A cylinder of 50 mm diameter is 250 mm high and has a semi-ball at the top as shown in the sketch (FIGURE 2) below. Calculate the centre of gravity of the body from the table on which it is placed.

The volume of the cylinder is $V_c = \pi r^2 h_c$.

$$V_c = \pi (25 \text{ mm})^2 (250 \text{ mm}) = 156250\pi \text{ mm}^3$$

The centroid of the cylinder from its base is $y_c = \frac{h_c}{2}$.

$$y_c = \frac{250 \text{ mm}}{2} = 125 \text{ mm}$$

The volume of the hemisphere is $V_h = \frac{2}{3} \pi r^3$.

$$V_h = \frac{2}{3} \pi (25 \text{ mm})^3 = \frac{2}{3} \pi (15625) \text{ mm}^3 = \frac{31250}{3}\pi \text{ mm}^3$$

The centroid of a hemisphere from its flat base is $\frac{3}{8}r$. Since the hemisphere is placed on top of the cylinder, its flat base is at $h_c = 250 \text{ mm}$ from the table.
So, the centroid of the hemisphere from the table is $y_h = h_c + \frac{3}{8}r$.

$$y_h = 250 \text{ mm} + \frac{3}{8}(25 \text{ mm}) = 250 \text{ mm} + 9.375 \text{ mm} = 259.375 \text{ mm}$$

$$\bar{Y} = \frac{(156250\pi \text{ mm}^3)(125 \text{ mm}) + (\frac{31250}{3}\pi \text{ mm}^3)(259.375 \text{ mm})}{156250\pi \text{ mm}^3 + \frac{31250}{3}\pi \text{ mm}^3}$$

We can cancel $\pi$ from all terms:

$$\bar{Y} = \frac{(156250)(125) + (\frac{31250}{3})(259.375)}{156250 + \frac{31250}{3}}$$ $$\bar{Y} = \frac{19531250 + 2701822.9166...}{156250 + 10416.6666...}$$ $$\bar{Y} = \frac{22233072.9166...}{166666.6666...}$$ $$\bar{Y} \approx 133.40 \text{ mm}$$

The centre of gravity of the body from the table is $\text{133.40 mm}$.