Circular motion

GeneralOJB
I have seen the derivation of the centripetal acceleration formula a=v^2/r by saying r= rcosθi+isinθj=rcosωti+rsinωtj and differentiating twice. Since ω is constant we get a=-ω$^{2}$r.

I've started looking at non-uniform circular motion where there is also the tangential acceleration vector component. I'm told that the component of acceleration directed towards the center of the circle has magnitude v^2/r, but I don't believe the original proof works because we assumed ω is constant, and now it isn't. Can this type of proof be made to work still?

Gold Member
2022 Award
Sure. Then you write
$$\vec{r}(t)=r \cos [\theta(t)] \vec{i} + r \sin[\theta(t)] \vec{j}$$
and differentiate twice (assuming $r=\text{const}$, because it's supposed to be a circular motion). Then you get
$$\vec{v}(t)=\dot{\vec{r}}(t) = -r \dot{\theta} \sin \theta \vec{i} + r \dot{\theta} \cos \theta \vec{j}=r \dot{\theta} \hat{\theta},$$
$$\vec{a}(t)=\dot{\vec{r}}(t)=-r \ddot{\theta} \sin \theta \vec{i} - r \dot{\theta}^2 \cos \theta \vec{i} + r \ddot{\theta} \cos \theta \vec{j} - r \dot{\theta}^2 \sin \theta \vec{j} =r \ddot{\theta} \hat{\theta}-r \dot{\theta}^2 \hat{r}.$$
Here I have used the orthonormal basis of the polar coordinates,
$$\hat{r}=\cos \theta \vec{i} + \sin \theta \vec{j}, \quad \hat{\theta}=-\sin \theta \vec{i}+\cos \theta \vec{j}.$$
It's clear that $\hat{r}$ points perpendicularly outward from the circle (normal of the curve) and $\hat{\theta}$ along the circle (tangent of the curve).

That shows that the tangential acceleration is
$$a_{\theta} = \hat{\theta} \cdot \vec{a}=r \ddot \theta$$
and the normal acceleration is indeed the momentaneous centripetal acceleration as in uniform circlar motion,
$$a_{r}=\hat{r} \cdot \vec{a} = -r \dot{\theta}^2.$$
It's easy to calculate that also then
$$a_{r}=-\frac{\vec{v}^2}{r}.$$

1 person
Tanya Sharma
In uniform circular motion ω is constant which means centripetal acceleration ac2R is constant .

In case of non uniform circular motion centripetal acceleration is still given by ac2R ,but as you have rightly said ω is changing ,which means centripetal acceleration is also varying (magnitude changing with time).

In addition there is also tangential acceleration aT = dv/dt .

The net acceleration of the object is given by the vector sum of the two i.e ac and aT .

Since they are at right angles the magnitude is given by aTotal = √(ac2 + aT2)

GeneralOJB
Thanks a lot!