Circular Motion: Proof for Non-Uniform Circular Motion Acceleration

[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?

 

[vanhees71]

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}.

 

[Tanya Sharma]

In uniform circular motion ω is constant which means centripetal acceleration a_c=ω²R is constant .

In case of non uniform circular motion centripetal acceleration is still given by a_c=ω²R ,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 a_T = dv/dt .

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

Since they are at right angles the magnitude is given by a_Total = √(a_c² + a_T²)

 

[GeneralOJB]

Thanks a lot!