Question

Show dimensionally that the centripetal force acting on a particle of mass m moving in a circle of radius r with a uniform speed of v
rotations per second is 4T v2 mr.
​

Answer 1

Hey Dear,

◆ Answer -

F = mv^2/r

◆ Explaination -

First, we'll write down known dimensions of given quantities.

[F] = [L1M1T-2]

[v] = [L1T-1]

[m] = [M]

[r] = [L]

Let x, y & z are numbers such that F = k.v^x.m^y.r^z

In dimensional form, this can be written as -

[F] = [v]^x.[m]^y. [r]^z

[L1M1T-2] = [L1T-1]^x.[M]^y.[L]^z

[L1M1T-2] = [L^(x+z).M^(y).T^(-x)

Comparing indexes on both sides -

x + z = 1

y = 1

-x = -2

Solving these equations,

x = 2

y = 1

z = -1

Hence,

F = k.ν^x.m^y.r^z

F = k × v^2 × m^1 × r^-1

F = k.v^2.m/r

F = mv^2/r ...(k=1)

Thus, centripetal force of the particle in UCM is mv^2/r.

Thanks for asking..