The centre of gravity (CG) of an object is the point through which the total weight of the object may be considered to act. In a uniform gravitational field the CG coincides with the centre of mass. (In a non‑uniform field the two points are not identical – this is outside the IGCSE syllabus but useful for deeper understanding.)
Core requirement – Experiment to determine the CG of an irregular‑shaped plane lamina
This is the experiment expected in the Cambridge IGCSE 0625 syllabus.
Place the lamina on a smooth table and secure it with a small pin so that it can rotate freely about the pin.
Hang a thin, light string from a point on the edge of the lamina and allow the lamina to swing until it comes to rest.
Use a plumb‑line (or a weight‑filled string) to verify that the hanging string is vertical. Draw a straight line on the lamina in the direction of the string – this is the line of action of the weight for that suspension point.
Repeat steps 2–3 with a different suspension point on the lamina.
The two lines drawn in step 3 intersect at the centre of gravity of the lamina.
Safety note: Use a lightweight string and a sharp but safe pin. Keep the work area clear of loose objects that could become entangled.
Qualitative effect of the CG position on the stability of simple objects
Low CG (close to the supporting surface) – the object is more stable and less likely to tip. Example: a wide, short glass or a car with a low chassis.
High CG (far above the supporting surface) – the object is less stable and will tip more easily. Example: a tall, narrow glass, a person standing on tip‑toes, or a ship with ballast removed.
When the object is tilted, the restoring torque is proportional to the horizontal distance between the line of action of the weight and the pivot point. The lower the CG, the larger this distance and therefore the larger the restoring torque, making the object more stable.
Simple methods to locate the CG (non‑experimental)
Symmetry method: For regular, symmetric objects the CG lies on the line(s) of symmetry (e.g., midpoint of a uniform rod, centre of a rectangle).
Balancing method: Rest the object on a sharp point or knife‑edge. The point at which it balances without rotating is the CG.
Principle of moments: Adjust the support point until the clockwise and anticlockwise moments of the weight are equal; the support point then coincides with the CG.
Only this scalar formula is required for IGCSE; the integral form for continuous bodies is beyond the syllabus.
Examples of CG positions for uniform objects
Object (uniform)
Position of CG
Thin rod of length \(L\)
Mid‑point, \(L/2\) from either end
Rectangular plate (width \(w\), height \(h\))
Intersection of the diagonals, \((w/2,\;h/2)\)
Solid sphere of radius \(R\)
Geometric centre
Irregular lamina (found by experiment)
Intersection of the two lines of action drawn in the experiment above
Common misconceptions
The CG does not have to lie inside the material. For hollow or oddly shaped objects it can be outside the physical body.
Do not confuse the centre of gravity with the centre of pressure (the latter relates to fluid forces, not weight).
Suggested diagrams: (a) a uniform rod balanced on a knife‑edge at its centre, showing equal torques on either side; (b) the experimental setup with a lamina suspended from two points and the intersecting lines indicating the CG.
Summary
The centre of gravity is the single point through which the total weight of an object may be considered to act.
Knowing the CG allows us to predict whether an object will balance or tip, and to design stable structures and vehicles.
The IGCSE‑required experiment locates the CG of any irregular plane lamina by intersecting two lines of action of the weight.
Qualitatively, a lower CG gives a larger restoring torque and therefore greater stability.
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