centre of gravity | ICSE Class 10 Physics Notes

This note covers the meaning of centre of gravity, weight and balance, the connection with centre of mass, the centres of regular uniform bodies, the triangular lamina, and the suspension method for locating the centre of gravity of an irregular lamina.

What is the centre of gravity of a body?

Definition: The centre of gravity of a body is the point through which its entire weight may be considered to act. It is also the point about which the total turning effect of the gravitational forces on the body is zero.

Weight is the gravitational force acting on a body. An extended body, meaning a body with finite size, contains matter spread over different positions. Gravity acts on all these parts. The centre of gravity lets us represent their combined effect by a single downward force.

The abbreviation CG means centre of gravity. The letter G is commonly used to label its position in a diagram. G is a point, not a force or a quantity of matter. The weight is the force represented as acting through that point.

What does the turning effect mean?

A moment, also called torque, is the turning effect of a force about a point or axis. An axis is a line about which rotation occurs. A body supported away from its centre of gravity may turn under its weight.

Different small parts of a body can have turning effects in opposite directions about G. Their total gravitational turning effect is zero there. This does not mean that gravity stops acting, that the body becomes weightless, or that the weights of its parts disappear.

A line of action is the straight line along which a force acts. Near Earth's surface, the line of action of the weight of a small body is vertical and passes through G. Vertical means the local upward-downward direction; horizontal means perpendicular to that direction.

Why is a representative point useful?

When considering balance, we can ask where G lies relative to the support. We do not need to draw a separate weight arrow for every small portion of the object. This makes the connection between the position of the support and the possibility of turning easier to see.

Note: Saying that the weight acts at G is a way of representing the combined gravitational effect. It does not mean that all the material of the body is physically collected at G.

How does balancing cardboard reveal its centre of gravity?

A thin, flat plate is called a lamina. An irregularly shaped piece of cardboard is a useful example. Its boundary does not provide an obvious geometrical centre, but its centre of gravity can be found by trying to balance it horizontally on a narrow pencil tip.

The pencil provides an upward reaction, meaning the support force exerted on the cardboard. When the cardboard balances at rest, this upward force is equal in magnitude and opposite in direction to its total weight. The support point identifies G for the thin cardboard.

What two conditions are satisfied?

Translational equilibrium means that the total external force is zero. An external force is one exerted on the body by something outside it. Here, the upward pencil reaction balances the downward weight, so there is no unbalanced force changing the motion of the cardboard as a whole.

Rotational equilibrium means that the total external turning effect is zero. In the balanced position, gravity does not produce an unbalanced moment about the support. If an unbalanced turning effect remained, the cardboard would tilt and fall from the pencil tip.

Mechanical equilibrium requires both conditions. Force balance alone does not establish that a body will remain without turning. In this demonstration, the horizontal cardboard is at rest because its forces balance and their turning effects balance as well.

How should the diagram be read?

For the diagram below, M denotes the cardboard's total mass, meaning its quantity of matter. The symbol g denotes acceleration due to gravity, the change of velocity per unit time caused by gravity. Velocity means speed in a specified direction. The product Mg denotes its total weight. The label R denotes the upward reaction of the pencil tip.

An irregular cardboard sheet rests on a pencil tip at G. An upward arrow labelled R and a downward arrow labelled Mg lie along the support line. Smaller downward arrows represent gravitational forces on separate parts of the cardboard.

See Fig. 6.24 in your NCERT textbook

The smaller arrows in the figure are labelled m₁g and m₂g, where m₁ and m₂ denote the masses of two individual parts. They illustrate that gravity acts throughout the cardboard, even though the total weight can be represented through G.

The trial-and-error method consists of changing the position of the tip under the sheet until the sheet balances horizontally. It is a direct demonstration of the centre of gravity as a balance point, rather than a rule for finding it from the outline alone.

How are centre of gravity and centre of mass related?

The centre of mass is the point representing the distribution of mass in a body. Mass is the quantity of matter in the body. The position of the centre of mass depends on how that mass is distributed, and its definition does not depend on gravity.

