Answer to Question #274199 in Mechanics | Relativity for Ibrahim

Question

Suppose we are told that the acceleration (a) of a particle moving with uniform speed (v)

in a circle of radius (r) is a=kr^ n v^ prime

Determine the values of n and m and write the simplest form of this equation.

Note: K is a dimensionless constant

"Show your work"

Answer 1

According to the principle of homogeneity in any physical formula, the dimensions of each term are the same.

Dimension of acceleration

$a = L’T^{-2}$

Dimension of velocity

$V=LT^{-1}$

Given, that acceleration of the particle moving along path is

$a=kr^nv^m =L^n(LT^{-1})^m = L^{m+m} T^{-m} \\ a =L’T^{-2} =L^{n+m} T^{-m}$

Then, according to the principle of homogeneity, this dimensional equation is balanced under the conditions

$1 = n+m \\ -2= -m \\ m=2 \\ n= 1-m = 1-2 = -1$

Then, we can write the acceleration expression as

$a=kr^nv^m = kr^{-1}v^2 =k\frac{v^2}{r}$

For uniform circular motion, k=1.

Then simplest form of the equation is,

$a = \frac{v^2}{r}$