Question

What are the units or dimensions of the shear rate dv/dy (English units)? Then, what are the dimensions of the shear stress τ= μ*dV/dy? Then, by dimensional analysis, show that the shear stress has the same units as momentum divided by (area*time).What are the unit or dimensions of viscosity?

Answer 1

Answer:

1) Dimensions of shear rate is $[T^{-1}]$ .

2)Dimensions of shear stress are $[ML^{-1}T^{-2}]$

Explanation:

Since the dimensions of velocity 'v' are $[LT^{-1}]$ and the dimensions of distance 'y'  are $[L]$ , thus the dimensions of $\frac{dv}{dy}$ become

$\frac{[LT^{-1}]}{[L]}=[T^{-1}]$ and hence the units become $s^{-1}$.

Now we know that the dimensions of coefficient of dynamic viscosity $\mu$ are $[ML^{-1}T^{-1}]$ thus the dimensions of shear stress can be obtained from the given formula as

$[\tau ]=[ML^{-1}T^{-1}]\times [T^{-1}]\\\\[\tau ]=[ML^{-1}T^{-2}]$

Now we know that dimensions of momentum are $[MLT^{-1}]$

The dimensions of $Area\times time$ are $[L^{2}T]$

Thus the dimensions of $\frac{Moumentum}{Area\times time}=\frac{MLT^{-1}}{L^{2}T}=[MLT^{-2}]$

Which is same as that of shear stress. Hence proved.