Question

find the centre of mass of 1/4 th of hollow hemisphere

• A. $(R/2, R/2, R/2)$
• B. $(R/8, R/8, R/8)$
• C. $(R/4, R/4, R/4)$
• D. $(R/2, 0, 0)$

Answer 1

Answer: (A)

The centre of mass of 1/4th of a hollow hemisphere is at $(\frac{R}{2}, \frac{R}{2}, \frac{R}{2})$.

Answer 2

The center of mass of 1/4th of a hollow hemisphere is found by integrating the position vector over the specified region.

Let the hollow hemisphere be defined in spherical coordinates by $R$ (constant radius), $0 \le \theta \le \pi/2$ (polar angle), and $0 \le \phi < 2\pi$ (azimuthal angle). The surface area of the full hemisphere is $A_{hemi} = 2\pi R^2$.

"1/4th of a hollow hemisphere" implies a portion with 1/4 of its surface area. This is achieved by restricting the azimuthal angle. We consider the region defined by $0 \le \theta \le \pi/2$ and $0 \le \phi \le \pi/2$. The surface area of this region is $A_{1/4} = R^2 \int_0^{\pi/2} \int_0^{\pi/2} \sin\theta d\theta d\phi = R^2 [-\cos\theta]_0^{\pi/2} [\phi]_0^{\pi/2} = R^2 (1)(\pi/2) = \frac{\pi R^2}{2}$. This is indeed $1/4$ of the hemisphere's area.

Let $\sigma$ be the uniform surface mass density. The mass element is $dm = \sigma dS = \sigma R^2 \sin\theta d\theta d\phi$. The total mass of this 1/4th part is $m = \sigma A_{1/4} = \sigma \frac{\pi R^2}{2}$.

The position vector in spherical coordinates is $\vec{r} = R \sin\theta \cos\phi \hat{i} + R \sin\theta \sin\phi \hat{j} + R \cos\theta \hat{k}$.

The coordinates of the center of mass $(\vec{r}_{COM})$ are given by: $\vec{r}_{COM} = \frac{1}{m} \int \vec{r} dm$

$x_{COM} = \frac{1}{m} \int_0^{\pi/2} \int_0^{\pi/2} (R \sin\theta \cos\phi) (\sigma R^2 \sin\theta d\theta d\phi)$ $y_{COM} = \frac{1}{m} \int_0^{\pi/2} \int_0^{\pi/2} (R \sin\theta \sin\phi) (\sigma R^2 \sin\theta d\theta d\phi)$ $z_{COM} = \frac{1}{m} \int_0^{\pi/2} \int_0^{\pi/2} (R \cos\theta) (\sigma R^2 \sin\theta d\theta d\phi)$

Substituting $m = \sigma \frac{\pi R^2}{2}$, we get $\frac{\sigma R^2}{m} = \frac{2}{\pi}$.

Calculating $x_{COM}$: $x_{COM} = \frac{2}{\pi} \int_0^{\pi/2} \int_0^{\pi/2} R^2 \sin^2\theta \cos\phi d\theta d\phi$ $x_{COM} = \frac{2R^2}{\pi} \left( \int_0^{\pi/2} \sin^2\theta d\theta \right) \left( \int_0^{\pi/2} \cos\phi d\phi \right)$ $\int_0^{\pi/2} \sin^2\theta d\theta = \frac{\pi}{4}$ and $\int_0^{\pi/2} \cos\phi d\phi = 1$. $x_{COM} = \frac{2R^2}{\pi} \left( \frac{\pi}{4} \right) (1) = \frac{R}{2}$

Calculating $y_{COM}$: $y_{COM} = \frac{2}{\pi} \int_0^{\pi/2} \int_0^{\pi/2} R^2 \sin^2\theta \sin\phi d\theta d\phi$ $y_{COM} = \frac{2R^2}{\pi} \left( \int_0^{\pi/2} \sin^2\theta d\theta \right) \left( \int_0^{\pi/2} \sin\phi d\phi \right)$ $\int_0^{\pi/2} \sin\phi d\phi = 1$. $y_{COM} = \frac{2R^2}{\pi} \left( \frac{\pi}{4} \right) (1) = \frac{R}{2}$

Calculating $z_{COM}$: $z_{COM} = \frac{2}{\pi} \int_0^{\pi/2} \int_0^{\pi/2} R^2 \cos\theta \sin\theta d\theta d\phi$ $z_{COM} = \frac{2R^2}{\pi} \left( \int_0^{\pi/2} \sin\theta \cos\theta d\theta \right) \left( \int_0^{\pi/2} d\phi \right)$ $\int_0^{\pi/2} \sin\theta \cos\theta d\theta = \frac{1}{2}$ and $\int_0^{\pi/2} d\phi = \frac{\pi}{2}$. $z_{COM} = \frac{2R^2}{\pi} \left( \frac{1}{2} \right) \left( \frac{\pi}{2} \right) = \frac{R}{2}$

Thus, the center of mass of 1/4th of a hollow hemisphere is located at $(\frac{R}{2}, \frac{R}{2}, \frac{R}{2})$.