Question

The centre of gravity of hemisphere with radius r lies at a distance of ............... form its base measured along the vertical radius.

• A. \(3r/8\)
• B. \(3/8r\)
• C. \(8/3r\)
• D. \(8r/3\)

Hint:

Be careful to distinguish between different geometries:
- Solid hemisphere centroid = \(\frac{3r}{8}\) from base.
- Hollow hemisphere centroid = \(\frac{r}{2}\) from base.
- Semicircular lamina centroid = \(\frac{4r}{3\pi}\) from diameter.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The center of gravity (or centroid for a body of uniform density) is the point where the entire weight of the body is assumed to be concentrated. For a solid of revolution like a hemisphere, symmetry dictates that the centroid lies along the axis of symmetry (the vertical radius).
Key Formula or Approach:
Using calculus, the position of the centroid coordinate (\(\bar{y}\)) along the axis of symmetry is: \[ \bar{y} = \frac{\int y \, dV}{\int dV} \]

Step 2: Detailed Explanation:
Let us derive the centroid position for a solid hemisphere of radius \(r\) with its flat base in the \(xz\)-plane centered at the origin:

The vertical axis of symmetry is the \(y\)-axis, ranging from \(y = 0\) at the base to \(y = r\) at the apex.

Consider a thin circular elemental disk of thickness \(dy\) at a height \(y\) from the base.

The radius of this disk, \(x\), is given by the equation of the sphere: \(x^2 + y^2 = r^2 \implies x^2 = r^2 - y^2\).

The volume of this thin disk is: \[ dV = \pi x^2 \, dy = \pi (r^2 - y^2) \, dy \]
The total volume (\(V\)) of the solid hemisphere is: \[ V = \frac{2}{3}\pi r^3 \]
Now, compute the integral of \(y \, dV\) from \(y = 0\) to \(y = r\): \[ \int_{0}^{r} y \, dV = \int_{0}^{r} y \pi (r^2 - y^2) \, dy = \pi \int_{0}^{r} (r^2 y - y^3) \, dy \] \[ \int_{0}^{r} y \, dV = \pi \left[ \frac{r^2 y^2}{2} - \frac{y^4}{4} \right]_{0}^{r} = \pi \left( \frac{r^4}{2} - \frac{r^4}{4} \right) = \frac{\pi r^4}{4} \]
Divide by the total volume to find \(\bar{y}\): \[ \bar{y} = \frac{\frac{\pi r^4}{4}}{\frac{2\pi r^3}{3}} = \frac{\pi r^4}{4} \times \frac{3}{2\pi r^3} = \frac{3r}{8} \] Thus, the centroid lies at a distance of \(\frac{3r}{8}\) from the flat base.

Step 3: Final Answer:
The distance is \(3r/8\), which corresponds to Option (A).