Question

Use dimensional analysis (Section 1-8) to obtain the form for the centripetal acceleration,\({a_R} = {v^2}/r\).

Short answer

The centripetal or radial acceleration is \(\left( {\frac{{{v^2}}}{r}} \right) = \;\frac{L}{{{T^2}}}\; = \left( {{a_R}} \right)\).

Step-by-step solution

Find the centripetal acceleration \({{\bf{a}}_{\bf{R}}} = \;{{\bf{v}}^{\bf{2}}}/{\bf{r}}\).

It is known that the dimension of the acceleration, which is the right-hand side of the given equation, is equal to:

\(\left( {{a_R}} \right) = \frac{L}{{{T^2}}}\)

The speed is given by:

\(v = \;\frac{{{\rm{Distance}}}}{{{\rm{Time}}}}\)

Find dimensions of the speed

Now, the dimension of the speed is\(\left( v \right) = \frac{L}{T}\), and the radius speed has a dimension of\(\left( r \right) = L\). So, the dimension of the left-hand side is equal to:

\(\begin{aligned}\left( {\frac{{{v^2}}}{r}} \right) &= \frac{{\frac{{{L^2}}}{{{T^2}}}}}{L}\\ &= \frac{L}{{{T^2}}}\end{aligned}\)

Therefore,

\(\left( {{a_R}} \right) = \;\left( {\frac{{{v^2}}}{r}} \right)\)

So, the equation is homogenous.