How can a person move with non-uniform centripetal acceleration along a circle?

I am just confused how a person or a particle can move with non uniform centripetal acceleration along a circle. Take the case of non uniform circular motion. The magnitude of centripetal acceleration is given by $$a=\frac{v^2}{r}$$. So if $$v$$ is changing, $$a$$ must also change, but $$a$$ is the rate of change of direction of velocity. If the rate is not constant, how can the object form circular path? Please explain.

• I am trying to say that the path should be a circle only when direction of velocity vector is changing at a constant rate. Then how can a particle or a person can move in a circular path if magnitude of centripetal acceleration vector is not constant? Jun 24 '20 at 20:52

The critical thing here is something you've said, in fact: $$a$$ is the magnitude of the acceleration, and $$v$$ is the magnitude of the velocity. But velocity and acceleration are vectors: their magnitudes do not capture everything about them.

For uniform circular motion, up to a constant added to $$t$$,

$$\vec{v} = (-r\omega \sin \omega t)\vec{i} + (r\omega \cos \omega t)\vec{j}$$

where $$\vec{i},\vec{j}$$ are unit vectors in the $$x,y$$ directions respectively. It's easy to compute the magnitude, $$v$$, of $$\vec{v}$$:

\begin{align} v &= \sqrt{\vec{v}\cdot\vec{v}}\\ &= \sqrt{r^2\omega^2\cos^2\omega t + r^2\omega^2\sin^2\omega t}\\ &= r\omega \end{align}

This is constant: $$\vec{v}$$ is not constant, but $$v$$, its magnitude, is.

The same holds for $$\vec{a}$$: it is not constant, but its magnitude, $$a$$, is.

In the case of non-uniform circular motion, $$v$$ is not constant, and neither in general is $$a$$.

As an example consider an object moving on a path such that $$\vec{p}(t) = (r\cos t^2)\vec{i} + (r \sin t^2)\vec{j}$$. This is a circle but the motion is not uniform. For this motion:

\begin{align} \vec{v}(t) &= -(2rt\sin t^2)\vec{i} + (2rt\sin t^2)\vec{j}\\ v(t) &= \sqrt{4r^2t^2}\\ &= 2rt\\ \vec{a}(t) &= (-4rt^2 \cos t^2 - 2r\sin t^2)\vec{i} + (2r\cos t^2 - 4rt\sin t^2)\vec{j}\\ a(t) &= 2r\sqrt{1 + 4t^4} \end{align}

Well, in this case none of $$\vec{v}, v, \vec{a}, a$$ are constant.

• r is constant the radius Jun 24 '20 at 21:49
• @Teaislife: yes.
– user107153
Jun 25 '20 at 12:30

If you work in polar coordinates ($$\bf \hat r,\hat{\theta}$$) then the position is defined by:

$${\bf \vec r}(t) = r(t){\bf \hat r}$$

The velocity is:

$${\bf \vec v}(t) = \dot r(t){\bf \hat r} + r(t)\omega(t){\bf \hat{\theta}}$$

The acceleration is:

$${\bf \vec a}(t) = (\ddot r(t)-r(t)\omega(t)^2){\bf \hat r} + (r(t)\dot{\omega}(t) +2\dot r(t)\omega(t)){\bf \hat{\theta}}$$

So the criterion for circular motion is $$\dot r(t)=0$$, or:

$${\bf \vec r}(t) = R{\bf \hat r}$$

then

$${\bf \vec v}(t) = R\omega(t){\bf \hat{\theta}}$$

which is tangential to the circle, while:

$${\bf \vec a}(t) = -R\omega(t)^2{\bf \hat r} + R\dot{\omega}(t){\bf \hat{\theta}}$$

is not radial for $$\dot{\omega} \ne 0$$.

Suppose you have a vector $$\vec{a}$$ having magnitude $$a$$. If you want to rotate it by $$\theta$$ without changing its magnitude then you would have to add to it a vector having magnitude $$\sqrt{2a^2 - 2a^2\cos{\theta}} = \sqrt{2a^2(1 - \cos{\theta})}$$ Note that this depends on the magnitude of $$\vec{a}$$. So to rotate two vectors having different magnitudes by the same angle you need to add a longer vector to rotate the longer of the two.

