Question

If an object of mass 'm' moves in uniform circular motion, a force F acts on it, whose direction is always towards the centre of the circular path. If F depends on m, speed v of the object and radius r of the circular path, Find an expression for F.​

Answer 1

Answer:

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

Explanation:

We are given That,

• F depends upon mass, m
• speed, v
• radius ,r of the circular path.

So, using DIMENSIONS of M,V,R,F

and DIMENSIONAL ANALYSIS the solution is in the adjoining figure

$\: \: \: \: \: !! \: ENJOY \: !!$

Answer 2

AnswEr:

$\huge{\boxed{\purple{\sf{F\:=\:\dfrac{mv^2}{r}}}}}$

ExplanaTion:

Let the expression be,

$\longrightarrow$ $\sf{F\:=\:am^{x} v^{y} r^{z}}$

Where, a is dimensionless constant.

• Power on m = x
• Power on v = y
• Power on r = z

Now, according to the principle of dimensional homogeneity,

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

: $\implies$ $\sf{[M^{1} L^{1} T^{-2}] = [M]^{x} [LT^{-1}]^{y} [L]^{2}}$

: $\implies$ $\sf{[M^{1} L^{1} T^{-2}] = [M^{x} L^{y+z} T^{-y}]}$

On equating dimensions, we get,

★ $\large{\boxed{\boxed{\sf{\blue{x\:=\:1}}}}}$

★ $\large{\boxed{\boxed{\sf{\pink{y\:=\:2}}}}}$

★ $\large{\boxed{\boxed{\sf{\green{z\:=\:-1}}}}}$

Expression for F :

: $\implies$ $\sf{F\:=\:amv^{2} r^{-1}}$

: $\implies$ $\sf{F\:=\:\dfrac{amv^2}{r}}$

Here, a is constant, say 1. Hence, our required expression is :

$\therefore$ $\huge{\boxed{\purple{\sf{F\:=\:\dfrac{mv^2}{r}}}}}$

The force F can be said as Centripetal force.