Question

expression on centripetal force using dimension method ​

Answer 1

Let the centripetal force F acting on the particle in uniform circular motion depend upon the mass of the particle (m), speed of the particle (v) and radius of the circular path (r) as follows:

$\quad$

$\sf{F\propto m^xv^yr^z}\\\\\\\sf{F=k\cdot m^xv^yr^z\quad\longrightarrow\quad (1)}$

$\quad$

Taking dimensions of each term,

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$\sf{[F]=[m]^x[v]^y[r]^z}\\\\\\\sf{M^1L^1T^{-2}=M^x\left(L^1T^{-1}\right)^yL^z}\\\\\\\sf{M^1L^1T^{-2}=M^xL^{y+z}T^{-y}}$

$\quad$

Equating similar powers on both sides of the equation,

$\quad$

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

$\quad$

Then (1) becomes,

$\quad$

$\sf{F=k\cdot m^1v^2r^{-1}}\\\\\\\large\boxed {\sf{F=k\cdot\dfrac {mv^2}{r}}}$

$\quad$

Experimentally, $\sf{k=1}.$ Then,

$\quad$

$\large\boxed {\sf{F=\dfrac {mv^2}{r}}}$