Question

check the dimensions of the equation a=v²/r where a is acceleration v is velocity r is radius
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Answer 1

Answer

Given equation is

$\rm{a = \dfrac{v^2}{r}}$

where a is acceleration, v is velocity and r is the radius.

We need to check the correctness of the given equation.

So,

the dimensions of acceleration 'a' are $\rm{[M^0 L^1 T^{-2}]}$

dimensions of velocity 'v' are $\rm{ [M^0 L^1 T^{-1}]}$

dimensions of the radius (length) 'r' are $\rm{[M^0 L^1 T^0]}$

Now, the given equation is

$\implies\rm{ a = \dfrac{v^2}{r}}$

Solving dimentionally

$\implies\rm{ [M^0 L^1 T^{-2}] = \dfrac{[M^0 L^1 T^{-1}]^2}{[M^0 L^1 T^0]}}$

$\implies\rm{ [M^0 L^1 T^{-2}] = \dfrac{[M^0 L^2 T^{-2}]}{[M^0 L^1 T^0]}}$

$\implies\rm{ [M^0 L^1 T^{-2}] = [M^{(0-0)} L^{(2-1)} T^{(-2-0)}]}$

$\implies\rm{ [M^0 L^1 T^{-2}] = [M^0 L^1 T^{-2}]}$

LHS = RHS

Hence, the equation $\rm{a = \dfrac{v^2}{r} }$ is correct.

Answer 2

Given equation:-

$a = \frac{v^2}{r}$ (centripetal acceleration)

LHS:-

Dimensions of 'a' = $M^0L^1T^{-2}$

RHS:-

Dimensions of 'v' = $\boxed{ M^0L^1T^{-1}}$

∴ Dimensions of 'v' squared:-

$(M^0L^1T^{-1})^2 = M^0L^2T^{-2}$

Dimensions of 'r' = $M^0L^1T^0$

∴ Dimensions of v²/r = $\frac{M^0L^2T^{-2}}{M^0L^1T^0}$

= $\boxed{M^0L^1T^{-2} }$ = LHS.

Hence, the dimensions are correct in the given equation and it is correct.

More:-

1. Dimensional units of,

• Mass = M
• Length = L
• Time = T
• Temperature = K
• Degrees = θ
• Current = A.

1. Centripetal force = $\boxed{\frac{mv^2}{r}}$.
2. Centripetal force helps in attraction whereas centrifugal, it's opposite.