Question

check correctness of equation F=mv²/r by by the method of dimensions where m is the mass of the body v is the linear velocity r the radius and F the centripetal force explain​

Answer 1

Answer:

The equation is 100% correct.

Explanation:

The working is as follows:

To check the correctness of the equation, both sides must have the same dimensions.

Basic Dimensions:

Mass = [M]

Length = [L]

Time = [T]

Dimensions of Force (L.H.S.):

By Newton's second law:

F = ma

[F] = [M] [LT^-2]

[F] = [MLT^-2]

Dimensions of mv^2/r (R.H.S.):

[m] = [M]

[v]^2 = [LT^-1]^2 = [L^2 T^-2]

[r] = [L]

[mv^2/r] = [M] [L^2 T^-2]/[L]

[mv^2/r] = [ML^2T^-2]/[L]

[mv^2/r] = [MLT^-2]

Since L.H.S. and R.H.S. are equal so, the formula is correct

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Answer 2

Answer:

The equation F=mv²/r is dimensionally correct by the method of dimensions.

Explanation:

By the principle of homogeneity, one equation is dimensionally correct only when the LHS and RHS of the equation have the same dimensions.

Basic dimensions  are,

Mass = [M]

Length = [L]

Time = [T]

Given equation is,

   $\mathrm{F}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$    Where F - Centrepital force

                                m - Mass of the body

                                 r - radius

                                 v- linear velocity

Consider LHS of the equation,

By Newton's first law,

Force =  mass  $\times$  acceleration

Or F = ma  

[m] = $[M^{1} L^{0} T^{0} ]$

acceleration,  a =  $\frac{Velocity}{Time}$ = $\frac{distance}{time} \times \frac{1}{Time}$  

So ,    

[a] = $\frac{L}{T} \times \frac{1}{T} = [M^{0} L^{1} T^{-2}]$

Dimension of LHS is,

[ LHS ]=   [F] = [$\mathrm{M^{1} L^{1} T}^{-2}$] ...(1)

Consider the RHS,

$\text { R.H.S }=\frac{m v^{2}}{r} , \[ [R . H . S]=\frac{[m] [v]^{2}}{[r]}$

[RHS] =  $\frac{\left[\mathrm{L}^{0} \mathrm{M}^{1} \mathrm{~T}^{0}\right]\left[\mathrm{L}^{1} \mathrm{M}^{0} \mathrm{~T}^{-1}\right]^{2}}{\left[\mathrm{L}^{1} \mathrm{M}^{0} \mathrm{~T}^{0}\right]}$         { Since, velocity = $\frac{distance}{time}$   [v] = $M^{0} LT^{-1}$ }

          =  $\left[L^{0} M^{1} T^{0}\right]\left[L^{2} M^{0} T^{-2}\right]\left[L^{-1} M^{0} T^{0}\right]$

          = $\left[\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{~T}^{-2}\right]$ ....(2)

From equation(1) and (2),  RHS = LHS

Hence by the principle of homogeneity, the given equation is dimensionally correct.