Question

Suppose we are told the acceleration of a particle moving in a circle of a radius r with uniform speed v is proportional to some power of r say n, some power of v, say m. Determine the powers of r and v?

Answer 1

Suppose we are told the acceleration of a particle moving in a circle of a radius $r$ with uniform speed $v$ is proportional to some power of $r$ say $n$, some power of $v$, say $m$. Determine the powers of $r$ and $v$?

Solution

We will use dimensional method.

The dimension of $\mathbf{v}$ is $[v] = \left[\frac{\text{meter}}{\text{second}}\right] = \left[\frac{m}{s}\right]$.

The dimension of $\mathbf{r}$ is $[r] = [meter] = [m]$.

The dimension of acceleration is $[a] = \left[\frac{\text{meter}}{\text{second}^2}\right] = \left[\frac{m}{s^2}\right]$

If $a \propto v^n r^m$, we have

$$[a] = \left[ v^n r^m \right] = \left[ \frac{m}{s} \right]^n [m]^m = \left[ \frac{m}{s^2} \right]$$

From whence we get the following system of equations:

$$\left\{ \begin{array}{l} n + m = 1 \\ n = 2 \end{array} \right. \Rightarrow$$ $$\left\{ \begin{array}{l} m = -1 \\ n = 2 \end{array} \right. \Rightarrow$$

From whence (using dimensional method we don't know the coefficient of proportionality):

$$a \propto \frac{v^2}{r}.$$

Answer:

$$\left\{ \begin{array}{l} m = -1 \\ n = 2 \end{array} \right.$$ $$a \propto \frac{v^2}{r}$$