Question

A solid hemisphere of radius R has a variable volume density $\rho = \rho_0 \frac{r}{R}$, where r is the distance from the center of hemisphere. The distance of center of mass of the hemisphere from the center is :-

• A. $\frac{R}{3}$
• B. $\frac{2R}{9}$
• C. $\frac{2R}{5}$
• D. $\frac{3R}{8}$

Answer 1

Answer: (C)

2R/5

Answer 2

To find the center of mass of the hemisphere, we will use spherical coordinates due to the spherical symmetry of the object and the density function.

Let the center of the hemisphere be at the origin (0,0,0). Assume the flat face of the hemisphere lies in the xy-plane and the curved surface extends into the positive z-axis (i.e., $z \ge 0$).

The density is given by $\rho = \rho_0 \frac{r}{R}$, where $r$ is the distance from the center.

In spherical coordinates, a differential volume element is $dV = r^2 \sin\theta dr d\theta d\phi$. The limits for a hemisphere with its flat face in the xy-plane and curved part in $z \ge 0$ are:

• Radius $r$: from 0 to R
• Polar angle $\theta$: from 0 to $\pi/2$ (from positive z-axis to xy-plane)
• Azimuthal angle $\phi$: from 0 to $2\pi$ (full circle around z-axis)

The differential mass element $dm$ is $\rho dV$: $dm = \left(\rho_0 \frac{r}{R}\right) (r^2 \sin\theta dr d\theta d\phi) = \frac{\rho_0}{R} r^3 \sin\theta dr d\theta d\phi$.

Due to symmetry, the center of mass will lie on the z-axis. So, we only need to calculate the z-coordinate of the center of mass, $Z_{CM}$. The formula for $Z_{CM}$ is: $Z_{CM} = \frac{\int z dm}{\int dm}$

1. Calculate the total mass (M): $M = \int dm = \int_0^{2\pi} \int_0^{\pi/2} \int_0^R \frac{\rho_0}{R} r^3 \sin\theta dr d\theta d\phi$ We can separate the integrals: $M = \frac{\rho_0}{R} \left( \int_0^R r^3 dr \right) \left( \int_0^{\pi/2} \sin\theta d\theta \right) \left( \int_0^{2\pi} d\phi \right)$

Evaluate each integral: $\int_0^R r^3 dr = \left[ \frac{r^4}{4} \right]_0^R = \frac{R^4}{4}$ $\int_0^{\pi/2} \sin\theta d\theta = \left[ -\cos\theta \right]_0^{\pi/2} = -\cos(\pi/2) - (-\cos(0)) = 0 - (-1) = 1$ $\int_0^{2\pi} d\phi = \left[ \phi \right]_0^{2\pi} = 2\pi$

Substitute these values back into the expression for M: $M = \frac{\rho_0}{R} \left( \frac{R^4}{4} \right) (1) (2\pi) = \frac{\rho_0 \pi R^3}{2}$

2. Calculate the moment of mass along the z-axis ($\int z dm$): In spherical coordinates, $z = r \cos\theta$. $\int z dm = \int_0^{2\pi} \int_0^{\pi/2} \int_0^R (r \cos\theta) \left( \frac{\rho_0}{R} r^3 \sin\theta \right) dr d\theta d\phi$ $\int z dm = \frac{\rho_0}{R} \left( \int_0^R r^4 dr \right) \left( \int_0^{\pi/2} \sin\theta \cos\theta d\theta \right) \left( \int_0^{2\pi} d\phi \right)$

Evaluate each integral: $\int_0^R r^4 dr = \left[ \frac{r^5}{5} \right]_0^R = \frac{R^5}{5}$ $\int_0^{\pi/2} \sin\theta \cos\theta d\theta$: Let $u = \sin\theta$, then $du = \cos\theta d\theta$. When $\theta=0, u=0$; when $\theta=\pi/2, u=1$. $\int_0^1 u du = \left[ \frac{u^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}$ $\int_0^{2\pi} d\phi = 2\pi$

Substitute these values back into the expression for $\int z dm$: $\int z dm = \frac{\rho_0}{R} \left( \frac{R^5}{5} \right) \left( \frac{1}{2} \right) (2\pi) = \frac{\rho_0 \pi R^4}{5}$

3. Calculate $Z_{CM}$: $Z_{CM} = \frac{\int z dm}{M} = \frac{\frac{\rho_0 \pi R^4}{5}}{\frac{\rho_0 \pi R^3}{2}}$ $Z_{CM} = \frac{\rho_0 \pi R^4}{5} \times \frac{2}{\rho_0 \pi R^3} = \frac{2R}{5}$

The distance of the center of mass of the hemisphere from the center is $\frac{2R}{5}$.