The centre of gravity concerns the combined effect of gravitational forces. These are different concepts, even though their positions coincide in a uniform gravitational field, meaning that gravitational acceleration has the same magnitude and direction throughout the body.

When can their positions be treated as the same?

For a body small enough that gravitational acceleration does not vary appreciably from one part to another, the same value of g can be used throughout it. Its centre of mass and centre of gravity then coincide. This is the condition used for the regular bodies discussed here.

To understand the connection qualitatively, consider equal masses at different positions in such a field. They have equal weights. A symmetrical arrangement, with matching masses about the centre, therefore produces the corresponding symmetrical arrangement of gravitational forces. The mass balance point is consequently also the gravitational balance point.

What qualification must be retained?

The phrase in a uniform gravitational field is part of the result. It should not disappear when writing a definition or explaining why the geometric centre, the centre identified from the shape, is also the centre of gravity of a small uniform object.

A homogeneous body has uniformly distributed mass. This is a property of the material distribution within the object. A uniform gravitational field is a property of gravity around it. The two conditions describe different things and must not be treated as interchangeable.

For the small cardboard used in the suspension experiment, locating the centre of gravity also locates the centre of mass. That conclusion uses the small-body condition; it does not change the meanings of the two terms.

Why is the centre of gravity of a uniform rod at its midpoint?

For a homogeneous rod of uniform cross-section in uniform gravity, the centre of gravity is at its geometric centre. A cross-section is the shape obtained by cutting across the rod. For a thin rod, its position along the length is the midpoint, halfway between the ends.

The result follows from symmetry, meaning a matching arrangement of parts about a point, line or plane. In this case, corresponding portions of the rod on opposite sides of the midpoint have equal masses and lie at equal distances from that midpoint.

Derivation: Why is the midpoint the centre of gravity of a uniform rod?

Take the midpoint as the origin and measure position along the rod by $x$. Let $M$ be its total mass and $\Delta m$ the mass of each small portion. Positions are measured in metres and masses in kilograms.

1. Pair a portion at $x$ with an equal mass at $-x$. Uniformity supplies such matching portions throughout the rod.
2. Each pair contributes zero to the mass-weighted sum: $\Delta m x+\Delta m(-x)=0$.
3. Adding every pair gives $\sum_i m_i x_i=0$, so the centre-of-mass coordinate is $X=\frac{\sum_i m_i x_i}{M}=0$. It is therefore at the chosen origin, the midpoint.
4. In uniform gravity, the centre of gravity coincides with the centre of mass. The equal weights on opposite sides also have equal and opposite moments about the midpoint.

Result: The centre of gravity is at the midpoint in uniform gravity.

The reasoning uses the distribution of mass, not simply the fact that the outline is straight. The matching arrangement allows the effects on opposite sides to balance. It is this physical correspondence that gives the midpoint its significance.

What should a location answer include?

A complete answer identifies both the position and the condition: the centre of gravity of a homogeneous rod with uniform cross-section is at its geometric centre in uniform gravity. For a thin rod, saying “at its midpoint” gives the required location along the length.

The same symmetry reasoning applies to homogeneous thicker rods with circular or rectangular cross-sections. Their centre lies within the body, centrally along the length and across the cross-section. The narrow-rod picture is a simpler way to see the pairing of corresponding mass elements.

Note: “Regular shape” and “uniform mass distribution” are not the same statement. Before using a geometric centre as G, check the mass-distribution condition as well as the shape and the assumption of uniform gravity.

If the uniformity condition is missing, the midpoint conclusion has not been established by this argument. It is better to state the required condition explicitly than to assume that any object called a rod has its centre of gravity halfway along its length.

Where is the centre of gravity of a uniform ring, disc or sphere?

A ring has material surrounding an open central region. A disc is a flat circular plate. A sphere is a round three-dimensional body whose surface points are equally distant from its centre. Their geometric centres can be identified from their shapes.

For homogeneous versions of these bodies in uniform gravity, the centre of gravity is at the geometric centre. Corresponding equal masses occupy opposite positions about that centre. Their gravitational effects therefore balance in the same way as corresponding portions of a uniform rod.

