Question

a particle of mass m is moving with a velocity v in a circle of radius r . find the expression for centripetal force by using dimensional analysis

Answer 1

Answer:

Let;

$\longrightarrow\sf [F] \propto [m]^x [v]^y [r]^z$

$\longrightarrow\sf [F] = k [m]^x [v]^y [r]^z \: \: \: ...(1)$

• k is dimensionless constant.

Dimensions;

•Centripetal Force =$\sf [M^1L^1{T}^{ - 2} ]$

•mass = $\sf [M^1L^0{T}^{0} ]$

•velocity = $\sf [M^0L^1{T}^{ - 1} ]$

•Radius = $\sf [M^0L^1{T}^{0} ]$

Writing the dimensions of RHS and LHS we have:

$\longrightarrow\sf [M^{1} L^{1} T^{-2}] = [M^1L^0T^0]^x [M^0L^1{T}^{ - 1}]^y [M^0L^1{T}^{0}]^z$

$\longrightarrow\sf [M^{1} L^{1} T^{-2}] = [M^xL^{y + z}T^{ - y} ]$

On comparing the powers of LHS and RHS we have :

$\longrightarrow\bf x = 1 \: \: ...(i)$

$\longrightarrow\sf - y= - 2$

$\longrightarrow\bf y= 2 \: \: ...(ii)$

$\longrightarrow\sf y + z= 1$

$\longrightarrow\sf 2 + z= 1$

$\longrightarrow\sf z= 1 - 2$

$\longrightarrow\bf z= - 1 \: \:... (iii)$

By plugging this three equations in equation (1) we have:

$\longrightarrow\sf [F] = k [m]^1 [v]^2 [r]^{ - 1}$

$\longrightarrow\sf [F] = \dfrac{[m]^1 [v]^2 }{[r]^{ 1} }$

$\longrightarrow \underline{ \boxed{\bf F = \dfrac{mv^2 }{r }}}$