Hard Further Maths Centres of mass: laminas and frameworks Questions

Challenging, exam-style Further Maths Centres of mass: laminas and frameworks questions with worked solutions. Stretch yourself on the hardest centre-of-mass, lamina-with-hole, rectangle, circle problems.

centre-of-masslamina-with-holerectanglecircleframework-of-rodssystem-of-particles
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
A uniform lamina is a sector of a circle of radius 99 and angle π3\frac{\pi}{3} at the centre. Which of the following is the distance of its centre of mass from the centre of the circle?
Show worked solution

Worked solution

  1. Quote the sector result

    2rsin⁡α3α\frac{2r\sin\alpha}{3\alpha}

    Here α\alpha is the half-angle of the sector.

  2. Substitute and simplify

    18π\frac{18}{\pi}

    Evaluate sin⁡α\sin\alpha exactly for this angle.

  3. State the definition of the 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}}

    The centre of mass is the mass-weighted mean position of the body.

  4. Note that a uniform lamina has mass proportional to area

    mi∝Aim_{i}\propto A_{i}

    For a lamina of constant surface density the mass may be replaced by the area throughout.

  5. Note that a uniform rod has mass proportional to length

    mi∝Lim_{i}\propto L_{i}

    For a framework of uniform wire the mass may be replaced by the length of each rod.

  6. Recall the centre of mass of a uniform rod

    at its midpoint\text{at its midpoint}

    By symmetry the centre of mass of a straight uniform rod is at its midpoint.

  7. Recall the centre of mass of a uniform rectangle

    at its centre\text{at its centre}

    The two diagonals are lines of symmetry, so the centre of mass is where they cross.

  8. Recall the centroid of a triangle

    xˉ=x1+x2+x33,yˉ=y1+y2+y33\bar{x}=\frac{x_{1}+x_{2}+x_{3}}{3},\qquad\bar{y}=\frac{y_{1}+y_{2}+y_{3}}{3}

    The centre of mass of a uniform triangular lamina is the mean of its vertices.

  9. Recall the standard result for a semicircular lamina

    4r3π\frac{4r}{3\pi}

    The centre of mass of a uniform semicircular lamina of radius rr lies this distance from the centre of the bounding diameter.

  10. Recall the standard result for a sector

    2rsin⁡α3α\frac{2r\sin\alpha}{3\alpha}

    For a sector of radius rr and half-angle α\alpha the centre of mass lies this distance from the centre, along the axis of symmetry.

  11. Take moments about the yy-axis

    xˉ∑Ai=∑Aixi\bar{x}\sum A_{i}=\sum A_{i}x_{i}

    The moment of the whole equals the sum of the moments of the parts.

  12. Take moments about the xx-axis

    yˉ∑Ai=∑Aiyi\bar{y}\sum A_{i}=\sum A_{i}y_{i}

    Moments about a horizontal axis fix the height of the centre of mass.

  13. Treat a removed region as negative area

    Ahole<0A_{\text{hole}}<0

    Cutting out a region subtracts both its area and its moment.

  14. Use symmetry to fix one coordinate

    xˉ=the axis of symmetry\bar{x}=\text{the axis of symmetry}

    If the body has an axis of symmetry the centre of mass lies on it.

  15. Set up a table of areas and coordinates

    (Ai, xi, yi)\left(A_{i},\ x_{i},\ y_{i}\right)

    A tidy table of the area and centre of each part makes the moment sums reliable.

  16. Select the distance

    18π\frac{18}{\pi}

    This is the exact distance from the centre.

Answer
18π\frac{18}{\pi}
Question 2
9 markschallenging
A uniform lamina is formed by removing the rectangle with vertices (9, 6)(9,\ 6), (16, 6)(16,\ 6), (16, 12)(16,\ 12) and (9, 12)(9,\ 12) from the rectangle with vertices (0, 0)(0,\ 0), (16, 0)(16,\ 0), (16, 12)(16,\ 12) and (0, 12)(0,\ 12) and is freely suspended from the point (0, 0)(0,\ 0). Which of the following is the angle, to the nearest degree, that the edge from (0, 0)(0,\ 0) to (0, 12)(0,\ 12) makes with the vertical?
Show worked solution

Worked solution

  1. Find the centre of mass of the lamina

    G=(33750, 12925)G=\left(\frac{337}{50},\ \frac{129}{25}\right)

    Combine the rectangle and the removed rectangle by moments.

  2. Form the tangent of the angle

    tan⁡θ=horizontal offsetvertical offset\tan\theta=\frac{\text{horizontal offset}}{\text{vertical offset}}

    The offsets are measured from the point of suspension.