How can the results be compared?

The table gives locations under stated conditions. It does not assert that every object with a circular boundary has the same mass distribution. A shape-based conclusion is justified here because the bodies are homogeneous and the gravitational field is uniform.

Must G be occupied by material?

No. The uniform circular ring is a useful example: its centre of gravity is at the circle's centre, where there is no ring material. G is a representative point determined by the gravitational effects of the material around it.

This does not mean that matter has moved into the opening. Nor does it mean that the weight is absent because the centre is empty. The ring retains its distributed material and its weight; the combined effect is represented through its centre.

The disc and ring share the same central location when each is uniform, despite the difference in whether the central region contains material. The reason is their symmetrical mass distribution about the centre, not the presence of a particle at that centre.

When labelling a drawing, mark G at the actual centre appropriate to the shape. For a ring, placing it on the circumference would confuse the location of material with the location of the centre of gravity.

How is the centre of gravity of a uniform triangular lamina located?

For a uniform triangular lamina in uniform gravity, the centre of gravity lies at the centroid, the point where the triangle's medians meet. A median is a straight line joining a vertex, or corner, to the midpoint of the opposite side.

This result concerns a thin triangular sheet with uniformly distributed mass. It is not a rule about arbitrary masses placed at the corners of a triangle. The shape of the sheet and the way its mass is spread over that shape both matter.

Derivation: Why do the medians determine the position?

1. Imagine dividing the lamina into very narrow strips parallel to one side, chosen as the base. Parallel lines lie in the same plane and do not meet. Each strip has its centre of mass at its midpoint because its mass is distributed symmetrically along its length.
2. The strip midpoints lie along the median from the opposite vertex. The centre of mass of the whole lamina must therefore lie on this median.
3. Repeating the reasoning with another side as the base places it on another median as well. The point common to these medians is the centroid.
4. In uniform gravity, the centre of mass is also the centre of gravity.

Result: The centre of gravity of a uniform triangular lamina in uniform gravity is at the centroid. This explanation establishes a position by using the distribution of the sheet's mass, without calculating the contribution from each separate strip.

What does the labelled figure show?

In the following figure, L, M and N label the triangle's vertices. Here M is a vertex label, not the mass symbol used in the cardboard diagram. P, Q and R label the midpoints of the sides opposite L, M and N respectively.

The triangle has L at the top and M and N at the ends of the base. Horizontal strips are drawn across it. The medians LP, MQ and NR meet at G, the centroid.

See Fig. 6.10 in your NCERT textbook

In this figure, R labels a side midpoint; it is not the reaction-force symbol used earlier. Diagram letters must be interpreted using the definitions belonging to that particular drawing. The point G identifies the intersection used to locate the centre.

To construct the position, mark the side midpoints and draw the corresponding medians. Their common intersection identifies G. The construction and the physical explanation belong together: the medians locate the centroid geometrically, while uniform mass distribution and uniform gravity justify identifying it as the centre of gravity.

How can the centre of mass be calculated from the mass distribution?

A mass-weighted mean locates the centre of mass. Choose an origin and measure all positions from it. In uniform gravity, the resulting point also locates the centre of gravity. For a flat arrangement, calculate both coordinates using the same masses.

The formulas are $M=\sum_i m_i$, $X=\frac{\sum_i m_i x_i}{M}$ and $Y=\frac{\sum_i m_i y_i}{M}$. Here $m_i$ is each mass, $M$ is total mass, and $x_i,y_i$ are its coordinates. The symbol $\sum$ means add all contributions.

Use consistent mass and length units: kilograms and metres give $X$ and $Y$ in metres. The mass unit cancels between numerator and denominator. A uniform component can be represented by its mass at its own centre when combining components.

How do unequal masses at a triangle's vertices affect its centre?

Worked example 1. Particles of masses 100 g, 150 g and 200 g occupy vertices O, A and B of an equilateral triangle of side 0.5 m. Find their centre of mass, taking O as the origin and OA as the horizontal axis.

