M2.3 Centres of mass
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This is an M2 (A2) sub-topic. Moments of single and parallel forces are on the M1 moments page, rigid bodies under non-parallel forces (ladders, hinged rods) are on the M2 statics page, and finding centres of mass by integration is not needed in M2.
What this page covers
7 points
- The centre of mass of particles on a line and in a plane, \(\bar{x}=\dfrac{\sum mx}{\sum m}\), including the vector form
- Using symmetry, and the standard results for a rectangle, triangle, semicircle, sector and circular arc
- Composite laminas, including holes and parts removed, using a table of areas and positions
- Uniform wire frameworks, where mass is proportional to length
- A lamina suspended from a point: the angle a given line makes with the vertical
- A lamina free to rotate about a fixed horizontal axis, with particles attached
- A lamina on a rough inclined plane: about to topple, and toppling versus sliding
In one minute
6 points
- The centre of mass of a body is the point where its whole weight can be taken to act, and it is found as a weighted mean of positions.
- Particles: \(\bar{x}=\sum m_i x_i/\sum m_i\) and \(\bar{y}=\sum m_i y_i/\sum m_i\).
- Uniform lamina: mass is proportional to area. Uniform wire: mass is proportional to length. A removed piece goes in the table with a negative area.
- If a uniform body has a line of symmetry, its centre of mass lies on that line.
- Hanging freely from a point, a body rests with its centre of mass vertically below that point.
- On a slope, a body is about to topple when the vertical through its centre of mass passes through the lowest point of contact.
Contents
14 sections
- Where is the centre of mass of particles on a straight line?
- How do you find the centre of mass of particles in a plane?
- Where are the centres of mass of the standard uniform shapes?
- How do you deal with a composite lamina or a part removed?
- How is a wire framework different from a lamina?
- How does a lamina hang when it is suspended from a point?
- What happens when a lamina can turn about a fixed horizontal axis?
- When does a lamina on an inclined plane topple?
- Exam tips for M2.3
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
Where is the centre of mass of particles on a straight line?
Imagine several particles fixed to a light rod lying along the \(x\)-axis. Each particle has a weight \(m_i g\) acting at its position \(x_i\). If we replace all of them by a single weight \(\left(\sum m_i\right)g\) at one point \(\bar{x}\), that single weight must have the same turning effect about the origin as all the separate weights together. Taking moments about the origin gives
\[\left(\sum m_i\right) g\,\bar{x}=\sum m_i g x_i .\]The \(g\) cancels on both sides, which is why the centre of mass does not depend on the strength of gravity. Dividing by the total mass gives the result that the whole topic is built on.
\[\bar{x}=\frac{\sum m_i x_i}{\sum m_i}\]Here \(m_i\) is the mass of particle \(i\) and \(x_i\) is its position measured from a chosen origin. You can choose any fixed point as the origin, but positions on the other side of it must be negative. This result is not in the formulae booklet: learn it, because every later idea on this page is the same equation in disguise.
It is often more useful in the rearranged form \(\left(\sum m\right)\bar{x}=\sum mx\): total mass times the position of the centre of mass equals the sum of the individual moments. When a question gives the centre of mass and asks for an unknown mass or position, this form gives a linear equation straight away.
Particles of mass 2 kg, 3 kg and 5 kg lie on the \(x\)-axis at the points where \(x=-1\), \(x=2\) and \(x=4\) respectively (distances in metres). (a) Find the position of the centre of mass of the three particles. (b) A fourth particle, of mass \(k\) kg, is added at the point where \(x=-1\). The centre of mass of the four particles is at \(x=1.5\). Find \(k\).
(a) Sum of moments: \(\sum mx=2(-1)+3(2)+5(4)=24\). Total mass: \(\sum m=10\). So \(\bar{x}=24\div 10=\) 2.4 m (the point where \(x=2.4\))
(b) Write the same equation with \(k\) included: \((10+k)\times 1.5=24+k(-1)\), so \(15+1.5k=24-k\), giving \(2.5k=9\) and \(k=3.6\)
Check: adding mass at \(x=-1\) should pull the centre of mass to the left of 2.4, and 1.5 is to the left.
How do you find the centre of mass of particles in a plane?
In two dimensions nothing new is needed. Apply the one-dimensional result twice: once using the \(x\)-coordinates and once using the \(y\)-coordinates. The two calculations are completely independent.
\[\bar{x}=\frac{\sum m_i x_i}{\sum m_i},\qquad \bar{y}=\frac{\sum m_i y_i}{\sum m_i}\qquad\text{or}\qquad \bar{\mathbf{r}}=\frac{\sum m_i\mathbf{r}_i}{\sum m_i}\]The vector form is the same thing written with position vectors \(\mathbf{r}_i\). If the positions are given as \(a\mathbf{i}+b\mathbf{j}\), work with the \(\mathbf{i}\) and \(\mathbf{j}\) components separately: it is exactly the coordinate calculation. None of these forms is in the formulae booklet, so learn them.
Because there are two separate equations, a question can contain two unknowns, such as an unknown mass and an unknown coordinate. The skill is spotting which equation contains only one of them.
Particles of mass 2 kg, 4 kg and 3 kg are at the points (1, 3), (5, 1) and (−2, 4) respectively. A fourth particle, of mass \(k\) kg, is placed at the point \((p, 5)\). The centre of mass of the four particles is at (2, 3). Find \(k\) and \(p\).
Step 1 Totals for the three known particles: \(\sum m=9\), \(\sum mx=2+20-6=16\), \(\sum my=6+4+12=22\).
Step 2 The \(y\)-equation contains only \(k\), so solve it first: \((9+k)\times 3=22+5k\), so \(27+3k=22+5k\) and \(k=2.5\)
Step 3 Now the \(x\)-equation: \((9+2.5)\times 2=16+2.5p\), so \(23=16+2.5p\) and \(p=2.8\)
When there are two unknowns, look for the equation with only one of them in it. Here the \(y\)-coordinate of the new particle was given, so the \(y\)-equation found \(k\) straight away, and then the \(x\)-equation gave \(p\).
Where are the centres of mass of the standard uniform shapes?
A lamina is a flat body whose thickness is ignored. Uniform means its mass per unit area is constant, so the mass of any piece is proportional to its area. That single fact lets us treat areas as if they were masses.
Symmetry does a lot of the work. If a uniform body has a line of symmetry, its centre of mass lies on that line, because every bit of mass on one side is matched by an equal bit on the other side. Two lines of symmetry fix the centre of mass at their intersection, so a uniform rectangle or circle has its centre of mass at its centre, and a uniform straight rod has it at its midpoint.
For a triangular lamina, the centre of mass is where the three medians meet, \(\tfrac{2}{3}\) of the way along each median from the vertex. With coordinates this is simply the mean of the three vertices:
\[\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right).\]Curved shapes have results that come from integration, and these are given in the formulae booklet. For a sector of radius \(r\) with angle \(2\alpha\) at the centre, the centre of mass lies on the axis of symmetry at a distance \(\dfrac{2r\sin\alpha}{3\alpha}\) from the centre O. A semicircle is the case \(\alpha=\tfrac{\pi}{2}\), which gives \(\dfrac{4r}{3\pi}\). For a circular arc (a curved wire) of radius \(r\) and angle \(2\alpha\), the distance is \(\dfrac{r\sin\alpha}{\alpha}\), and a semicircular arc gives \(\dfrac{2r}{\pi}\). In every case \(\alpha\) is half the angle at the centre and must be in radians.
| Uniform body | Position of the centre of mass | In the formulae booklet? |
|---|---|---|
| Rectangle, circle (lamina); straight rod | at the centre / midpoint, by symmetry | No — use symmetry |
| Triangular lamina | \(\tfrac{2}{3}\) along each median from the vertex; mean of the vertices | Yes |
| Sector of a circle, radius \(r\), angle \(2\alpha\) | \(\dfrac{2r\sin\alpha}{3\alpha}\) from the centre, on the axis of symmetry | Yes |
| Semicircular lamina, radius \(r\) | \(\dfrac{4r}{3\pi}\) from the centre, on the axis of symmetry | Yes — the sector with \(\alpha=\pi/2\) |
| Circular arc (wire), radius \(r\), angle \(2\alpha\) | \(\dfrac{r\sin\alpha}{\alpha}\) from the centre, on the axis of symmetry | Yes |
| Semicircular arc (wire), radius \(r\) | \(\dfrac{2r}{\pi}\) from the centre, on the axis of symmetry | Yes — the arc with \(\alpha=\pi/2\) |
A uniform lamina OAB is a sector of a circle with centre O, radius 12 cm and angle AOB = 60°. G is its centre of mass. (a) Find the exact distance OG. (b) Find the distance of G from the edge OA, correct to 3 significant figures.
(a) The angle at the centre is \(2\alpha=60°\), so \(\alpha=\tfrac{\pi}{6}\). Then \(OG=\dfrac{2\times 12\times\sin(\pi/6)}{3\times\pi/6}=\dfrac{12}{\pi/2}=\) \(\dfrac{24}{\pi}\) cm (7.64 cm)
(b) G lies on the line of symmetry, at 30° to OA. Distance from OA \(=OG\sin 30°=\dfrac{12}{\pi}=\) 3.82 cm
Using the whole angle at the centre (60° here) as \(\alpha\), or putting \(\alpha\) in degrees. With \(\alpha=60\) in the denominator the answer is wildly too small; with \(\alpha=\pi/3\) it is the centre of mass of a 120° sector.
How do you deal with a composite lamina or a part removed?
Most lamina questions use a shape made from standard pieces. The method is to split the lamina into pieces whose centres of mass you know. Each piece then behaves like a particle placed at its own centre of mass, with "mass" proportional to its area. After that it is the particle calculation again.
Organise the working in a table: one row for each piece (area, \(x\), \(y\)) and a final row for the whole lamina (total area, \(\bar{x}\), \(\bar{y}\)). Then write one moment equation for each direction you need.
A hole or a removed piece is handled by thinking of it as (whole) − (removed) = (remaining). In the table the removed piece gets a negative area, so the moment equation reads: moment of remainder = moment of whole − moment of removed piece. Before any of this, use symmetry: it often gives one coordinate with no working at all.
If the lamina has mass \(M\), each piece has mass \(M\times\) (its area ÷ total area). For the position of the centre of mass only the areas matter, but you will need actual masses later when weights appear in a moments equation.
A uniform rectangular lamina ABCD has AB = 20 cm and AD = 16 cm. A semicircle of radius 6 cm, whose diameter lies along AD and whose centre is the midpoint of AD, is removed. Find the exact distance of the centre of mass of the remaining lamina from AD, and give it to 3 significant figures.
Step 1 Symmetry about the line through the midpoints of AD and BC gives \(\bar{y}=8\) cm (measured from AB). Take \(x\) as the distance from AD.
| Piece | Area (cm²) | Distance from AD (cm) |
|---|---|---|
| rectangle | 320 | 10 |
| semicircle (removed) | \(-\tfrac12\pi(6)^2=-18\pi\) | \(\dfrac{4(6)}{3\pi}=\dfrac{8}{\pi}\) |
| remainder | \(320-18\pi\) | \(\bar{x}\) |
Step 2 Moments about AD: \((320-18\pi)\bar{x}=320\times 10-18\pi\times\dfrac{8}{\pi}=3200-144=3056\).
So \(\bar{x}=\dfrac{3056}{320-18\pi}=\) \(\dfrac{1528}{160-9\pi}\) cm \(=11.6\) cm (3 s.f.)
