First slide

Dimensions of physical quantities

Question

The centripetal force $F$ acting on a particle moving uniformly in a circle may depend upon mass $m$ , velocity $v$ and radius $r$ of the circle. Derive the formula for $F$ using the method of dimensions.

Moderate

A

$F = \frac{m v^{3}}{r}$
B

$F = \frac{m v^{2}}{r}$
C

$F = \frac{m v^{- 2}}{r}$
D

$F = \frac{m v^{4}}{r}$

Solution

$L e t F = k (m)^{x} (v)^{y} (r)^{z} \dots (i)$
Here, k is a dimensionless constant of proportionality. Writing the dimensions of RHS and LHS in Eq. (i), we have
$[{MLT}^{- 2}] = [M]^{x} {[{LT}^{- 1}]}^{y} [L]^{z} = [M^{x} L^{y + 2} T^{- y}]$
Equating the powers of M, L and T of both sides, we have,
$x = 1, y = 2 and y + z = 1 or z = 1 - y = - 1$
Putting the values in Eq. (i), we get
$F = k m v^{2} r^{- 1} = k \frac{m v^{2}}{r}$
or
$F = \frac{m v^{2}}{r} (where, k = 1)$

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