Formula: $M=m_1+m_2+m_3$, $X=\frac{m_1x_1+m_2x_2+m_3x_3}{M}$, $Y=\frac{m_1y_1+m_2y_2+m_3y_3}{M}$.

Substitute: The coordinates in metres are O $(0,0)$, A $(0.5,0)$ and B $(0.25,0.25\sqrt{3})$. The total mass is $M=100+150+200=450\,\mathrm{g}$.

$X=\frac{100(0)+150(0.5)+200(0.25)}{450}=\frac{125}{450}=\frac{5}{18}\,\mathrm{m}$.

$Y=\frac{100(0)+150(0)+200(0.25\sqrt{3})}{450}=\frac{\sqrt{3}}{9}\,\mathrm{m}$.

Answer: The centre is approximately 0.278 m along OA and 0.192 m above it. It is not the triangle's geometric centre because the masses at its vertices are unequal.

How can a composite lamina be divided into simpler shapes?

Worked example 2. A uniform L-shaped lamina has mass 3 kg and consists of three squares of side 1 m. Taking the lower left outer corner as origin, their centres are at $(0.5,0.5)$, $(1.5,0.5)$ and $(0.5,1.5)$, in metres. Find its centre of mass.

Formula: $X=\frac{m_1x_1+m_2x_2+m_3x_3}{M}$, $Y=\frac{m_1y_1+m_2y_2+m_3y_3}{M}$.

Substitute: Each equal square has mass $m=3/3=1\,\mathrm{kg}$. Treat these masses as acting at the square centres.

$X=\frac{1(0.5)+1(1.5)+1(0.5)}{3}=\frac{2.5}{3}=\frac{5}{6}\,\mathrm{m}$.

$Y=\frac{1(0.5)+1(0.5)+1(1.5)}{3}=\frac{5}{6}\,\mathrm{m}$.

Answer: Both coordinates are approximately 0.833 m from the chosen origin. In uniform gravity, this point is also the centre of gravity.

Where is the centre of mass between two unequal atomic masses?

Worked example 3. The nuclei in an HCl molecule are separated by about 1.27 Å, where $1\,\text{Å}=10^{-10}\,\mathrm{m}$. Chlorine is about 35.5 times as massive as hydrogen. Taking nearly all each atom's mass at its nucleus, find the approximate centre of mass.

Formula: $X=\frac{m_1x_1+m_2x_2}{m_1+m_2}$. Choose the hydrogen nucleus as origin and measure towards chlorine. Let the hydrogen mass be $m$; the chlorine mass is $35.5m$.

Substitute: $X=\frac{m(0)+35.5m(1.27)}{m+35.5m}=\frac{45.085}{36.5}\,\text{Å}$. Thus $X\approx1.2352\,\text{Å}$.

The distance from chlorine is $d=1.27-1.2352\approx0.0348\,\text{Å}$.

Answer: The centre is about 0.000000000124 m from hydrogen towards chlorine, or $1.24\times10^{-10}\,\mathrm{m}$. It lies much closer to the heavier chlorine nucleus, about 3.48 × 10⁻¹² m from it.

How is the centre of gravity of an irregular lamina found by suspension?

An irregular lamina is a thin plate without a regular geometrical outline. For such a sheet, looking for an obvious geometric centre is unreliable. The suspension method finds its centre of gravity by observing the directions of vertical lines through different points from which it hangs.

A point of suspension is the point from which the body is hung. When a small lamina hangs freely at rest, the vertical through its suspension point passes through its centre of gravity. Each suspension therefore supplies a line on which G must lie.

What is the experimental procedure?

Use an irregular cardboard lamina and a support from which it can turn freely. A plumb line is a thread carrying a small weight, called a plumb bob, which indicates the vertical when hanging at rest. It can be used to mark the required direction.

1. Suspend the lamina freely from a point near its edge. Allow it to settle at rest before marking a line.
2. Use the vertical through that suspension point to draw a line on the lamina. A plumb line hanging from the same support provides the vertical reference.
3. Suspend the same lamina from a different point. Once it is at rest, mark the new vertical through that point on the sheet.
4. Identify the intersection of the marked lines. Both pass through the centre of gravity, so their intersection locates G.
5. Repeat the suspension from another point and mark its vertical. This provides an additional check on the location already found.