Now suppose this vector is the velocity vector. If you want to rotate it by an amount $$\Delta\theta$$ then you would add to it a vector having magnitude $$\Delta v = v\sqrt{2(1 - \cos{\Delta\theta})}$$. Dividing by $$\Delta t$$, the time over which this rotation occurred, we get,
$$\frac{\Delta v}{\Delta t} = \frac{ v\sqrt{2(1 - \cos{\Delta\theta})}}{\Delta t}$$

And if you take limits, the left hand side becomes the magnitude of acceleration responsible for this rotation and the right hand side becomes the familiar $$\frac{v^2}{r}$$,

$$a_c = \lim_{\Delta t \to 0} \frac{v\sqrt{2(1-(1-\frac{(\Delta\theta)^2}{2!} + \frac{(\Delta\theta)^4}{4!}.....))}}{\Delta t}$$ $$a_c = \lim_{\Delta t \to 0}\frac{v\Delta\theta}{\Delta t} = \frac{v^2}{r}$$

You are having the same doubt that I had a while back. In circular motion, there are two types of acceleration :

1. $$a_{centripetal}$$ - which gives rate of change of direction in velocity
2. $$a_{tangential}$$ - which gives rate of change of magnitude of velocity

We can also define a net change in acceleration which is noting but vector sum of above two accelerations.

Now to answer your question (and I am quoting my professor's words):

It depends on what controls the radial distance. If you have a body at the end of a rigid rod, pivoted at one end, then the radial distance cannot change so tangential acceleration just makes it circle faster/slower with the same radius but, if you have a body moving in a gravitational field, then the tangential acceleration will make it spiral in or out.

I hope it helps.

• But it didn't answer my question Jun 24 '20 at 21:12
• @ayushsharma what's the problem? Jun 24 '20 at 21:13
• I am saying that circular motion is possible only when the direction of velocity vector changes at a constant rate. So if speed is not constant, how can the motion be circular in mathematical sense? Ok you answered my question. But I am not able to figure out where you have answered this query. Can you please illustrate in a more clear way? Jun 24 '20 at 21:17
• @ayushsharma If velocity vector only changes it's direction or as per u if its speed is constant it will be a case of uniform circular motion. Now read quoted part in my answer for the case when the speed is not constant. Jun 24 '20 at 21:20
• @ayushsharma, "circular motion is possible only when the direction of velocity vector changes at a constant rate." is not correct. That would be true for uniform circular motion. Jun 24 '20 at 21:35

While all the other answers are correct (+1 to all), they appear not to address the source of your confusion.

the path should be a circle only when direction of velocity vector is changing at a constant rate.

This is incorrect. It is true for the simplest case, when the speed is constant.

Suppose you and I are in a car with no steering wheel on a circular track. You control the gas and brake, the force parallel to the direction the car is traveling. I control a rocket mounted on the side of the car. Its force is perpendicular to the direction the car is traveling. My goal is to keep the car on the track no matter what you do.

I can do this because I control the curvature of the path. If I turn the rocket up, the car turns more sharply.

For any constant speed, I can of course stay on the track. If you let the car travel at a constant speed $$v_1$$, I adjust the rocket so the sideways force is $$v_1^2/r$$.

If you smoothly accelerate from $$v_1$$ to $$v_2$$, at any instant you are traveling at some speed $$v$$. So I can adjust the rocket so the acceleration is $$v^2/r$$. I will have to smoothly increase the power so the acceleration changes from $$v_1^2/r$$ to $$v_2^2/r$$ in sync with your change.

So if v is changing, a must also change, but a is the rate of change of direction of velocity. If the rate is not constant, how can the object form circular path?

If you go around a circle at a constant speed, the velocity vector changes direction at a constant rate. A constant sideways acceleration provides this change in direction.

If you step on the gas and the car travels in circles at an ever increasing velocity, the velocity vector changes direction at an ever increasing rate. An increasing sideways acceleration is needed to provide this change in direction.