Questions on centre of mass of a discrete distribution

Question 1

Explain why the centre of mass is a weighted average rather than a plain average of the positions.

Accepted 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.

Question 2

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.

Accepted 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.

Question 3

A light rod carries a mass of 2 kg at x = 0 and a mass of 6 kg at x = 4. The rod is to be rested on a single knife edge so that it balances horizontally. Explain where the knife edge must go, and find that point.

Accepted Answer

A body balances when supported at its centre of mass, because the moments of the weights about that point cancel and there is no tendency to turn. So the knife edge goes at x(G) = (2 × 0 + 6 × 4)/(2 + 6) = 24/8 = x = 3. B1 for the knife edge at the centre of mass, M1 for the weighted average, A1 for x = 3. Check the moments about x = 3: 2g × 3 anticlockwise against 6g × 1 clockwise, equal, so the rod stays level. The edge is three times as far from the light mass as from the heavy one, the inverse of the mass ratio.

Question 4

Masses of 2 kg and 3 kg are placed at (1, 2) and (3, 6). Find the centre of mass, and show that it lies on the line joining the two particles.

Accepted Answer

Total mass 5 kg. x(G) = (2 × 1 + 3 × 3)/5 = 11/5 = 2.2 and y(G) = (2 × 2 + 3 × 6)/5 = 22/5 = 4.4. Both particles lie on y = 2x, and 4.4 = 2 × 2.2, so (2.2, 4.4) lies on that line too. M1 for the weighted averages, A1 for (2.2, 4.4), B1 for checking the point against y = 2x. The centre of mass of two particles is always on the segment joining them, dividing it in the inverse ratio of the masses: here it is 3/5 of the way from the 2 kg mass towards the 3 kg mass.

Question 5

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.

Accepted 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.

Question 6

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.

Accepted 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.

Question 7

State two physical facts about a rigid body that follow directly from knowing where its centre of mass is.

Accepted 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.

Question 8

Masses of 6 kg at (2, 1) and 4 kg at (5, 3) are to be balanced by a 10 kg counterweight at (a, b), so that the centre of mass of the three particles is at the origin. Find a and b.

Accepted Answer

The moments about each axis must close on zero. For x: 6(2) + 4(5) + 10a = 0, so 12 + 20 + 10a = 0 and a = −3.2. M1 for the x moment equation, A1 for a. For y: 6(1) + 4(3) + 10b = 0, so 6 + 12 + 10b = 0 and b = −1.8. M1 for the y equation, A1 for b. The counterweight must sit on the opposite side of the origin from the pair it cancels, which the negative signs confirm.

Question 9

Particles of mass m, m, 3m and m are placed at the corners (0, 0), (2, 0), (2, 2) and (0, 2) of a square. Find the centre of mass, using symmetry to shorten the working.

Accepted Answer

Total mass 6m. x(G) = (0 + 2m + 6m + 0)/6m = 8/6 = 4/3. The arrangement is symmetric about the diagonal y = x (the 3m mass sits on it, and the two m masses at (2, 0) and (0, 2) are mirror images), so y(G) = x(G) without a second calculation: the centre of mass is at (4/3, 4/3), about (1.33, 1.33). M1 for the x moment, A1 for 4/3, B1 for the symmetry argument, A1 for the point. Direct calculation gives Σmy = 0 + 0 + 6m + 2m = 8m and the same 4/3, which confirms the symmetry. The m cancels throughout, so the answer is a matter of the ratios alone.

Question 10

Particles of mass 2 kg, 3 kg and 5 kg have position vectors 3i + j, −i + 4j and i − 2j metres. Find the position vector of their centre of mass, and its distance from the origin.

Accepted Answer

Σm r = 2(3i + j) + 3(−i + 4j) + 5(i − 2j) = (6 − 3 + 5)i + (2 + 12 − 10)j = 8i + 4j. The total mass is 10 kg, so the centre of mass is at 0.8i + 0.4j. Its distance from the origin is √(0.8² + 0.4²) = √0.8 = 0.894 m. M1 for the mass-weighted sum of the position vectors, A1 for 8i + 4j, A1 for 0.8i + 0.4j, B1 for the distance. The vector form is the two coordinate calculations written on one line; the components never mix, and a negative component stays negative inside the sum.