How are the lines labelled?

In the diagram below, A, B and C identify different suspension points on the lamina. A₁, B₁ and C₁ identify points on the corresponding marked lines. The subscripts distinguish these labels; they do not represent measurements or powers.

An irregular outline hangs from the upper suspension arrangement. The dashed line through A and A₁ passes through G. Dashed lines through B and B₁, and through C and C₁, also meet at G.

See Fig. 6.25 in your NCERT textbook

Only the line for the current suspension is vertical in the room. When the lamina is hung from another point, it turns into a different orientation. The earlier pencil line turns with the cardboard, while a new vertical is marked through the new support.

The lines remain on the same sheet, allowing their intersection to be compared after the suspensions. This is why the method can locate a single point even though the sheet has been held in different orientations during the observations.

One suspension establishes a line, not a unique point. Different suspension points are needed to produce distinct lines. In an accurate construction they pass through the same G; the extra suspension checks the consistency of the result.

Why does the suspension method work, and how should its result be checked?

The suspension method rests on rotational equilibrium. Consider the lamina after it has settled at rest. Its support force acts at the suspension point, so it has no turning effect about that point. The weight must also have zero turning effect about the support.

For this to happen, the line of action of the weight passes through the suspension point. Since the weight acts vertically through G, the support and G lie on the same vertical line. This explains the observation used to locate the centre.

Why is one marked vertical insufficient?

A single line contains many points, so it does not specify which point is G. A second distinct suspension gives another line containing G. Their intersection supplies the location common to both observations, provided the marked lines are distinct and cross.

The additional line from a third suspension is a check, not a different centre of gravity. Changing the support changes the orientation of the sheet. It does not change the mass distribution of the unchanged sheet or select a new material point as its centre.

Which precautions follow from the explanation?

• Allow free suspension: contact with another object can introduce an additional force and turning effect, so the simple weight-and-support explanation would no longer describe the arrangement.
• Wait for rest: mark the line after the sheet and plumb line settle, since the reasoning concerns equilibrium rather than a moving sheet.
• Mark the vertical: use the actual hanging direction through the support, rather than a guessed centre line based on the outline.
• Compare distinct lines: check whether the lines from different suspension points identify a common intersection instead of assuming the first mark is sufficient.

How do the two experimental methods compare?

The pencil method locates a balance point by supporting the cardboard horizontally from below. The suspension method locates intersecting lines by hanging it from different points. Both depend on gravitational turning effects balancing at equilibrium, although their arrangements and observations differ.

A clear experimental conclusion names the method, identifies the observed intersection or balance point, and labels it G. The conclusion should describe the unchanged lamina tested. It should not claim that a guessed geometric centre was established merely because the outline appeared nearly symmetrical.

If marked lines fail to identify a common point, recheck free suspension, rest and line marking before reporting a location. A disagreement between construction lines is a reason to examine the observations, not evidence that the same unchanged lamina has several centres of gravity.

How are balance problems solved using the centre of gravity?

Replace each body's weight by a downward force through its centre of gravity. Use both force balance and moment balance. For a force perpendicular to its arm, $\tau=Fd$, where $F$ is force in newtons, $d$ is the perpendicular distance in metres and $\tau$ is moment in newton metres.

Write the weight as $W=mg$, where $m$ is mass in kilograms and $g$ is gravitational acceleration in metres per second squared. In the numerical force calculations below, use $g=9.8\,\mathrm{m\,s^{-2}}$.

Choose the point about which moments are taken to simplify the calculation. A support force through that point has zero moment. At equilibrium, the clockwise and anticlockwise moments balance, while the upward support forces equal the total downward weight.

How is a loaded bar supported by two knife edges?

Worked example 4. A uniform metal bar is 70 cm long and has mass 4.00 kg. Two knife edges support it 10 cm from each end. A 6.00 kg load hangs 30 cm from the left end. Find the support reactions.

Formula: $W=mg$, $P=m_Lg$, $R_1+R_2=W+P$. Here $P$ is the load's weight and $R_1,R_2$ are the left and right upward reactions.

Substitute: The bar's centre is 35 cm from the left end. Each support is 0.25 m from this centre; the load is 0.05 m to its left. The weights are $W=4.00(9.8)=39.2\,\mathrm{N}$ and $P=6.00(9.8)=58.8\,\mathrm{N}$.

Force balance gives $R_1+R_2=98.0\,\mathrm{N}$. Taking anticlockwise moments as positive about the centre gives $-0.25R_1+0.05(58.8)+0.25R_2=0$, with moments in newton metres.

Rearranging, $R_1-R_2=\frac{0.05(58.8)}{0.25}=11.76\,\mathrm{N}$. Adding the force equations gives $2R_1=109.76\,\mathrm{N}$, so $R_1=54.88\,\mathrm{N}$ and $R_2=98.0-54.88=43.12\,\mathrm{N}$.

Answer: The left reaction is 54.88 N and the right reaction is 43.12 N, both upward. Their sum equals the combined weight of 98.0 N.

How is a car's weight shared between its wheels?

Worked example 5. A car has mass 1800 kg. Its front and rear axles are 1.8 m apart, and its centre of gravity is 1.05 m behind the front axle. Find the upward force from level ground on each front and rear wheel.

Formula: $W=mg$, $R_f+R_b=W$. Here $R_f$ and $R_b$ are the total reactions at the front and rear axles. Taking moments about the front axle gives $1.8R_b=1.05W$, with distances in metres.

Substitute: $W=1800(9.8)=17640\,\mathrm{N}$, so $R_b=\frac{1.05(17640)}{1.8}=10290\,\mathrm{N}$. Hence $R_f=17640-10290=7350\,\mathrm{N}$.

Sharing each axle's reaction equally between its two wheels gives $F_f=7350/2=3675\,\mathrm{N}$ and $F_b=10290/2=5145\,\mathrm{N}$.

Answer: Each front wheel carries 3675 N and each rear wheel 5145 N. These per-wheel values assume equal sharing across each axle; the four forces add to 17640 N.

How can an unknown mass be found with a metre stick?

Worked example 6. A metre stick balances at its centre on a knife edge. Two coins, each of mass 5 g, are placed together at the 12.0 cm mark. The loaded stick balances at 45.0 cm. Find the stick's mass.

Formula: Balance clockwise and anticlockwise moments about the knife edge: $m_cgd_c=Mgd_s$. Here $m_c$ is total coin mass, $M$ is stick mass, and $d_c,d_s$ are their distances from the support.

Substitute: The unchanged stick's centre of gravity remains at 50.0 cm. The coin mass is $m_c=2(5)=10\,\mathrm{g}$. The opposite moment arms are $d_c=45.0-12.0=33.0\,\mathrm{cm}$ and $d_s=50.0-45.0=5.0\,\mathrm{cm}$.

Cancel the common gravitational acceleration. Using grams and centimetres consistently gives $M=\frac{m_cd_c}{d_s}=\frac{10(33.0)}{5.0}=66\,\mathrm{g}$.

Answer: The metre stick has mass 66 g. The shorter arm of its weight balances the longer arm of the lighter coins.

Glossary

• Centre of gravity — The point through which the total weight of a body may be considered to act.
• Weight — The gravitational force acting on a body because of the gravitational field around it.
• Centre of mass — The point representing a body's mass distribution, whose definition does not depend on gravity.
• Uniform gravitational field — A field in which gravitational acceleration has the same magnitude and direction throughout the region considered.
• Homogeneous body — A body whose mass is uniformly distributed throughout its material.
• Lamina — A thin, flat plate, such as the cardboard used in a balancing or suspension experiment.
• Moment — The turning effect of a force about a specified point or axis.
• Line of action — The straight line along which a force acts on a body.
• Translational equilibrium — The condition in which the total external force acting on a body is zero.
• Rotational equilibrium — The condition in which the total external turning effect acting on a body is zero.
• Median — A straight line joining a triangle's vertex to the midpoint of its opposite side.
• Centroid — The point where the medians of a triangle meet one another.
• Point of suspension — The point from which a body is hung during a suspension experiment.
• Plumb line — A thread carrying a small weight that indicates the vertical direction when hanging at rest.