Notice how the \(\pi\) cancels in the moment of the semicircle: area × distance \(=18\pi\times 8/\pi=144\). That always happens with a semicircle, so exact answers stay tidy.
For "show that" and "find the exact distance" questions, keep \(\pi\) in the working until the very last line. A table with a heading row is clear method even if an arithmetic slip follows.
How is a wire framework different from a lamina?
A framework is made from uniform wire or thin uniform rods. The mass of each piece is proportional to its length, not to any area, because there is no material inside the outline. Each straight piece acts at its midpoint, and each curved piece acts at the arc result \(\dfrac{r\sin\alpha}{\alpha}\) from its centre.
This changes the answer. A triangular framework does not have its centre of mass at the centroid of the triangle unless the triangle is equilateral: the longest side carries the most mass and pulls the centre of mass towards itself. Pieces described as "light" have no mass and are left out, and a particle fixed to a framework is just another row in the table.
A uniform wire of length 120 cm is bent into the shape of a triangle OAB with OA = 40 cm, OB = 30 cm, AB = 50 cm and angle AOB = 90°. Taking O as origin, OA along the \(x\)-axis and OB along the \(y\)-axis, find the coordinates of the centre of mass of the framework, and compare it with the centre of mass of a uniform triangular lamina OAB.
Step 1 Each side acts at its midpoint, with mass proportional to its length.
| Side | Length (cm) | Midpoint |
|---|---|---|
| OA | 40 | (20, 0) |
| OB | 30 | (0, 15) |
| AB | 50 | (20, 15) |
Step 2 Moments: \(120\bar{x}=40(20)+30(0)+50(20)=1800\), so \(\bar{x}=15\). \(120\bar{y}=40(0)+30(15)+50(15)=1200\), so \(\bar{y}=10\). The framework's centre of mass is at (15, 10)
Step 3 The lamina's centre of mass is the mean of the vertices: \(\left(\tfrac{0+40+0}{3},\tfrac{0+0+30}{3}\right)=(13\tfrac13, 10)\). The framework's centre of mass is \(1\tfrac23\) cm further from OB than the lamina's: a different body, a different answer.
How does a lamina hang when it is suspended from a point?
A lamina hanging freely from a point P has only two forces on it: its weight, acting at G, and the force at P. Two forces can only be in equilibrium if they are equal, opposite and act along the same line. The weight is vertical, so the line PG must be vertical: G is vertically below P. The same rule fixes how a lamina hangs from a string attached at P.
To find the angle that a line PQ of the lamina makes with the vertical, draw the line PG, which is vertical, and find the angle between PQ and PG. Usually this is one right-angled triangle built from the coordinates of G. If particles are attached to the lamina, use the centre of mass of the whole system, lamina plus particles.
The lamina from worked example 4 is freely suspended from A and hangs in equilibrium. Find the angle between AB and the vertical.
Step 1 G is 11.60 cm from AD (so 11.60 cm along AB from A) and 8 cm from AB.
Step 2 AG is vertical, so the required angle is angle GAB: \(\tan\theta=\dfrac{8}{11.60}\), giving \(\theta=34.6°\)
Sketch first: AB goes down to one side of the vertical, and G must lie inside the angle between AB and AD.
Writing the tangent upside down. The angle at A between AB and AG has opposite = distance of G from AB and adjacent = distance of G along AB. Mark both on a sketch before writing \(\tan\theta\).
What happens when a lamina can turn about a fixed horizontal axis?
Questions often say that a lamina is "free to rotate about a fixed smooth horizontal axis through P, perpendicular to the lamina". With no other forces, this behaves exactly like a lamina hanging from P: it rests with the centre of mass of the system vertically below P.
