MG4 - Centres of mass of composite bodies

MG4 - Centres of mass of composite bodies

A complicated uniform body can often be split into simple shapes whose centres of mass you already know. Treat each part as a particle at its own centre, using area, length or volume to find the appropriate mass ratios.

Add simple uniform areas

Can cutting a hole move a body's centre of mass into the hole itself? To understand why removing material can do that, first build a body by adding material and watch its centre move.

You already know

For particles, MrG=∑imiriM\mathbf r_G=\sum_i m_i\mathbf r_i. A uniform rod has its centre at its midpoint; uniform rectangles and triangles have their standard centres. Symmetry may remove a coordinate calculation.

For a composite rigid body, replace each constituent part by its mass concentrated at its own centre of mass. Choose parts whose interiors do not overlap, so no material is counted twice.

Combining uniform laminas

For a lamina of uniform thickness and density, mass is proportional to area. If non-overlapping parts have areas AiA_i and centres (xi,yi)(x_i,y_i), then

xˉ=∑iAixi∑iAi,yˉ=∑iAiyi∑iAi\bar x=\frac{\sum_i A_i x_i}{\sum_i A_i},\qquad \bar y=\frac{\sum_i A_i y_i}{\sum_i A_i}

The common mass per unit area cancels. If actual part masses are given, use those masses instead.

Not in the formula booklet: learn it.

Make a table of area, centre and area times coordinate before combining anything. A sketch should show which physical area each row represents.

[DIAGRAM: asset_name: A composite rectangle and triangle; asset_slug: h645_b067_addition; caption: Replace each non-overlapping area by its area at its own centre; description: A rectangle from x equals zero to six and y equals zero to four joins a right triangle with vertices six zero, nine zero and six four. Their centres three two and seven four thirds are marked, together with the combined centre nineteen fifths twenty-eight fifteenths.]
Replace each non-overlapping area by its area at its own centre

Worked example 1: a rectangle with a triangular extension

Question. A uniform lamina consists of the rectangle 0≤x≤6, 0≤y≤40\le x\le6,\ 0\le y\le4 and the triangle with vertices (6,0)(6,0), (9,0)(9,0), (6,4)(6,4). Coordinates are in centimetres. Find its centre of mass.

Plan. Use the rectangle and triangle as two non-overlapping parts.

PartAreaCentre
Rectangle2424(3,2)(3,2)
Triangle66(7,4/3)(7,4/3)
30xˉ=24(3)+6(7)=114horizontal area moments30yˉ=24(2)+6(43)=56vertical area momentsG=(195,2815)\begin{aligned} 30\bar x&=24(3)+6(7)=114 && \text{horizontal area moments}\\ 30\bar y&=24(2)+6\left(\frac43\right)=56 && \text{vertical area moments}\\ G&=\left(\frac{19}{5},\frac{28}{15}\right) \end{aligned}

Why this step? The rectangle has four times the triangle's area, so it must have four times its weight in the average.

Answer. G=(19/5,28/15)G=(19/5,28/15), in centimetres.

Check. Each coordinate lies between the corresponding coordinates of the two part centres, and GG is closer to the larger rectangle's centre.

Change only the triangle's horizontal reach, from 33 to 6 cm6\,\mathrm{cm}. Its area and horizontal centre both change, so update both entries.

Worked example 2: extending the triangular part

Question. Find GG when the triangle instead has vertices (6,0)(6,0), (12,0)(12,0), (6,4)(6,4), with the rectangle unchanged.

Plan. Recalculate the triangle's area and centre, then combine with the same rectangle.

AT=12(6)(4)=12,GT=(8,43)36xˉ=24(3)+12(8)=16836yˉ=24(2)+12(43)=64G=(143,169)\begin{aligned} A_T&=\tfrac12(6)(4)=12,\qquad G_T=\left(8,\frac43\right)\\ 36\bar x&=24(3)+12(8)=168\\ 36\bar y&=24(2)+12\left(\frac43\right)=64\\ G&=\left(\frac{14}{3},\frac{16}{9}\right) \end{aligned}

Why this step? Keeping the original triangle's area while moving its centre would model a different mass distribution.

Answer. G=(14/3,16/9)G=(14/3,16/9), in centimetres.

Check. Compared with example 1, GG moves right and slightly down, towards the enlarged triangular part.