Question 11

A uniform rod AB has mass 2 kg and length 1.2 m. Particles of mass 1 kg and 3 kg are fixed to it at A and B. Find the distance of the centre of mass of the loaded rod from A. A further particle of mass m kg is then fixed at A so that the centre of mass moves to the midpoint of the rod. Find m.

Accepted Answer

The rod is uniform, so its own 2 kg acts at its midpoint, 0.6 m from A. Moments about A: Σmx = 1 × 0 + 2 × 0.6 + 3 × 1.2 = 4.8, total mass 6 kg, so the centre of mass is 4.8/6 = 0.8 m from A. With m kg added at A the moment is unchanged and the mass becomes 6 + m: 4.8/(6 + m) = 0.6, so 6 + m = 8 and m = 2. B1 for the rod's mass at its midpoint, M1 for the moments about A, A1 for 0.8 m, M1 for the equation with the added mass, A1 for m = 2. The added mass sits at the point moments are taken about, so it enters the denominator only; that is what makes A the convenient origin.

Question 12

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.

Accepted Answer

Take moments about each axis in turn, keeping the negative coordinates negative.
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. Cancel before multiplying out; expanding to a quadratic is longer and more error-prone.
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.

Question 13

A light rod carries a mass of 5 kg at x = 0 and a mass of 3 kg at x = 8, and hangs horizontally in equilibrium from a single vertical string. Explain where the string must be attached, and find that point.

Accepted Answer

A body suspended from a point hangs with its centre of mass directly below that point; for the rod to hang horizontally, the string must therefore be attached at the centre of mass itself, where the moments of the two weights about the attachment cancel. B1 for the suspension argument. So the string goes at x(G) = Σmx/Σm. M1 for the weighted average. x(G) = (5 × 0 + 3 × 8)/(5 + 3) = 24/8 = 3. A1. The point lies within the spread of the masses and nearer the heavier one, at 3 rather than the midpoint 4, which is the check to make. B1 for a stated check against the spread or the midpoint. Attaching the string at the midpoint would leave a net moment of 5g × 4 − 3g × 4 about it, and the rod would turn.

Question 14

Particles of mass 3 kg, 2 kg and m kg are placed at (1, 1), (4, 3) and (p, 0). The centre of mass of the three particles is at (3, 1.5). Explain why the y coordinate determines m on its own, and find m and p.

Accepted Answer

The third particle lies on the x-axis, so it contributes to the mass but not to the y moment: the y equation contains m but not p, and can be solved first.
y: (3 × 1 + 2 × 3 + m × 0)/(5 + m) = 1.5, so 9 = 7.5 + 1.5m and m = 1.
x: (3 × 1 + 2 × 4 + 1 × p)/6 = 3, so 11 + p = 18 and p = 7.
B1 for the explanation, M1 for the y moment equation, A1 for m = 1, M1 for the x equation with m substituted, A1 for p = 7. Taking the x equation first gives (11 + mp)/(5 + m) = 3, one equation in two unknowns, which cannot be solved until m is known; the order of the two equations is the decision that keeps the algebra linear.

Question 15

Three particles of equal mass are placed at points with position vectors a, b and c, the vertices of a triangle. Prove that their centre of mass lies on the median from the vertex at a, two thirds of the way from that vertex to the midpoint of the opposite side.

Accepted Answer

With equal masses m, the centre of mass is at (ma + mb + mc)/(3m) = (a + b + c)/3. The midpoint of the opposite side has position vector d = (b + c)/2. The point two thirds of the way from a to d is a + (2/3)(d − a) = a + (2/3)((b + c)/2 − a) = a + (b + c)/3 − (2/3)a = (a + b + c)/3, which is the centre of mass. B1 for (a + b + c)/3, B1 for the midpoint (b + c)/2, M1 for the point dividing the median in the ratio 2 : 1, A1 for the algebra reducing it to (a + b + c)/3, B1 for the conclusion that the same holds for every median, since the expression is symmetric in a, b and c. This is why the three medians of a triangle meet at one point: each passes through (a + b + c)/3.