Question

Centripetal force 'F' acting on a particle of mass 'm' moving in a circle of radius 'r' with velocity 'v' depends upon m1v and r, find expression for F.​

Answer 1

Answer:

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Answer 2

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Given :- }}}}$

• Centripetal force 'F' acting on a particle of mass 'm' moving in a circle of radius 'r' with velocity 'v' depends upon m , v and r

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ To Find :- }}}}$

• The expression for force .

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Solution :- }}}}$

Given that , force acting on a particle depends upon mass of the particle , radius of the track and the velocity of the particle . And we need to find the expression of the force .

Let the expression ,

$\sf\longrightarrow Force = a [ m^x v^y r^z ] \\\\\\\sf\longrightarrow Force =[m^x] [v^y ] [r^z] </p><p>$

where ,

• a is dimensionless constant.
• x , y and z are the powers on m , v and r respectively .

We know that the Dimensional Formula of Force is ,

$\sf\longrightarrow\red{ Force = [ M^1 L^1 T^{-2}] }$

Now , by the Principal of Homogeneity ,

$\sf\longrightarrow [ M^1 L^1 T^{-2}] = [m^x] [v^y ] [r^z] \\\\\\\sf\longrightarrow [M^1L^1T^{-2}] = [ M]^x [ LT^{-1}]^y [L]^z \\\\\\\sf\longrightarrow [M^1L^1T^{-2}] = [ M^x L^{(y+z)} T^{-y} ]$

On comparing ,

$\sf\longrightarrow x = 1$

$\sf\longrightarrow y + z = 1$

$\sf\longrightarrow -y = -2$

On solving ,

$\longrightarrow\textsf{\textbf{ x = 1}}$

$\longrightarrow\textsf{\textbf{ y = 2}}$

$\longrightarrow\textsf{\textbf{ z= -1}}$

Therefore , the expression for the Force is ,

$\sf\longrightarrow Force = a [ m^x v^y r^z ] \\\\\\\sf\longrightarrow Force = a [ m^1 v^2 r^{-1} ] \\\\\\\sf\longrightarrow Force = 1 [ m^1 v^2 r^{-1} ] \qquad \bigg\{ \red{\sf The \ Dimensionless \ constant \ here \ is \ 1 .}\bigg\} \\\\\\\sf\longrightarrow \underset{\blue{\sf Centripetal\ Force }}{\underbrace{\boxed{\pink{\frak{ Force_{(Centripetal)} =\dfrac{mv^2}{r}}}}}}$

$\rule{200}2$