Uniform Circular Motion Derivation of Radial Acceleration

[hermes1908]

Homework Statement

I am trying to derive the formula a_r=\frac{v^2}{r} for uniform circular motion (for personal understanding, this is not an assignment). But am having some difficulty. I have seen other proofs, but I want to know why my approach is wrong.

The Attempt at a Solution

Since we are considering uniform circular motion, we assume the tangential and angular accelerations are 0, and the magnitude of the velocity of the particle at any point is the same. Adding the two vectors shown in attached picture, gives us the magnitude of the change in velocity (sqrt(2)*v) over a distance of ∏/2. Since the angular displacement of the particle is ∏/2 we can find the amount of time that has passed by \frac{\frac{\pi}{2} r}{v}. Now simply taking the velocity and dividing it by the derived time expression should yield the magnitude of the radial acceleration, but as you can see it gives \frac{2\sqrt{2}v^2}{\pi r}. My answer is close to what is expected so I suspect my mistake is mathematical in nature, but I can't seem to figure out what it is.

 

[TSny]

Your method gives the correct value for the average acceleration over the finite time interval of a quarter of a circle. The acceleration is not constant because it is continually changing its direction. So, there is a difference between average acceleration and instantaneous acceleration.

Note that if you took a full circle for your time interval, then your calculation would give the correct answer of zero average acceleration.

You need to consider infinitesimal time intervals to get the instantaneous acceleration.

 

[Chestermiller]

You made a valiant attempt, but you used too large an angle. Try it with a smaller angle, and see what you get. Another way to do it is using vector calculus. In terms of a unit vector in the θ direction \vec{i_\theta}, the velocity vector \vec{V} is given by:
\vec{V}=V\vec{i_\theta} The unit vector in the θ direction \vec{i_\theta} varies with θ according to:
\frac{\partial \vec{i_\theta}}{\partial \theta}=-\vec{i_r}
If we take the derivative of the velocity vector \vec{V} with respect to time, we obtain:\frac{d\vec{V}}{dt}=-V\vec{i_r}\frac{d \theta}{dt}
But,
r\frac{d\theta}{dt}=V
Therefore, \frac{d\vec{V}}{dt}=-\frac{V^2}{r}\vec{i_r}