Question #274199

Suppose we are told that the acceleration (a) of a particle moving with uniform speed (v)


in a circle of radius (r) is a=kr^ n v^ prime


Determine the values of n and m and write the simplest form of this equation.


Note: K is a dimensionless constant


"Show your work"


1
Expert's answer
2021-12-02T10:05:47-0500

According to the principle of homogeneity in any physical formula, the dimensions of each term are the same.

Dimension of acceleration

a=LT2a = L’T^{-2}

Dimension of velocity

V=LT1V=LT^{-1}

Given, that acceleration of the particle moving along path is

a=krnvm=Ln(LT1)m=Lm+mTma=LT2=Ln+mTma=kr^nv^m =L^n(LT^{-1})^m = L^{m+m} T^{-m} \\ a =L’T^{-2} =L^{n+m} T^{-m}

Then, according to the principle of homogeneity, this dimensional equation is balanced under the conditions

1=n+m2=mm=2n=1m=12=11 = n+m \\ -2= -m \\ m=2 \\ n= 1-m = 1-2 = -1

Then, we can write the acceleration expression as

a=krnvm=kr1v2=kv2ra=kr^nv^m = kr^{-1}v^2 =k\frac{v^2}{r}

For uniform circular motion, k=1.

Then simplest form of the equation is,

a=v2ra = \frac{v^2}{r}


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