A circular hollow of 50 mm diameter is cut from a circular disc of 100 mm diameter as shown in Figure Q5b. Calculate the center of gravity of the section from A.

• Large Disc (Component 1):

  • Diameter $D_1 = 100 \text{ mm}$, so Radius $R_1 = 50 \text{ mm}$.
  • Area $A_1 = \pi R_1^2 = \pi (50 \text{ mm})^2 = 2500\pi \text{ mm}^2$.
  • If point A is at $y=0$, the center of the large disc is at $y_1 = -R_1 = -50 \text{ mm}$ (since it's below A).

• Small Hollow (Component 2):

  • Diameter $D_2 = 50 \text{ mm}$, so Radius $R_2 = 25 \text{ mm}$.
  • Area $A_2 = \pi R_2^2 = \pi (25 \text{ mm})^2 = 625\pi \text{ mm}^2$.
  • The hollow is tangent to the bottom of the large disc. The distance from the bottom of the large disc to the center of the small hollow is $R_2 = 25 \text{ mm}$.
  • The total height of the large disc is $100 \text{ mm}$.
  • The distance from A to the bottom of the large disc is $100 \text{ mm}$.
  • Therefore, the center of the small hollow is at $y_2 = -(100 \text{ mm} - R_2) = -(100 \text{ mm} - 25 \text{ mm}) = -75 \text{ mm}$ from A.

The y-coordinate of the center of gravity ($\bar{y}$) is given by:

The negative sign indicates that the center of gravity is below point A. The distance from A is the magnitude of $\bar{y}$.

The center of gravity of the section from A is: