Centres of mass flashcards

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Centre of mass
The point at which the whole weight of a body or system can be taken to act.
Lamina
A flat body whose thickness is ignored; for a uniform lamina, mass is proportional to area.
Uniform (lamina or wire)
Mass per unit area (lamina) or per unit length (wire) is the same everywhere.
Framework
A body made of uniform wire or thin rods; each piece's mass is proportional to its length.
Freely suspended
Hanging from a single point (or string) with no other support, so it can turn until it is in equilibrium.
About to topple (on a plane)
The position in which the vertical through the centre of mass passes through the lowest point of contact.
Centre of mass of particles on a line
x̄ = Σmx / Σm (positions measured from a chosen origin, with signs). (No — learn it)
Centre of mass of particles in a plane (vector form)
r̄ = Σm r / Σm; in components x̄ = Σmx/Σm, ȳ = Σmy/Σm. (No — learn it)
Centre of mass of a uniform triangular lamina
⅔ of the way along each median from the vertex; with coordinates, the mean of the three vertices. (Yes)
Centre of mass of a uniform sector, radius r, angle 2α
On the axis of symmetry, 2r sin α / 3α from the centre (α in radians). (Yes)
Centre of mass of a uniform semicircular lamina, radius r
4r/3π from the centre, on the axis of symmetry (sector with α = π/2). (Yes — the sector with α = π/2)
Centre of mass of a uniform circular arc, radius r, angle 2α
On the axis of symmetry, r sin α / α from the centre. (Yes)
Centre of mass of a uniform semicircular arc, radius r
2r/π from the centre, on the axis of symmetry (arc with α = π/2). (Yes — the arc with α = π/2)
Critical angle for toppling on a plane
tan α = d/h: d = distance of G up the slope from the lowest contact point, h = height of G above the plane. (No — derive it)
Sketch for a lamina suspended from P
Draw PG vertical; the angle a line PQ makes with the vertical is angle QPG, found in a right-angled triangle.
Table for a composite lamina
One row per piece (area, x, y), a negative area for any removed piece, and a final row for the whole (total area, x̄, ȳ).
Diagram of a lamina about to topple on a slope
The vertical line through G passes exactly through the lower corner of the base.
Where are the centres of mass of a triangular frame and a triangular lamina?
Different points: the frame's is pulled towards the longest side (midpoints weighted by length); the lamina's is the mean of the vertices.
How does symmetry help?
A uniform body's centre of mass lies on every line of symmetry; two lines fix it at their intersection.
Why is a removed piece given a negative area?
Whole = remainder + removed piece, so the remainder's moment = whole's moment − removed piece's moment.
Why does a freely suspended body hang with G below the point of suspension?
Only two forces act (weight at G, force at P); for equilibrium they must act along the same vertical line.
How does a particle attached to a lamina change how it hangs?
Use the centre of mass of the whole system (lamina + particle); that combined point hangs below the support.
Does the centre of mass have to lie on the body?
No. It is a weighted mean position; for an arc, ring or L-shape it can be in empty space.
Topple or slide first on a rough slope?
Compare tan α = μ (sliding) with tan α = d/h (toppling); whichever angle is smaller happens first.
Why do you take moments about a smooth axis?
The unknown force from the axis then has no moment, leaving one equation in the weights.
2 kg at x = 0, 3 kg at x = 4, 5 kg at x = 8. Centre of mass?
Σmx = 0 + 12 + 40 = 52; Σm = 10; x̄ = 5.2.
Uniform sector, radius 9 cm, angle 90°. Distance of G from the centre?
α = π/4: 2(9) sin(π/4) ÷ (3π/4) = 12√2/π ≈ 5.40 cm. (Yes)
Uniform semicircular lamina, radius 6 cm. Distance of G from the centre?
4r/3π = 24/3π = 8/π ≈ 2.55 cm. (Yes)
A 10 cm square has a 3 cm square hole whose centre is 7.5 cm from the left edge (on the horizontal line of symmetry). Distance of G from the left edge?
(100 × 5 − 9 × 7.5) ÷ (100 − 9) = 432.5 ÷ 91 ≈ 4.75 cm.
A uniform 12 cm × 8 cm rectangle ABCD (AB = 12) hangs from A. Angle between AB and the vertical?
G is 6 along AB and 4 from AB: tan θ = 4/6, θ ≈ 33.7°.
A uniform 6 cm wide, 10 cm tall rectangle stands on a rough slope. Toppling angle (no sliding)?
G is 5 cm above the base and 3 cm from the lower corner: tan α = 3/5, α ≈ 31.0°.
Wire frame: equilateral triangle of side 20 cm. Where is its centre of mass?
At the centroid of the triangle, by symmetry: for an equilateral frame the frame and lamina results agree.
A lamina of mass M has G at x = 8/3. It turns about an axis at x = 2. Mass at x = 0 to put the system's G below the axis?
M(8/3 − 2) = m(2 − 0), so m = M/3.
Sector or arc formula with angle 2α
α is HALF the angle at the centre and must be in radians.
“Find the exact distance”
Keep π and surds to the last line; with semicircles, area × distance often makes π cancel.
Measuring positions in a composite lamina
State the origin or reference edge (“distance from AD”) before calculating.
tan θ for a suspended lamina
Opposite = distance of G from the given line; adjacent = distance along the line from the point of suspension. Sketch first.
Total mass with particles attached
Include the lamina's own mass: e.g. 4m lamina + m + 2m = 7m.

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