ddot\theta\hat\theta$$

Since, by Newton's laws, the acceleration vector is proportional to the force via the mass $m$ of our particle, we see that we need a radial force magnitude of $$F_r=-mr\dot\theta^2$$ and a tangential force magnitude of $$F_\theta=mr\ddot\theta$$

Because we are confined to move around a single circle we can determine the role of each force. The radial force must only be responsible for changing the direction of the velocity, since if it could effect the speed then this means the object would have to change its $r$ coordinate, hence knocking us off the circle. Similarly, the tangential force must only be responsible for changing the speed of the particle as it moves around the circle, since if it could effect the direction of the velocity then it would do so by knocking us off the circle.

As a quick aside, you can probably tell by now that if we want to stay on the circle, these forces are required to be "linked", in a sense. Indeed, if you take the time derivative of $F_r$ and use the requirement that $\dot r=0$, you will find that $$\frac{\text dF_r}{\text d t}=-2\dot\theta F_\theta$$ showing that the presence of a tangential force requires a change in the magnitude of the radial force in order for the particle to remain on the circle (or, on the flip side, a change in the magnitude of the radial force must be accompanied by a tangential force).

If we additionally require uniform circular motion, then $\ddot\theta=0$, and we can only have a radial force without a tangential force. i.e. $$\mathbf F=-mr\dot\theta^2\hat r$$

and, by the preceding aside, this also means that the radial force must remain constant in magnitude (which makes sense, since $r$ nor $\dot\theta$ is allowed to change for uniform circular motion).$^{**}$

Therefore, the answer to your question is no. You always need a radial force (technically this specific radial force. Not any radial force will do) for circular motion (uniform or not). However, you can have the tangential force as zero (uniform circular motion) or non zero (non-uniform circular motion).


$^*$ dots represent a rate of change with respect to time if you are not familiar with calculus. for example, $\dot r$ is the rate of change of the variable $r$ with respect to time. The derivation of this equation can be found here.

$^{**}$ Note that this is what you usually encounter in your introductory physics classes as $F_{r}=mv^2/r$, since for motion along a circle of radius $r$, $\dot\theta=v/r$. The negative sign in this answer is to keep track of the direction of increasing/decreasing $r$, but if you are working a question where you only care about the magnitude of the radial force then this is irrelevant, hence why you usually don't see the negative sign.

$\endgroup$
7
$\begingroup$

No. Radial acceleration is necessary in order to change the direction of the velocity.

$\endgroup$
3
$\begingroup$

In circular motion, $$\vec{a}_{\text{radial}}\neq0,$$ but $\vec{a}_{\text{tangential}}$ may or may not be equal to zero.

$\endgroup$
1
$\begingroup$

"Tangential force" and "radial force" are merely other names for "force in the same direction as the object's current movement", and "force in a direction perpendicular to the object's current movement" respectively.

In other words, your question is equivalent to asking "If the only forces being applied to an object are in the same direction that the object is traveling, will it change direction?"

The answer is "No, it won't".


Semantic notes:
1) It is possible to apply forces that are neither in the same direction, nor perpendicular. But those forces can be decomposed and treated like two separate forces, one that is tangential, and one that is perpendicular, so we don't have to worry about that.

2) An object moving in a circle can be seen as having three directions: radial, tangential, and a direction perpendicular to both. But a force in that third direction will cause the object to curve in a new direction, changing the circle the object is moving around in such a way that the sum of the new force and the old radial force are in the radial direction of the new circle.

3) An object only moves in a circle when it is subject to a force that continually adjusts itself to always point in the radial direction (and there's a magnitude requirement as well, but that's not important right now). But you can calculate the instantaneous 'circle' that an object that is changing direction is moving around, much in the same way you can calculate the instantaneous slope of a function that is non-linear.

$\endgroup$
  • $\begingroup$ That's true if the object's velocity is perpendicular to its displacement from the center of the circle, and that is in turn implied by circular motion, but it is not true in general. $\endgroup$ – Acccumulation Jun 26 at 16:25
-1
$\begingroup$

I think the premise of your question is false, because the accelerated tangential motion will lead to a change in the centrifugal force on the particle, which in turn will cause radial acceleration. You can't have one kind of acceleration without causing the other in the first place.

But of course, when the acceleration and the resulting adjustment of the radius are finished, you can end up with a new circular orbit at a the new radius. Perhaps this is what you meant?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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