  3. State the definition of the 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}}

    The centre of mass is the mass-weighted mean position of the body.

  4. Note that a uniform lamina has mass proportional to area

    mi∝Aim_{i}\propto A_{i}

    For a lamina of constant surface density the mass may be replaced by the area throughout.

  5. Note that a uniform rod has mass proportional to length

    mi∝Lim_{i}\propto L_{i}

    For a framework of uniform wire the mass may be replaced by the length of each rod.

  6. Recall the centre of mass of a uniform rod

    at its midpoint\text{at its midpoint}

    By symmetry the centre of mass of a straight uniform rod is at its midpoint.

  7. Recall the centre of mass of a uniform rectangle

    at its centre\text{at its centre}

    The two diagonals are lines of symmetry, so the centre of mass is where they cross.

  8. Recall the centroid of a triangle

    xˉ=x1+x2+x33,yˉ=y1+y2+y33\bar{x}=\frac{x_{1}+x_{2}+x_{3}}{3},\qquad\bar{y}=\frac{y_{1}+y_{2}+y_{3}}{3}

    The centre of mass of a uniform triangular lamina is the mean of its vertices.

  9. Recall the standard result for a semicircular lamina

    4r3π\frac{4r}{3\pi}

    The centre of mass of a uniform semicircular lamina of radius rr lies this distance from the centre of the bounding diameter.

  10. Recall the standard result for a sector

    2rsin⁡α3α\frac{2r\sin\alpha}{3\alpha}

    For a sector of radius rr and half-angle α\alpha the centre of mass lies this distance from the centre, along the axis of symmetry.

  11. Take moments about the yy-axis

    xˉ∑Ai=∑Aixi\bar{x}\sum A_{i}=\sum A_{i}x_{i}

    The moment of the whole equals the sum of the moments of the parts.

  12. Take moments about the xx-axis

    yˉ∑Ai=∑Aiyi\bar{y}\sum A_{i}=\sum A_{i}y_{i}

    Moments about a horizontal axis fix the height of the centre of mass.

  13. Treat a removed region as negative area

    Ahole<0A_{\text{hole}}<0

    Cutting out a region subtracts both its area and its moment.

  14. Use symmetry to fix one coordinate

    xˉ=the axis of symmetry\bar{x}=\text{the axis of symmetry}

    If the body has an axis of symmetry the centre of mass lies on it.

  15. Select the angle

    53∘53^{\circ}

    Rounding the inverse tangent gives this angle to the nearest degree.

Answer
53∘53^{\circ}
Question 3
9 markschallenging
A uniform triangular lamina has vertices A(0, 0)A(0,\ 0), B(14, 0)B(14,\ 0) and C(6, 9)C(6,\ 9). It is freely suspended from the vertex CC and hangs in equilibrium. Which of the following points of the lamina hangs vertically lowest?
Show worked solution

Worked solution

  1. Apply the suspension principle

    \text{GG hangs directly below CC}

    The lowest point is the one furthest below CC in the vertical direction.

  2. Compare the vertical drops of the candidate points

    project each point onto the downward vertical\text{project each point onto the downward vertical}

    The point with the greatest downward projection hangs lowest.

  3. State the definition of the 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}}

    The centre of mass is the mass-weighted mean position of the body.

  4. Note that a uniform lamina has mass proportional to area

    mi∝Aim_{i}\propto A_{i}

    For a lamina of constant surface density the mass may be replaced by the area throughout.

  5. Note that a uniform rod has mass proportional to length

    mi∝Lim_{i}\propto L_{i}

    For a framework of uniform wire the mass may be replaced by the length of each rod.

  6. Recall the centre of mass of a uniform rod

    at its midpoint\text{at its midpoint}

    By symmetry the centre of mass of a straight uniform rod is at its midpoint.

  7. Recall the centre of mass of a uniform rectangle

    at its centre\text{at its centre}

    The two diagonals are lines of symmetry, so the centre of mass is where they cross.

  8. Recall the centroid of a triangle

    xˉ=x1+x2+x33,yˉ=y1+y2+y33\bar{x}=\frac{x_{1}+x_{2}+x_{3}}{3},\qquad\bar{y}=\frac{y_{1}+y_{2}+y_{3}}{3}

    The centre of mass of a uniform triangular lamina is the mean of its vertices.

  9. Recall the standard result for a semicircular lamina

    4r3π\frac{4r}{3\pi}

    The centre of mass of a uniform semicircular lamina of radius rr lies this distance from the centre of the bounding diameter.

