Using the method of dimensions find the acceleration of a particle mov

1.	Using the method of dimensions find the acceleration of a particle moving with a constant speed v in a circle of radius r.
Answer» \(a \propto v-r\) \(a = kvr\) where a = acceleration v = velocity r = radius \(a = kv^ar^b\) ....(1) \([M^0LT^{-2}] = k[v]^a [r]^b\) \([M^0LT^{-2}] = k[LT^{-1}]^a [L]^b\) \([M^0LT^{-2}] = k[L^{a + b} T^{-a}]\) Comparing power \(a + b = 1\) \(a = 2\) \(b = 1 -2\) \(b= -1\) Then these value put in equation (1) \(a = kv^2r^{-1}\) \(a =\frac{v^2}r\)

Using the method of dimensions find the acceleration of a particle moving with a constant speed v in a circle of radius r.

Answer»

\(a \propto v-r\)

\(a = kvr\)

where

a = acceleration

v = velocity

r = radius

\(a = kv^ar^b\) ....(1)

\([M^0LT^{-2}] = k[v]^a [r]^b\)

\([M^0LT^{-2}] = k[LT^{-1}]^a [L]^b\)

\([M^0LT^{-2}] = k[L^{a + b} T^{-a}]\)

Comparing power

\(a + b = 1\)

\(a = 2\)

\(b = 1 -2\)

\(b= -1\)

Then these value put in equation (1)

\(a = kv^2r^{-1}\)

\(a =\frac{v^2}r\)