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!

 

[PJC01]

Yes, the proof for non-uniform circular motion can still be derived using the same method as for uniform circular motion. However, we need to take into account the changing angular velocity (ω) in our calculations.

In non-uniform circular motion, the tangential acceleration (at) is given by the equation at = αr, where α is the angular acceleration. This means that the tangential acceleration is directly proportional to the distance from the center of the circle (r) and the angular acceleration.

Using the same approach as before, we can write the position vector as r = rcos(ωt)i + rsin(ωt)j, where ω is the angular velocity. By differentiating twice, we get the following equations for the velocity and acceleration vectors:

v = -rsin(ωt)ωi + rcos(ωt)ωj
a = -rcos(ωt)ω^2i - rsin(ωt)ω^2j + rcos(ωt)αi + rsin(ωt)αj

Now, we can break down the acceleration vector into its radial and tangential components. The radial component (ar) is given by -rcos(ωt)ω^2i - rsin(ωt)ω^2j, which is equivalent to -ω^2r. This is the same as the formula for uniform circular motion. The tangential component (at) is given by rcos(ωt)αi + rsin(ωt)αj.

Therefore, the total acceleration vector is given by a = -ω^2r + rcos(ωt)αi + rsin(ωt)αj. This shows that the component of acceleration directed towards the center of the circle is still given by v^2/r, but now the angular velocity (ω) is not constant. Instead, it is changing with time due to the angular acceleration (α).

In conclusion, the proof for non-uniform circular motion can be derived using the same method as for uniform circular motion, but we need to take into account the changing angular velocity in our calculations. This shows that the formula a = v^2/r is still valid for non-uniform circular motion, but the angular velocity must be taken into consideration.