Further Maths Centres of mass: laminas and frameworks Practice Questions

Free Further Maths Centres of mass: laminas and frameworks practice questions with full step-by-step worked solutions. Covers centre-of-mass, system-of-particles, standard-lamina, triangle. Practise exam-style problems and check your method.

centre-of-masssystem-of-particlesstandard-laminatriangleframework-of-rodssemicircle
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Three particles of mass 2 kg2\text{ kg}, 3 kg3\text{ kg} and 5 kg5\text{ kg} are placed at the points (1, 2)(1,\ 2), (4, 0)(4,\ 0) and (−2, 3)(-2,\ 3) respectively, referred to a fixed origin OO. Find the coordinates of the centre of mass of the system.
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Worked solution

  1. Take moments about the yy-axis

    xˉ=2×1+3×4+5×−210=25\bar{x}=\frac{2\times 1+3\times 4+5\times -2}{10}=\frac{2}{5}

    The xx-coordinate is the mass-weighted mean of the xx-coordinates.

  2. Take moments about the xx-axis

    yˉ=2×2+3×0+5×310=1910\bar{y}=\frac{2\times 2+3\times 0+5\times 3}{10}=\frac{19}{10}

    The yy-coordinate is the mass-weighted mean of the yy-coordinates.

  3. State the centre of mass

    G=(25, 1910)G=\left(\frac{2}{5},\ \frac{19}{10}\right)

    This is the centre of mass of the system of particles.

Answer
(25, 1910)\left(\frac{2}{5},\ \frac{19}{10}\right)
Question 2
2 markseasy
A piece of uniform wire is bent to form a framework consisting of a rod from (0, 0)(0,\ 0) to (0, 8)(0,\ 8) and a rod from (0, 0)(0,\ 0) to (6, 0)(6,\ 0). Which of the following is the xx-coordinate of the centre of mass of the framework?
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Worked solution

  1. Weight each midpoint by rod length

    xˉ=∑Lixi∑Li\bar{x}=\frac{\sum L_{i}x_{i}}{\sum L_{i}}

    Longer rods carry more of the mass of the framework.

  2. Evaluate the length-weighted mean

    97\frac{9}{7}

    An unweighted average of the midpoints would be wrong.

  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. Select the coordinate

    97\frac{9}{7}

    The rods must be weighted by their lengths, not counted equally.

Answer
97\frac{9}{7}
Question 3
4 marksintermediate
A uniform lamina is formed by removing the rectangle with vertices (3, 3)(3,\ 3), (6, 3)(6,\ 3), (6, 6)(6,\ 6) and (3, 6)(3,\ 6) from the rectangle with vertices (0, 0)(0,\ 0), (6, 0)(6,\ 0), (6, 6)(6,\ 6) and (0, 6)(0,\ 6) 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, 6)(0,\ 6) makes with the vertical?
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Worked solution

  1. Find the centre of mass of the lamina

    G=(52, 52)G=\left(\frac{5}{2},\ \frac{5}{2}\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. Select the angle

    45∘45^{\circ}

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

Answer
45∘45^{\circ}
Question 4
6 markshard
A uniform lamina is formed by removing the rectangle with vertices (0, 0)(0,\ 0), (4, 0)(4,\ 0), (4, 6)(4,\ 6) and (0, 6)(0,\ 6) from the rectangle with vertices (0, 0)(0,\ 0), (12, 0)(12,\ 0), (12, 6)(12,\ 6) and (0, 6)(0,\ 6). Which of the following is the xx-coordinate of the centre of mass of the lamina?
Show worked solution

Worked solution

  1. Treat the hole as a negative area

    xˉ=A1x1−A2x2A1−A2\bar{x}=\frac{A_{1}x_{1}-A_{2}x_{2}}{A_{1}-A_{2}}

    The moment of the removed rectangle is subtracted.

  2. Evaluate the coordinate

    88

    Adding the moment instead of subtracting it gives a wrong value.

  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. Select the coordinate

    88

    The hole must be subtracted from both the area and the moment.

Answer
88
Question 5
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}

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