Common errors and misconceptions

• Misconception: All the matter of a body is physically concentrated at its centre of gravity. Correct: G represents the combined gravitational effect of matter distributed throughout the body; it is not a collection of all that matter.
• Misconception: Centre of gravity and centre of mass mean the same thing. Correct: They are different concepts. Their positions coincide when the gravitational field is uniform over the body.
• Misconception: Every rod has its centre of gravity at its midpoint. Correct: The midpoint result requires the appropriate uniform mass distribution and uniform gravity. A straight outline alone does not establish that result.
• Misconception: A centre of gravity must lie on material. Correct: A homogeneous circular ring in uniform gravity has its centre of gravity in its central opening, where there is no ring material.
• Misconception: The centre of a triangular lamina can be found by drawing arbitrary lines from its corners. Correct: For a uniform triangular sheet in uniform gravity, draw medians to the opposite side midpoints and identify their intersection.
• Misconception: One suspension gives the exact point G. Correct: One suspension gives a vertical line containing G. Another distinct suspension line is needed to locate their intersection.
• Misconception: All the marked suspension lines must be vertical in the final drawing. Correct: Each line was vertical when its own suspension point was used. Earlier lines turn with the sheet when it is rehung.
• Misconception: Equal upward and downward forces are sufficient to prove complete equilibrium. Correct: The total turning effect must also be zero. Mechanical equilibrium requires both force balance and moment balance.

Exam-style questions with model answers

Q1. Define centre of gravity and state the total gravitational turning effect about it. [2 marks]

1. The centre of gravity is the point through which the entire weight of a body may be considered to act.
2. The total turning effect of the gravitational forces about this point is zero.

Q2. An irregular cardboard lamina balances horizontally at rest on a narrow pencil tip. State the location of its centre of gravity and explain its translational and rotational equilibrium. [3 marks]

1. The centre of gravity is at the balance point where the pencil tip supports the thin cardboard sheet.
2. The upward reaction of the pencil equals the downward weight, so the total external force is zero. This gives translational equilibrium.
3. The total turning effect about the support is also zero. This gives rotational equilibrium, so the sheet has no unbalanced turning effect making it tilt.

Q3. State the location of the centre of gravity of each of these homogeneous bodies in a uniform gravitational field: a thin rod of uniform cross-section, a circular ring, a circular disc and a sphere. [4 marks]

1. The uniform thin rod has its centre of gravity at its geometric centre, halfway along its length.
2. The circular ring has its centre of gravity at the circle's centre, in its open central region.
3. The circular disc has its centre of gravity at the centre of the flat circular plate.
4. The homogeneous sphere has its centre of gravity at the sphere's geometric centre, inside the body.

Q4. Explain why a uniform triangular lamina in a uniform gravitational field has its centre of gravity at the intersection of its medians. Include the meaning of median in your answer. [5 marks]

1. A median joins a vertex of the triangle to the midpoint of the opposite side. The medians meet at the centroid.
2. Imagine the uniform triangular sheet divided into narrow strips parallel to one side, taken as the base of the triangle.
3. Each strip has its centre of mass at its midpoint because its mass is symmetrically distributed along its length.
4. The strip midpoints lie on the median from the opposite vertex, so the whole sheet's centre of mass lies on that median. Repeating with another base gives another median.
5. Their intersection is therefore the centre of mass. In the uniform gravitational field, the centre of gravity coincides with this point.

