Acceleration of Particle moving in Circle

Theorem
Let $P$ be a particle moving in a circular path $C$.

Then the acceleration of $P$ is given as:


 * $\mathbf a = -\dfrac {\size {\mathbf v}^2 \mathbf r} {\size {\mathbf r}^2}$

where:
 * $\mathbf v$ is the instantaneous velocity of $P$
 * $\mathbf r$ is the vector whose magnitude equals the length of the radius of $C$ and whose direction is from the center of $C$ to $P$
 * $\size {\, \cdot \,}$ denotes the magnitude of a vector.

Proof
We use Lagrangian mechanics to derive the result. First we remember that $\mathbf r = \mathbf r(t)$ is a function of time $t$.

Let $R = \size{r} = \text{ const}$ be the constant radius of the circle $C$ for clarity.

The constraint can then be written as


 * $\displaystyle f(\mathbf r) := \size{\mathbf r}^2-R^2 = 0\quad\forall t$

The principle of stationary action then states that the first variation of the action $S$ must vanish.


 * $\displaystyle \delta S = \delta \int L\d t = 0$

The Lagrangian $L$ is given by the kinetic energy alone because the constraint force is handled implicitly by the constraint and there are no other forces acting on the particle.
 * $\displaystyle L = \dfrac{1}{2}m\size{\mathbf v}^2$

To incorporate the constraint we use a scalar Lagrange multiplier $\lambda$.

Because we are dealing with functionals and not functions, $\lambda$ is also a function of time $t$, as is $\mathbf r$
 * $\lambda = \lambda(t)$

The augmented action $S^+$ is given by $S^+ = S + \lambda f$

Then


 * $\displaystyle \delta S^+ = \left[-\dfrac{\d}{\d t}\left(m \mathbf{v}\right) + \lambda\underbrace{(2\mathbf{r})}_{=\frac{\d f}{\d \mathbf r}}\right]\cdot\delta \mathbf{r}+ f\delta\lambda = 0$

This means that both the term in the brackets as well as $f$ must vanish.

We eliminate $\lambda$ from the term in the brackets by taking its scalar product with $\mathbf r$ and simplifying $\frac{\d}{\d t}(m\mathbf v)$ to $m\mathbf a$


 * $\displaystyle -\dfrac{\d}{\d t}\left(m \mathbf{v}\right)\cdot \mathbf r + \lambda(2\mathbf{r})\cdot \mathbf r = 0$

This gives


 * $\displaystyle 2\lambda = \dfrac{m\mathbf a \cdot \mathbf r}{R^2}$

We have to replace $\mathbf a\cdot \mathbf r$.

To do this, we take the second time derivative of the constraint, which must vanish too


 * $\displaystyle \dfrac{\d^2 f}{\d t^2} = 2 \dfrac{\d}{\d t}\left(\mathbf r \cdot \mathbf v\right) = 2 \left(\mathbf a \cdot \mathbf r + \size{\mathbf v}^2\right)$

So
 * $\displaystyle 2\lambda = \dfrac{-m\size{\mathbf v}^2}{R^2}$

Plugging in $2\lambda$ in the term with the big brackets yields


 * $\displaystyle -m \mathbf a - \dfrac{m\size{\mathbf v}^2}{R^2} \mathbf r = 0$

Division by $m$ completes the proof