To make it rest in a chosen position, for example with a given edge horizontal, a particle is attached so that the combined centre of mass moves onto the vertical through P. There are two equivalent ways to write the equation. Either find the combined centre of mass and say it lies on the vertical through P, or take moments about the axis: the moment of the lamina's weight must balance the moment of the particle's weight. Both give the same equation.
The force from a smooth axis passes through the axis, so it has no moment about it. If only vertical forces act, the axis supplies a vertical force equal to the total weight.
A uniform L-shaped lamina OABCDE, of mass \(M\), has vertices O(0, 0), A(\(6a\), 0), B(\(6a\), \(2a\)), C(\(2a\), \(2a\)), D(\(2a\), \(5a\)) and E(0, \(5a\)). (a) Show that the centre of mass of the lamina is at \(\left(\dfrac{7a}{3},\dfrac{11a}{6}\right)\). (b) The lamina is freely suspended from E and hangs in equilibrium. Find the angle between EO and the vertical. (c) The lamina is now free to rotate in a vertical plane about a fixed smooth horizontal axis through D, perpendicular to the lamina. A particle of mass \(m\) is attached at E, and the lamina rests in equilibrium with DE horizontal. Find \(m\) in terms of \(M\).
(a) Split into two rectangles, with F = (0, \(2a\)).
| Piece | Area | \(x\) | \(y\) |
|---|---|---|---|
| OABF, \(6a\times 2a\) | \(12a^2\) | \(3a\) | \(a\) |
| FCDE, \(2a\times 3a\) | \(6a^2\) | \(a\) | \(3.5a\) |
| lamina | \(18a^2\) | \(\bar{x}\) | \(\bar{y}\) |
\(18a^2\bar{x}=36a^3+6a^3=42a^3\), so \(\bar{x}=\dfrac{7a}{3}\). \(18a^2\bar{y}=12a^3+21a^3=33a^3\), so \(\bar{y}=\dfrac{11a}{6}\), as required
(b) G is vertically below E. Relative to E, G is \(\dfrac{7a}{3}\) across and \(5a-\dfrac{11a}{6}=\dfrac{19a}{6}\) down (along EO). So \(\tan\theta=\dfrac{7a/3}{19a/6}=\dfrac{14}{19}\), giving \(\theta=36.4°\)
(c) With DE horizontal the lamina is in the position shown, and the combined centre of mass must be vertically below D, at \(x=2a\). Moments about the vertical through D: \(Mg\left(\dfrac{7a}{3}-2a\right)=mg(2a-0)\), so \(\dfrac{Ma}{3}=2ma\) and \(m=\dfrac{M}{6}\)
Check that the position really hangs: the combined centre of mass is at height \(\left(M\times\tfrac{11a}{6}+\tfrac{M}{6}\times 5a\right)\div\tfrac{7M}{6}=\tfrac{16a}{7}\), below D (at \(5a\)). The axis then supplies a vertical force of \(\tfrac{7Mg}{6}\).
When does a lamina on an inclined plane topple?
Picture a lamina standing in a vertical plane with one edge on a rough plane inclined at \(\alpha\). Its weight acts vertically downwards through G. While the vertical line through G passes inside the edge in contact, the weight has a moment that holds the lamina against the plane, and it stays upright (as long as it does not slide). Once the vertical through G passes outside the lowest point of contact, the weight has a moment that turns the lamina over that point.
So the lamina is about to topple about the lowest point of contact P when the vertical through G passes through P. If G is a distance \(h\) from the plane and a distance \(d\) up the slope from P (both measured in the lamina), the critical angle is given by
\[\tan\alpha=\frac{d}{h}.\]The lamina might slide before it topples. From M1, a block on the point of sliding has \(F=\mu R\), which on a plane gives \(\tan\alpha=\mu\). It slides when \(\tan\alpha>\mu\). The question "slide or topple?" is therefore a comparison of two critical angles: whichever is smaller is reached first as the plane is tilted.