  10. Recall the standard result for a sector

    2rsin⁡α3α\frac{2r\sin\alpha}{3\alpha}

    For a sector of radius rr and half-angle α\alpha the centre of mass lies this distance from the centre, along the axis of symmetry.

  11. Take moments about the yy-axis

    xˉ∑Ai=∑Aixi\bar{x}\sum A_{i}=\sum A_{i}x_{i}

    The moment of the whole equals the sum of the moments of the parts.

  12. Take moments about the xx-axis

    yˉ∑Ai=∑Aiyi\bar{y}\sum A_{i}=\sum A_{i}y_{i}

    Moments about a horizontal axis fix the height of the centre of mass.

  13. Treat a removed region as negative area

    Ahole<0A_{\text{hole}}<0

    Cutting out a region subtracts both its area and its moment.

  14. Use symmetry to fix one coordinate

    xˉ=the axis of symmetry\bar{x}=\text{the axis of symmetry}

    If the body has an axis of symmetry the centre of mass lies on it.

  15. Set up a table of areas and coordinates

    (Ai, xi, yi)\left(A_{i},\ x_{i},\ y_{i}\right)

    A tidy table of the area and centre of each part makes the moment sums reliable.

  16. Add the moments of the parts

    ∑Aixi\sum A_{i}x_{i}

    The total moment is additive because moment is a linear function of position.

  17. Select the lowest point

    BB

    This vertex has the greatest vertical drop below the suspension point.

Answer
BB
Question 4
9 markschallenging
A uniform lamina occupies the region formed by removing the rectangle with vertices (9, 6)(9,\ 6), (13, 6)(13,\ 6), (13, 11)(13,\ 11) and (9, 11)(9,\ 11) from the rectangle with vertices (0, 0)(0,\ 0), (13, 0)(13,\ 0), (13, 11)(13,\ 11) and (0, 11)(0,\ 11). Find the coordinates of the centre of mass of the lamina.
Show worked solution

Worked solution

  1. Record the area and centre of the whole rectangle

    A1=143,G1=(132, 112)A_{1}=143,\quad G_{1}=\left(\frac{13}{2},\ \frac{11}{2}\right)

    The centre of a rectangle is at the midpoint of its diagonals.

  2. Record the area and centre of the removed rectangle

    A2=20,G2=(11, 172)A_{2}=20,\quad G_{2}=\left(11,\ \frac{17}{2}\right)

    The removed region is treated as a negative area.

  3. Take moments about the yy-axis

    xˉ=143×132−20×11143−20=47382\bar{x}=\frac{143\times \frac{13}{2}-20\times 11}{143-20}=\frac{473}{82}

    Subtract the moment of the hole from the moment of the whole.

  4. Take moments about the xx-axis

    yˉ=143×112−20×172143−20=41182\bar{y}=\frac{143\times \frac{11}{2}-20\times \frac{17}{2}}{143-20}=\frac{411}{82}

    Do the same with the yy-coordinates of the two centres.

  5. State the definition of the 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}}

    The centre of mass is the mass-weighted mean position of the body.

  6. Note that a uniform lamina has mass proportional to area

    mi∝Aim_{i}\propto A_{i}

    For a lamina of constant surface density the mass may be replaced by the area throughout.

  7. Note that a uniform rod has mass proportional to length

    mi∝Lim_{i}\propto L_{i}

    For a framework of uniform wire the mass may be replaced by the length of each rod.

  8. Recall the centre of mass of a uniform rod

    at its midpoint\text{at its midpoint}

    By symmetry the centre of mass of a straight uniform rod is at its midpoint.

  9. Recall the centre of mass of a uniform rectangle

    at its centre\text{at its centre}

    The two diagonals are lines of symmetry, so the centre of mass is where they cross.

  10. Recall the centroid of a triangle

    xˉ=x1+x2+x33,yˉ=y1+y2+y33\bar{x}=\frac{x_{1}+x_{2}+x_{3}}{3},\qquad\bar{y}=\frac{y_{1}+y_{2}+y_{3}}{3}

    The centre of mass of a uniform triangular lamina is the mean of its vertices.

  11. Recall the standard result for a semicircular lamina

    4r3π\frac{4r}{3\pi}

    The centre of mass of a uniform semicircular lamina of radius rr lies this distance from the centre of the bounding diameter.

  12. Recall the standard result for a sector

    2rsin⁡α3α\frac{2r\sin\alpha}{3\alpha}

    For a sector of radius rr and half-angle α\alpha the centre of mass lies this distance from the centre, along the axis of symmetry.

