A liquid of density `rho` is in a bucket that spins with angular velocity `'omega'` as shown in figure. Prove that the free surface of the liquid has a parabolic shape. Find equation of this.
.

Answer

A

Explanation

In the figure shown, suppose the coordinates of point `M` are `(x,y)` with respect to the coordinates axes.

Then, `MN=y, NQ=x`
Points `M` and `Q` are open to atmosphere.
`:. p_(M)=p_(Q)=`atmosphric pressure p_(0)`
Now, p_(M)-p_(N)=-rhogy`
`implies p_(0)-p_(N)=-rhogy` ...(i)
`p_(N)-p_(Q)=(rho omega^(2)x^(2))/(2)`
`:. p_(N)-p_(0)=(rho omega^(2)x^(2))/(2) `...(ii)
Adding Eqs. (i) and (ii), we have
`0=-rhogy+(rho omega^(2)x^(2))/(2)`
`y=(omega^(2)x^(2))/(2g)`
This is required equation of free surface of the liquid and we can see that this is an equation of a parabola.