Question

A solid object is generated by the rotation of a parabola as shown in the figure. Assuming that the height of object is ℎ as shown in figure, the location of centre of mass of such a paraboloid (from 𝑂) of uniform density formed by rotating a parabola $y=k{x}^2$ about 𝑥−𝑎𝑥𝑖𝑠 is

• A. $y_{COM}=\frac{h}{6}$
• B. $y_{COM}=\frac{h}{2}$
• C. $y_{COM}=\frac{h}{3}$
• D. $y_{COM}=\frac{2h}{3}$

Answer 1

The correct option is D $y_{COM}=\frac{2h}{3}$

Given, parabolic equation
$y=k{x}^2\Rightarrow x^2=\frac{y}{k}$
Let ρ be the volume density of the paraboloid.
Consider an infinitesimal disc of radius $x$, width $dy$, mass $dm$ at a distance $y$ from O.


$\Rightarrow dm=\rho \left(\pi x^2\right)dy=\pi \rho (\frac{y}{k})dy$

$\Rightarrow y_{COM}=\frac{\int ydm}{\int dm}=\frac{{\int }_0^h\pi \rho (\frac{y^2}{k}dy)}{{\int }_0^h\pi \rho (\frac{y}{k})dy}=\frac{[\frac{y^3}{3}{]}_0^h}{[\frac{y^2}{2}{]}_0^h}$

$\Rightarrow y_{COM}=\frac{2h}{3}$