Question

Using dimensional analysis, derive an expression for centripetal force which may depend on mass m of a body moving with uniform velocity v in circular path of radius r?

Answer 1

Let centripetal force depend on mass m, velocity v and radius r as follows,

$\longrightarrow\sf{F\propto m^x\ v^y\ r^z\quad\quad\dots(1)}$

Taking dimensions of each term,

$\longrightarrow\sf{[F]=[m]^x\ [v]^y\ [r]^z}$

$\longrightarrow\sf{MLT^{-2}=M^x\ (LT^{-1})^y\ L^z}$

$\longrightarrow\sf{MLT^{-2}=M^x\ L^{y+z}\ T^{-y}}$

Equating the corresponding powers

$\longrightarrow\sf{\underline{\underline{x=1}}}$

$\longrightarrow\sf{-y=-2\ \implies\ \underline{\underline{y=2}}}$

$\longrightarrow\sf{y+z=2+z=1\ \implies\ \underline{\underline{z=-1}}}$

Then (1) becomes,

$\longrightarrow\sf{F\propto m^1\ v^2\ r^{-1}}$

$\longrightarrow\sf{\underline{\underline{F\propto\dfrac{mv^2}{r}}}}$

$\longrightarrow\sf{F=k\cdot\dfrac{mv^2}{r}}$

where 'k' is a constant. But practically, k = 1. Hence,

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

Answer 2

Answer:

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