Q5. Describe how to locate the centre of gravity of a small irregular cardboard lamina using free suspension, a support, a pencil and a plumb line. Explain the principle and include an additional check. [6 marks]

1. Suspend the cardboard freely from a point near its edge, allowing it to turn and settle at rest without touching another object.
2. Hang the plumb line from the same support. When it is at rest, mark its vertical direction through the suspension point on the cardboard.
3. Suspend the same cardboard from a different point and mark the new vertical after it settles, retaining the first line on the sheet.
4. The intersection of the distinct marked verticals locates the centre of gravity because that point lies on each of the suspension lines.
5. The principle is that, at equilibrium, the weight has no turning effect about the support. Its vertical line of action must therefore pass through the suspension point.
6. Suspend the cardboard from another point and mark a further vertical. Its passage through the same intersection checks the location found.

Q6. A small unchanged lamina is freely suspended at rest from one point and a vertical is marked through that point. Explain why this line does not locate its centre of gravity uniquely and how a second distinct suspension line helps. [3 marks]

1. The centre of gravity lies somewhere on the first marked vertical because the weight's line of action passes through the suspension point at equilibrium.
2. A line contains many possible positions, so the first observation alone does not specify which point on it is the centre of gravity.
3. A second distinct suspension line also contains the same centre of gravity. The intersection of the two lines identifies its location on the unchanged sheet.

Q7. A homogeneous circular ring is in a uniform gravitational field. Where is its centre of gravity, and what does this show about whether G must be occupied by material? [2 marks]

1. The centre of gravity is at the geometric centre of the circular ring, in its central opening.
2. There is no ring material there, showing that a body's centre of gravity need not be occupied by material.

Q8. In a suspension experiment using a plumb line and an unchanged irregular lamina, several marked verticals fail to meet at one point. Give four practical checks before reporting its centre of gravity. [4 marks]

1. Check that the lamina hangs freely and does not touch another object that could exert an extra force.
2. Check that the sheet and the plumb line have settled at rest before the vertical direction is marked.
3. Check that each marked line passes through its suspension point and follows the actual vertical shown by the plumb line.
4. Repeat using distinct suspension points and compare the intersections of the newly checked lines before reporting a single location.

Key takeaways

• The centre of gravity represents the point through which the total weight of a body may be considered to act.
• The total gravitational turning effect about the centre of gravity is zero, although gravity continues to act on the body's parts.
• Centre of mass and centre of gravity describe different ideas, but their positions coincide in a uniform gravitational field.
• Homogeneous rods, circular rings, circular discs and spheres have their centres of gravity at their geometric centres in uniform gravity.
• A ring shows that the centre of gravity need not lie on material belonging to the body.
• A uniform triangular lamina has its centre of gravity at the intersection of its medians in uniform gravity.
• For a freely suspended lamina at rest, the vertical through the suspension point also passes through its centre of gravity.
• Distinct suspension lines locate the centre of gravity at their intersection, while another suspension checks the result.

Test yourself

Why does zero gravitational turning effect about G not mean that the body has no weight?

The turning effects of gravitational forces balance about G. The gravitational forces themselves still act on the body's distributed material, giving the body its weight.

Under what condition do centre of mass and centre of gravity coincide?

They coincide when the gravitational field is uniform throughout the body, so gravitational acceleration has the same magnitude and direction at all its parts.

Why is uniform mass distribution important when identifying a rod's midpoint as G?

It supplies matching equal masses at equal distances on opposite sides of the midpoint. In uniform gravity, their gravitational turning effects balance about that point.

Which regular body demonstrates that G can lie where the body has no material?

A homogeneous circular ring in uniform gravity has G at the circle's centre, within the empty central opening.

What lines locate G in a uniform triangular lamina in uniform gravity?

The medians, joining vertices to opposite side midpoints, meet at the centroid, which is also the lamina's centre of gravity.

Why should a suspended lamina settle before its vertical is marked?

The method uses the equilibrium condition: the weight's line of action passes through the support when the freely suspended lamina is at rest.

Why do the old marked lines turn when the cardboard is rehung?

The marks are attached to the cardboard and turn with it. Each line represents the vertical direction during its own suspension, not during every suspension.

What does an additional suspension check after an intersection has been found?

It checks whether another independently marked vertical passes through the same point, supporting the identification of that intersection as G.