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Centre of mass of a discrete distribution questions
The centre of mass is the point at which a system of masses balances. For separate masses it is a weighted average, and each coordinate is worked out on its own.
6 original questions · 22 marks · the centre of mass of a discrete distribution notes · Further Mechanics 2
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Explain why the centre of mass is a weighted average rather than a plain average of the positions.
Worked answer
Taking moments about the origin, each particle contributes its mass times its distance, and the total must equal the whole mass placed at the centre of mass. The masses are therefore the weights, so the point is pulled towards the heavier particles. B1 for the moment argument, B1 for the pull towards the heavier particles. Only when all the masses are equal does it reduce to the plain average.Masses of 3 kg, 5 kg and 2 kg lie at x = 0, 4 and 10 on a light rod. Find the centre of mass.
Worked answer
Σmx = 3(0) + 5(4) + 2(10) = 0 + 20 + 20 = 40, and the total mass is 10 kg. So x(G) = 40/10 = 4. M1 for the sum of the moments, A1 for 40 over a total mass of 10, A1 for the position. It lies between the outer masses as it must.Masses of 1, 2, 3 and 4 kg are placed at (0, 0), (2, 0), (2, 3) and (0, 3). Find the centre of mass.
Worked answer
Total mass 10 kg. Σmx = 0 + 4 + 6 + 0 = 10, so x(G) = 1. Σmy = 0 + 0 + 9 + 12 = 21, so y(G) = 2.1. M1 A1 for the x coordinate, M1 A1 for the y coordinate. The point (1, 2.1) sits towards the top of the rectangle, pulled there by the two heavier masses at y = 3.Masses of 4 kg and m kg are placed at x = 2 and x = 8. The centre of mass is at x = 5. Find m.
Worked answer
The formula gives (4 × 2 + 8m)/(4 + m) = 5, so 8 + 8m = 20 + 5m and 3m = 12, giving m = 4 kg. M1 for the moment equation, A1 for the linear equation in m, M1 for solving, A1 for the mass. Checking: (8 + 32)/8 = 5. The centre is halfway between them, which is what equal masses always produce.State two physical facts about a rigid body that follow directly from knowing where its centre of mass is.
Worked answer
Supported at that point it balances, because the moments of the weights about it cancel. And suspended freely from any point, it hangs with the centre of mass vertically below that point, since otherwise the weight would have a moment about the pivot and the body would turn. B1 for balancing at that point, B1 B1 for hanging with the centre of mass below the point of suspension, with a reason.Particles of mass 2 kg, 3 kg and 5 kg are attached to a light rectangular framework at the points (1, 4), (−2, 0) and (3, −2) respectively. A fourth particle, of mass m kg, is attached at (5, 6). Given that the centre of mass of the four particles lies on the line y = x, find m.
Worked answer
Take moments about each axis in turn, keeping the negative coordinates negative. Dropping a sign here is the error the question is built around.
For the three original particles: Σmx = 2(1) + 3(−2) + 5(3) = 2 − 6 + 15 = 11, and Σmy = 2(4) + 3(0) + 5(−2) = 8 + 0 − 10 = −2, with total mass 10 kg.
With the fourth particle included, the total mass is 10 + m and
x(G) = (11 + 5m)/(10 + m), y(G) = (−2 + 6m)/(10 + m).
The condition y(G) = x(G) has the same denominator on both sides, so it cancels. That cancellation is the method mark; multiplying out and solving a quadratic wastes time and invites errors.
So 11 + 5m = −2 + 6m, giving m = 13 kg.
M1 A1 for the moments of the first three particles, M1 for the two coordinates with the fourth included, M1 for setting them equal and cancelling the denominator, A1 for the linear equation, A1 for the mass.
Check: the total mass is 23 kg, and both coordinates come to 76/23 ≈ 3.30, so the centre of mass is at (3.30, 3.30), which does lie on y = x.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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