A uniform lamina is a rectangle 10 cm wide and 16 cm tall, with an isosceles triangle of base 10 cm and height 6 cm fixed along its top edge (a "house" shape). It stands in a vertical plane with its 10 cm base on a rough plane, along a line of greatest slope. The coefficient of friction is 0.6. The angle of the plane is slowly increased from zero. (a) Find the distance of the centre of mass from the base. (b) Determine whether the lamina slides or topples first.
(a) Symmetry puts G on the central vertical line. The centroid of the triangle is \(\tfrac13\) of its height above its base, at \(16+6/3=18\) cm. Moments about the base: \((160+30)\bar{y}=160\times 8+30\times 18=1280+540=1820\), so \(\bar{y}=\dfrac{1820}{190}=\dfrac{182}{19}=\) 9.58 cm
(b) About to topple when G is vertically above the lower corner, which is 5 cm along the base from G's line: \(\tan\alpha=\dfrac{5}{182/19}=\dfrac{95}{182}=0.522\), so \(\alpha=27.6°\).
Sliding needs \(\tan\alpha>0.6\), that is \(\alpha>31.0°\). Since \(0.522<0.6\), the toppling angle is reached first: the lamina topples (at about 27.6°)
Exam tips for M2.3
Pieces in rows, with area (or length, or mass) and the coordinates of each piece's centre of mass, then a final row for the whole. It organises the working and shows clear method even when a number goes wrong.
State the origin and axes ("distance from AD", "taking O as origin") before the calculation. A correct number measured from the wrong edge is still a wrong answer.
Write "by symmetry, \(\bar{y}=8\)" as a line of working. It is quicker than a second moment equation, and it counts as working in its own right.
Mark the point of suspension, G, and the line whose angle you want. Then the angle is in one right-angled triangle, and the sketch tells you which length is opposite and which is adjacent.
Find both critical values (\(\tan\alpha=d/h\) for toppling, \(\tan\alpha=\mu\) for sliding), compare them, and finish with a sentence saying which happens first.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A hole can be ignored if it is small. | Any removed area changes the moment and the total area. | Enter the removed piece with a negative area; subtract its moment and its area. |
| A triangular frame has the same centre of mass as a triangular lamina. | A frame's mass is in its sides, proportional to their lengths. | Frame: weight the midpoints by length. Lamina: mean of the vertices. |
| In the sector formula, \(\alpha\) is the angle of the sector. | The formula is written for a sector of angle \(2\alpha\). | Use half the angle at the centre, in radians. |
| The centre of mass must be inside the body. | It is a weighted mean position; it need not be on the material (an L-shape, a ring, an arc). | Calculate it; never assume it lies on the lamina. |
| A hanging lamina rests with an edge vertical. | Only the line from the point of suspension to G is vertical. | Find the angle between the given edge and the line PG. |
| A body on a slope topples as soon as it starts to slide. | Sliding depends on \(\mu\); toppling depends on the position of G. | Compare \(\tan\alpha=\mu\) with \(\tan\alpha=d/h\); the smaller angle wins. |
How this topic connects
M1 Moments
The formula \(\bar{x}=\sum mx/\sum m\) is a moments equation, and a non-uniform rod with an unknown centre of mass is the one-dimensional version of this topic.
M2.6 Statics of rigid bodies
Ladders, hinged rods and laminas held by strings all need the position of G before you can take moments: the weight always acts there.
