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M3.5 - Statics of rigid bodies

Syllabus
2019
Topic
M3.5
Level
A2

Find the centre of mass of a rigid body

The centre of mass is the point at which the body's total mass may be treated as concentrated when calculating moments. For a uniform body, symmetry locates every coordinate that lies on an axis or plane of symmetry; integration is needed for a remaining coordinate when the shape varies.

xˉ=1M∫x dm,yˉ=1M∫y dm,M=∫dm\bar x=\frac{1}{M}\int x\,\mathrm dm,\qquad \bar y=\frac{1}{M}\int y\,\mathrm dm,\qquad M=\int\mathrm dm

Uniform body Suitable mass element
thin rod or wire dm=λ ds\mathrm dm=\lambda\,\mathrm ds
lamina dm=ρA dA\mathrm dm=\rho_A\,\mathrm dA
solid dm=ρV dV\mathrm dm=\rho_V\,\mathrm dV
solid of revolution about the xx-axis dV=πy2 dx\mathrm dV=\pi y^2\,\mathrm dx

Choose axes that exploit symmetry. Write both the total mass and its first moment using the same density model and limits, then divide first moment by total mass. For a simple composite body, treat each part as a point mass at its own centre of mass:xˉ=∑mixi∑mi,yˉ=∑miyi∑mi.\bar x=\frac{\sum m_i x_i}{\sum m_i},\qquad \bar y=\frac{\sum m_i y_i}{\sum m_i}.A removed piece is entered with negative mass (or negative area or volume when density is common).

A uniform 6×46\times4 rectangular lamina has a 2×22\times2 square removed from its top-right corner. Taking the lower-left corner as the origin, use the full rectangle minus the hole:xˉ=24(3)−4(5)24−4=2.6,yˉ=24(2)−4(3)24−4=1.8.\bar x=\frac{24(3)-4(5)}{24-4}=2.6,\qquad \bar y=\frac{24(2)-4(3)}{24-4}=1.8.The centre shifts left and down, away from the removed mass, which checks the result.

Use mass, not geometric area or volume, unless uniform density makes them proportional. Keep every coordinate relative to one origin, use the perpendicular distance required by the chosen moment, and never assume that the centre of mass lies inside a hollow or concave body.

Test equilibrium, suspension and toppling

A rigid body is in equilibrium only when the resultant force is zero and the resultant moment about any point is zero. Its weight acts vertically through its centre of mass, so the position of that vertical line controls suspension and toppling.

∑F=0,∑MP=0\sum \mathbf F=\mathbf0,\qquad \sum M_P=0

Situation Equilibrium condition
freely suspended from a fixed point the centre of mass lies vertically below the suspension point in stable equilibrium, so the weight has zero moment about the point
body on a horizontal plane the vertical through the centre of mass must meet the contact base; at limiting toppling it passes through an edge
body on an inclined plane provided sliding is prevented, the same vertical line must meet the contact base; increasing inclination moves its intersection towards the downhill edge

First find the centre of mass in body-fixed coordinates. For suspension, rotate the body until the line joining the suspension point to the centre of mass is vertical. For a supported body, identify the possible pivot edge and take moments about it. At the instant of toppling the reaction at the opposite side has fallen to zero and the resultant contact force acts through the pivot edge.

A uniform rectangular block has base length bb measured up the line of greatest slope and its centre of mass is a perpendicular distance hh above the plane. On a rough plane inclined at θ\theta, the vertical through the centre of mass meets the plane htan⁡θh\tan\theta downhill from the middle of the base. It is stable against toppling whilehtan⁡θ<b2,h\tan\theta<\frac b2,and is on the point of toppling when htan⁡θ=b/2h\tan\theta=b/2. This test assumes the available friction prevents sliding first.

Do not align the centre of mass with the plane's normal: weight is vertical. Do not set friction equal to μR\mu R unless sliding is limiting, and do not assume toppling occurs before sliding; both conditions must be checked when the friction information is available.