Question

Find centre of mass of an object (Paraboloid) which is formed by rotating a parabola $x=ky^2$ about x-axis and height of object is $h$ as shown in figure-4.38. Assume the object is of uniform density.

Answer 1

The center of mass of the paraboloid is located at $(\frac{2h}{3}, 0, 0)$.

Answer 2

The object is a paraboloid formed by rotating the parabola $x = ky^2$ about the x-axis, from $x=0$ to $x=h$. We consider a thin disc of thickness $dx$ at a position $x$. The radius of this disc is $y$. From the equation of the parabola, $y^2 = \frac{x}{k}$. The area of the cross-section at $x$ is $A(x) = \pi y^2 = \pi \frac{x}{k}$. The volume of the elemental disc is $dV = A(x) dx = \pi \frac{x}{k} dx$. Assuming uniform density $\rho$, the mass of the elemental disc is $dm = \rho dV = \rho \pi \frac{x}{k} dx$. The x-coordinate of the center of mass ($X_{cm}$) is given by the formula: $X_{cm} = \frac{\int x dm}{\int dm}$ The integration is performed from $x=0$ to $x=h$.

Numerator: $\int x dm = \int_{0}^{h} x \left( \rho \pi \frac{x}{k} dx \right) = \frac{\rho \pi}{k} \int_{0}^{h} x^2 dx = \frac{\rho \pi}{k} \left[ \frac{x^3}{3} \right]_{0}^{h} = \frac{\rho \pi}{k} \frac{h^3}{3}$

Denominator: $\int dm = \int_{0}^{h} \left( \rho \pi \frac{x}{k} dx \right) = \frac{\rho \pi}{k} \int_{0}^{h} x dx = \frac{\rho \pi}{k} \left[ \frac{x^2}{2} \right]_{0}^{h} = \frac{\rho \pi}{k} \frac{h^2}{2}$

Calculating $X_{cm}$: $X_{cm} = \frac{\frac{\rho \pi}{k} \frac{h^3}{3}}{\frac{\rho \pi}{k} \frac{h^2}{2}} = \frac{h^3/3}{h^2/2} = \frac{2h}{3}$

Due to the symmetry of the object formed by rotation about the x-axis, the center of mass will lie on the x-axis. Therefore, $Y_{cm} = 0$ and $Z_{cm} = 0$.