M1 Friction, P3/P4 Integration and IAL Physics
The sliding half of a topple-or-slide question is the M1 limiting-friction condition \(F=\mu R\), which on a plane gives \(\tan\alpha=\mu\). The standard results in the formulae booklet come from integrating, and in Physics the same point is the centre of gravity, used for the stability and toppling of real objects.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(\bar{x}=\sum mx/\sum m,\ \bar{y}=\sum my/\sum m\) | particles; pieces of a composite body | No — learn it |
| \(\bar{\mathbf{r}}=\sum m\mathbf{r}/\sum m\) | particles given by position vectors | No — learn it |
| triangle: \(\tfrac23\) along median from vertex; \(\left(\tfrac{x_1+x_2+x_3}{3},\tfrac{y_1+y_2+y_3}{3}\right)\) | triangular laminas | Yes (the \(\tfrac23\) result); the vertex mean follows from it |
| sector: \(\dfrac{2r\sin\alpha}{3\alpha}\) from centre | sectors; semicircle \(\dfrac{4r}{3\pi}\) | Yes |
| arc: \(\dfrac{r\sin\alpha}{\alpha}\) from centre | curved wires; semicircular arc \(\dfrac{2r}{\pi}\) | Yes |
| suspended from P: PG vertical, \(\tan\theta=\) (distance of G from the line) ÷ (distance along the line) | angle of a line with the vertical | No — derive it |
| about to topple: \(\tan\alpha=d/h\) | lamina on an inclined plane | No — derive it |
| about to slide: \(\tan\alpha=\mu\) | comparing with toppling | No — derive it from \(F=\mu R\) |
Check your understanding
1. Particles of mass 1 kg, 2 kg and 3 kg are at \(x=0\), 3 and 6 on a light rod. Where is their centre of mass?
\(\sum mx=0+6+18=24\) and \(\sum m=6\), so \(\bar{x}=4\).
2. A uniform wire is bent into a semicircular arc of radius 10 cm. How far is its centre of mass from the centre of the circle?
\(\dfrac{2r}{\pi}=\dfrac{20}{\pi}\approx 6.37\) cm, on the axis of symmetry — not on the wire itself.
3. Find the centre of mass of a uniform triangular lamina with vertices (0, 0), (6, 0) and (3, 9).
The mean of the vertices: \(\left(\tfrac{0+6+3}{3},\tfrac{0+0+9}{3}\right)=(3, 3)\).
4. A uniform rectangular lamina PQRS has PQ = 10 cm and QR = 6 cm. It hangs freely from P. What angle does PQ make with the vertical?
G is 5 cm along PQ from P and 3 cm from PQ, so \(\tan\theta=\tfrac35\) and \(\theta\approx 31.0°\).
5. A uniform lamina is a sector of radius 6 cm with angle 90° at the centre. Find the exact distance of its centre of mass from the centre.
\(2\alpha=\tfrac{\pi}{2}\), so \(\alpha=\tfrac{\pi}{4}\). Distance \(=\dfrac{2\times 6\times\sin(\pi/4)}{3\times\pi/4}=\dfrac{6\sqrt2}{3\pi/4}=\dfrac{8\sqrt2}{\pi}\approx 3.60\) cm, along the axis of symmetry.
6. Why does a disc with a hole cut off-centre have its centre of mass on the opposite side of the centre from the hole?
The full disc balances about its centre. Removing mass from one side leaves more mass on the other side, so the weighted mean position moves away from the hole — in the table the hole has a negative area.
Specification checklist
Tick every box before the exam. Wording is ours; references are to the M2 specification sections.
- (M2 2.1) I can find the centre of mass of a set of particles on a line or in a plane, using \(\sum mx/\sum m\) or position vectors, and find an unknown mass or position.
- (M2 2.2) I can use symmetry and the standard results for a rectangle, triangle, sector, semicircle and circular arc, with \(\alpha\) as half the angle in radians.
- (M2 2.2) I can find the centre of mass of a composite lamina, including one with a part removed, using a table of areas and positions.
- (M2 2.2) I can find the centre of mass of a uniform wire framework, with mass proportional to length, including frameworks with particles attached.
- (M2 2.3) I can find the angle a given line makes with the vertical when a lamina hangs freely from a point.
- (M2 2.3) I can solve problems with a lamina free to rotate about a fixed horizontal axis, including finding a particle that makes it rest in a given position.
- (M2 2.3) I can find when a lamina on an inclined plane is about to topple, and decide whether it slides or topples first.