Centre of mass of composite bodies — A/L Combined Mathematics

How do I practise Centre of mass of composite bodies?

1 question with a worked solution

Check before you answer: Common Axis Origin Alignment: Measuring all component distances from the exact same reference origin or base edge.

1. From a uniform circular lamina of radius 2a and center O, a circular hole of radius a touching the outer boundary is removed. How far from O is the center of mass of the remaining lamina?
   1. A) a / 2 in the direction opposite to the hole
   2. B) a / 3 in the direction opposite to the hole
   3. C) a / 4 in the direction of the hole
   4. D) 2a / 3 in the direction opposite to the hole
   Check my answer

   Correct answer: B) a / 3 in the direction opposite to the hole

   Let O be the origin (0, 0). Mass is proportional to area: Original disk M₁ ∝ π(2a)² = 4πa² at (0, 0). Removed hole M₂ ∝ πa² with center at (a, 0). Remaining mass M_rem = 4πa² - πa² = 3πa². Center of mass x_cm: M_rem · x_cm = M₁ · 0 - M₂ · a => 3πa² · x_cm = -πa³ => x_cm = -a / 3.

   Show worked solution

   Step 1: Areas: Whole = 4πa², Cutout = πa², Remainder = 3πa². Step 2: Moments about O: 3πa² · x_cm + πa² · a = 4πa² · 0 = 0. Step 3: x_cm = -πa³ / (3πa²) = -a / 3. Answer: a / 3 in the direction opposite to the hole.
   Modelled on: the A/L Combined Mathematics examination pattern

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Where does Centre of mass of composite bodies fit in the syllabus?

Year 2 · Term 2