Three balls have masses and positions as follows:

• Ball 1: m₁ = 1.2 kg, x₁ = 0.4 m, y₁ = 0.7 m.
• Ball 2: m₂ = 2.1 kg, x₂ = -0.4 m, y₂ = 0.7 m.
• Ball 3: m₃ = 1.7 kg, x₃ = 0.1 m, y₃ = 0.2 m.

The center-of-mass of this system is given by:

X_cm = (x₁m₁ + x₂m₂ + x₃m₃) / (m₁ + m₂+ m₃ )

X_cm = (0.4 x 1.2 + (-0.4) x 2.1 + 0.1 x 1.7) / (1.2 + 2.1 + 1.7)

X_cm = -0.038

Y_cm = (y₁m₁ + y₂m₂ + y₃m₃) / (m₁ + m₂+ m₃ )

Y_cm = (0.7 x 1.2 + 0.7 x 2.1 + 0.2 x 1.7) / (1.2 + 2.1 + 1.7)

Y_cm = 0.53

A square sheet with a uniform mass per unit area has a circular hole cut out of it. The square extends from x=1 to x=5, and from y=0 to y=4. The circle has a radius of 1 and is centered at x=2.5, y=2.5. Where is the center of mass of the sheet with the hole?

The center-of mass equation can be written:

X_cm = ( x₁m₁ + x₂m₂ ) / M

X_cm M = x₁m₁ + x₂m₂

Here X_cm and M represent the x-coordinate of the center-of-mass position, and the mass, of the sheet without the hole. The subscript 1 represents the sheet with the hole, and the 2 represents the piece cut out to make the hole. We're looking for x₁.

x₁ = ( X_cm M - x₂m₂ ) / m₁

X_cm = 3, and x₂ = 2.5.

How do we handle the masses? Because the sheet is uniform, the mass of a piece of sheet is proportional to the area. In fact:

m = sigma A, where sigma is the mass/unit area of the sheet.

M = 16, m₂ = pi, and m₁ = (16-pi)

Substituting these values into the expression for x₁ gives:

x₁ = ( 48 - 2.5 pi ) / (16-pi)

A right-angled triangle has sides of length 5, 12, and 13. Where is its center of mass?