Question

The expression for centripetal force $(F)$ depends upon mass of body $(m)$, speed $(v)$ of the body and the radius ( $r$ ) of circular path will be expression for centripetal force

• A. $F=\frac{m{v}^2}{2r^3}$
• B. $F=\frac{m{v}^2}{r}$
• C. $F=\frac{m{v}^2}{r^2}$
• D. $F=\frac{m^2{v}^2}{2r}$

Answer 1

Answer: (B)

According to question,
$F\propto m^a{v}^b{r}^c$
$F=k{m}^a{v}^b{r}^c$
$k$, being a dimensionless constant.
From homogeneity of dimensions,
$LHS=RHS$
$\left[ML{T}^{-2}\right]=[M{]}^a{\left[L{T}^{-1}\right]}^b[L{]}^c$
Or $\left[ML{T}^{-2}\right]=\left[M^a{L}^{b+c}{T}^{-b}\right]$
Comparing the powers, we obtain
$a=1$
$b+c=1$
$-b=-2\Rightarrow b=2$
$\therefore 2+c=1$
$\Rightarrow c=-1$
Therefore, $F=km{v}^2{r}^{-1}=\frac{km{v}^2}{r}$.
The experimental value of $k$ is found to be here
$\therefore F=\frac{m{v}^2}{r}$