The centripetal force $F$ acting on a particle moving uniformly in a circle may depend upon mass $m$, velocity $v$ and radius $r$ of the circle. Derive the formula for $F$ using the method of dimensions.

$$\begin{gathered} Let \\ F=k\left(m{)}^x\right(v{)}^y(r{)}^z\dots \text{ (i) } \end{gathered}$$

Here, k is a dimensionless constant of proportionality. Writing the dimensions of RHS and LHS in Eq. (i), we have

$\left[{\mathrm{MLT}}^{-2}\right]=\left[\mathrm{M}{]}^x{\left[{\mathrm{LT}}^{-1}\right]}^y\right[ \mathrm{L}{]}^z=\left[{\mathrm{M}}^x{ \mathrm{L}}^{y+2}{ \mathrm{T}}^{-y}\right]$

Equating the powers of M, L and T of both sides, we have, 

$$\begin{gathered} x=1,y=2 \\ \text{ and } \\ y+z=1 \\ \text{ or } \\ z=1-y=-1 \end{gathered}$$

Putting the values in Eq. (i), we get

$F=km{v}^2{r}^{-1}=k\frac{m{v}^2}{r}$

or

$F=\frac{m{v}^2}{r}\text{ (where, }k=1\text{ ) }$