Splitting a trapezium into a rectangle and triangles is one useful choice, but it is not compulsory. Any complete partition into familiar shapes gives the same answer.

Watch out

Wrong: average the two part centres with equal weights.

Right: use the parts' masses, or their areas when the lamina has common uniform thickness and density. Two shapes do not automatically mean two equal masses.

Extend the same triangular part once more.

Remove a hole by subtracting its moments

To cut out a hole, start with the complete body and subtract the mass and moments of the removed part. A negative entry in the calculation represents removal; there is no physical negative mass.

A uniform lamina with a hole

If an original lamina has area AA and centre (xA,yA)(x_A,y_A), and a hole has area H<AH<A and centre (xH,yH)(x_H,y_H), then the remainder has

xˉ=AxA−HxHA−H,yˉ=AyA−HyHA−H\bar x=\frac{Ax_A-Hx_H}{A-H},\qquad \bar y=\frac{Ay_A-Hy_H}{A-H}

The original lamina and the removed piece must have the same uniform mass per unit area.

Not in the formula booklet: learn it.

The subtraction follows by writing “original equals remainder plus removed piece”. Removing mass on the right generally moves the centre of the remainder to the left.

Worked example 3: an off-centre circular hole

Question. A circular hole of radius 1 cm1\,\mathrm{cm}, centred at (4,2)(4,2), is cut from the uniform rectangle 0≤x≤6, 0≤y≤40\le x\le6,\ 0\le y\le4. Find the centre of mass of the remainder.

Plan. Subtract the hole's area and its area moments from those of the complete rectangle.

AR=24,GR=(3,2),AH=π,GH=(4,2)(24−π)xˉ=24(3)−π(4)=72−4π(24−π)yˉ=24(2)−π(2)=2(24−π)\begin{aligned} A_R&=24,\quad G_R=(3,2),\qquad A_H=\pi,\quad G_H=(4,2)\\ (24-\pi)\bar x&=24(3)-\pi(4)=72-4\pi\\ (24-\pi)\bar y&=24(2)-\pi(2)=2(24-\pi) \end{aligned}

Why this step? The removed circle has its own centre, (4,2)(4,2), rather than the rectangle's centre.

Answer.

G=(72−4π24−π,2)≈(2.85,2) cmG=\left(\frac{72-4\pi}{24-\pi},2\right) \approx(2.85,2)\,\mathrm{cm}

Check. Reflection in y=2y=2 remains a symmetry. Also, xˉ−3=−π/(24−π)<0\bar x-3=-\pi/(24-\pi)<0, consistent with removing material from the right.

Move only the hole's centre upwards by 1 cm1\,\mathrm{cm}. Its area stays the same, but the vertical symmetry is lost.

Worked example 4: moving the hole

Question. Find the centre of mass when the same circular hole is centred at (4,3)(4,3) instead.

Plan. Keep the horizontal equation, but update the removed vertical moment.

(24−π)xˉ=72−4π(24−π)yˉ=24(2)−π(3)=48−3πG=(72−4π24−π,48−3π24−π)\begin{aligned} (24-\pi)\bar x&=72-4\pi\\ (24-\pi)\bar y&=24(2)-\pi(3)=48-3\pi\\ G&=\left(\frac{72-4\pi}{24-\pi},\frac{48-3\pi}{24-\pi}\right) \end{aligned}

Why this step? Moving the hole changes where mass is removed, rather than the amount removed.

Answer. G≈(2.85,1.85) cmG\approx(2.85,1.85)\,\mathrm{cm}, with exact coordinates above.

Check. Both coordinate shifts from (3,2)(3,2) are −π/(24−π)-\pi/(24-\pi): the hole centre is one unit right and one unit above the original centre.

There is a striking limiting example. Take a disc of radius 3a3a, and remove a circular hole of radius 2a2a whose centre is distance aa to the right of the disc's centre. The hole fits, touching the outer rim. Along that diameter,

xˉ=9πa2(0)−4πa2(a)5πa2=−45a\bar x=\frac{9\pi a^2(0)-4\pi a^2(a)}{5\pi a^2}=-\frac45a

This point is only 9a/5<2a9a/5<2a from the hole's centre, so the remaining crescent has its centre of mass inside the empty hole. Cutting away material really can produce the situation in the opening question.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account