Question

Show dimensionally that the centripetal force acting on a particle of mass m moving in a circle of radius r moving with a velocity v is F=mv²/r

Answer 1

Answer:

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. Derivation of the formula for 'F' using the method of dimensions is based on the principle of comparing the dimesions of the. Fundamental unit and the derived formula.

Answer 2

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HERE IS UR ANSWER::-

Hey Dear,

◆ Answer -

F = mv^2/r

◆ Explaination -

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

[F] = [L1M1T-2]

[F] = [L1M1T-2][v] = [L1T-1]

[F] = [L1M1T-2][v] = [L1T-1][m] = [M]

[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

[F] = [v]^x.[m]^y. [r]^z[L1M1T-2] = [L1T-1]^x.[M]^y.[L]^z

[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 -

Comparing indexes on both sides -x + z = 1

Comparing indexes on both sides -x + z = 1y = 1

Comparing indexes on both sides -x + z = 1y = 1-x = -2

Solving these equations,

Solving these equations,x = 2

Solving these equations,x = 2y = 1

Solving these equations,x = 2y = 1z = -1

Hence,

Hence,F = k.ν^x.m^y.r^z

Hence,F = k.ν^x.m^y.r^zF = k × v^2 × m^1 × r^-1

Hence,F = k.ν^x.m^y.r^zF = k × v^2 × m^1 × r^-1F = k.v^2.m/r

Hence,F = k.ν^x.m^y.r^zF = k × v^2 × m^1 × r^-1F = k.v^2.m/rF = mv^2/r ...(k=1)

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

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