  13. Take moments about the yy-axis

    xˉ∑Ai=∑Aixi\bar{x}\sum A_{i}=\sum A_{i}x_{i}

    The moment of the whole equals the sum of the moments of the parts.

  14. Take moments about the xx-axis

    yˉ∑Ai=∑Aiyi\bar{y}\sum A_{i}=\sum A_{i}y_{i}

    Moments about a horizontal axis fix the height of the centre of mass.

  15. Treat a removed region as negative area

    Ahole<0A_{\text{hole}}<0

    Cutting out a region subtracts both its area and its moment.

  16. State the centre of mass

    G=(47382, 41182)G=\left(\frac{473}{82},\ \frac{411}{82}\right)

    This is the centre of mass of the remaining lamina.

Answer
(47382, 41182)\left(\frac{473}{82},\ \frac{411}{82}\right)
Question 5
9 markschallenging
A uniform lamina occupies the region formed by removing the rectangle with vertices (8, 5)(8,\ 5), (16, 5)(16,\ 5), (16, 11)(16,\ 11) and (8, 11)(8,\ 11) from the rectangle with vertices (0, 0)(0,\ 0), (22, 0)(22,\ 0), (22, 14)(22,\ 14) and (0, 14)(0,\ 14). Find the coordinates of the centre of mass of the lamina.
Show worked solution

Worked solution

  1. Record the area and centre of the whole rectangle

    A1=308,G1=(11, 7)A_{1}=308,\quad G_{1}=\left(11,\ 7\right)

    The centre of a rectangle is at the midpoint of its diagonals.

  2. Record the area and centre of the removed rectangle

    A2=48,G2=(12, 8)A_{2}=48,\quad G_{2}=\left(12,\ 8\right)

    The removed region is treated as a negative area.

  3. Take moments about the yy-axis

    xˉ=308×11−48×12308−48=70365\bar{x}=\frac{308\times 11-48\times 12}{308-48}=\frac{703}{65}

    Subtract the moment of the hole from the moment of the whole.

  4. Take moments about the xx-axis

    yˉ=308×7−48×8308−48=44365\bar{y}=\frac{308\times 7-48\times 8}{308-48}=\frac{443}{65}

    Do the same with the yy-coordinates of the two centres.

  5. State the definition of the 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}}

    The centre of mass is the mass-weighted mean position of the body.

  6. Note that a uniform lamina has mass proportional to area

    mi∝Aim_{i}\propto A_{i}

    For a lamina of constant surface density the mass may be replaced by the area throughout.

  7. Note that a uniform rod has mass proportional to length

    mi∝Lim_{i}\propto L_{i}

    For a framework of uniform wire the mass may be replaced by the length of each rod.

  8. Recall the centre of mass of a uniform rod

    at its midpoint\text{at its midpoint}

    By symmetry the centre of mass of a straight uniform rod is at its midpoint.

  9. Recall the centre of mass of a uniform rectangle

    at its centre\text{at its centre}

    The two diagonals are lines of symmetry, so the centre of mass is where they cross.

  10. Recall the centroid of a triangle

    xˉ=x1+x2+x33,yˉ=y1+y2+y33\bar{x}=\frac{x_{1}+x_{2}+x_{3}}{3},\qquad\bar{y}=\frac{y_{1}+y_{2}+y_{3}}{3}

    The centre of mass of a uniform triangular lamina is the mean of its vertices.

  11. Recall the standard result for a semicircular lamina

    4r3π\frac{4r}{3\pi}

    The centre of mass of a uniform semicircular lamina of radius rr lies this distance from the centre of the bounding diameter.

  12. Recall the standard result for a sector

    2rsin⁡α3α\frac{2r\sin\alpha}{3\alpha}

    For a sector of radius rr and half-angle α\alpha the centre of mass lies this distance from the centre, along the axis of symmetry.

  13. Take moments about the yy-axis

    xˉ∑Ai=∑Aixi\bar{x}\sum A_{i}=\sum A_{i}x_{i}

    The moment of the whole equals the sum of the moments of the parts.

  14. Take moments about the xx-axis

    yˉ∑Ai=∑Aiyi\bar{y}\sum A_{i}=\sum A_{i}y_{i}

    Moments about a horizontal axis fix the height of the centre of mass.

  15. State the centre of mass

    G=(70365, 44365)G=\left(\frac{703}{65},\ \frac{443}{65}\right)

    This is the centre of mass of the remaining lamina.

Answer
(70365, 44365)\left(\frac{703}{65},\ \frac{443}{65}\right)

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