Hard Further Maths Centres of mass: solids and calculus Questions

Challenging, exam-style Further Maths Centres of mass: solids and calculus questions with worked solutions. Stretch yourself on the hardest centres-of-mass, solid-of-revolution, integration, composite-solids problems.

centres-of-masssolid-of-revolutionintegrationcomposite-solidsmomentssuspension
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
A uniform solid spinner is formed by rotating the region RR bounded by the curve y=3xy=3\sqrt{x}, the xx-axis and the lines x=1x=1 and x=4x=4 through 360∘360^{\circ} about the xx-axis, where the units are centimetres. Find the distance of the centre of mass of the spinner from the yy-axis. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Model the spinner as a uniform solid of revolution

    ρ=constant\rho=\text{constant}

    The spinner is uniform, so its centre of mass depends only on its shape.

  2. Write down the centre-of-mass formula for revolution about the xx-axis

    xˉ=∫14xy2 dx∫14y2 dx\bar{x}=\frac{\int_{1}^{4} x y^{2}\,dx}{\int_{1}^{4} y^{2}\,dx}

    The constant factor πρ\pi\rho cancels between the moment and the volume.

  3. Substitute the equation of the curve

    y2=(3x)2=9xy^{2}=\left(3\sqrt{x}\right)^{2}=9 x

    Squaring the ordinate gives the integrand for both integrals.

  4. Evaluate the denominator (proportional to the volume)

    ∫14y2 dx=67.5\int_{1}^{4} y^{2}\,dx=67.5

    This integral is proportional to the mass of the solid.

  5. Evaluate the numerator (proportional to the moment)

    ∫14xy2 dx=189\int_{1}^{4} x y^{2}\,dx=189

    This is the first moment of the solid about the yy-axis.

  6. Divide the moment by the volume

    xˉ=18967.5=2.8\bar{x}=\frac{189}{67.5}=2.8

    The quotient is the xx-coordinate of the centre of mass.

  7. Recall the model of a uniform solid

    ρ=constant\rho=\text{constant}

    A uniform solid has constant density, so its centre of mass depends only on its shape.

  8. Recall the centre-of-mass formula for a solid of revolution about the xx-axis

    xˉ=∫xy2 dx∫y2 dx\bar{x}=\frac{\int x y^{2}\,dx}{\int y^{2}\,dx}

    The factor πρ\pi\rho appears in the moment and the mass, so it cancels.

  9. Recall the centre-of-mass formula for a solid of revolution about the yy-axis

    yˉ=∫yx2 dy∫x2 dy\bar{y}=\frac{\int y x^{2}\,dy}{\int x^{2}\,dy}

    By symmetry the centre of mass of such a solid lies on the axis of rotation.

  10. Note that the centre of mass lies on the axis of symmetry

    yˉ=0\bar{y}=0

    Every plane through the axis is a plane of symmetry, so the centre of mass is on the axis.

  11. Recall the volume of a solid of revolution

    V=π∫y2 dxV=\pi\int y^{2}\,dx

    This is the mass, divided by the density, up to the constant π\pi.

  12. Recall the standard result for a uniform solid cone

    xˉ=34h from the vertex\bar{x}=\tfrac{3}{4}h\ \text{from the vertex}

    A solid cone has its centre of mass three quarters of the way from the vertex to the base.

  13. Recall the standard result for a uniform solid hemisphere

    xˉ=38r from the centre\bar{x}=\tfrac{3}{8}r\ \text{from the centre}

    A solid hemisphere has its centre of mass three eighths of the radius from the flat face.

  14. Recall the standard result for a uniform hemispherical shell

    xˉ=12r from the centre\bar{x}=\tfrac{1}{2}r\ \text{from the centre}

    A hollow hemisphere has its centre of mass at the midpoint of the radius to the pole.

  15. Recall the volume of a cylinder

    V=πr2lV=\pi r^{2} l

    This is needed to weight the cylinder in the moment equation.

  16. Recall the volume of a cone

    V=13πr2hV=\tfrac{1}{3}\pi r^{2} h

    A cone has one third of the volume of the cylinder on the same base.

  17. Recall the volume of a hemisphere

    V=23πr3V=\tfrac{2}{3}\pi r^{3}

    A hemisphere has half the volume of the sphere of the same radius.

  18. Take moments about the reference plane

    xˉ∑Vi=∑Vixi\bar{x}\sum V_i=\sum V_i x_i

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

  19. State the required distance

    xˉ=2.8 cm\bar{x}=2.8\ \text{cm}

    This is the distance of the centre of mass from the stated plane.

Answer
xˉ=2.8 cm\bar{x}=2.8\ \text{cm}
Question 2
9 markschallenging
A uniform solid uniform cuboid whose base is a square of side 6 cm6\text{ cm} and whose height is 25 cm25\text{ cm} stands on its square base on a rough horizontal plane. The plane is slowly tilted about a horizontal axis until the bung is on the point of moving and the coefficient of friction between the bung and the plane is μ=0.3\mu=0.3. Find the angle of inclination, in degrees to 3 significant figures, at which the bung first moves (whether by sliding or by toppling). Select the correct value from the options given.
Show worked solution

Worked solution

  1. Model the bung as a uniform solid on the point of moving

    ρ=constant\rho=\text{constant}

    On a rough tilted plane a solid either slides or topples, whichever needs the smaller angle.

  2. State the toppling condition

    tan⁡θtopple=312.5=0.24\tan\theta_{\text{topple}}=\frac{3}{12.5}=0.24

    It topples when the vertical through the centre of mass reaches the edge of the base; the half-width is 33 cm and the centre of mass is 12.512.5 cm high.

  3. State the sliding condition

    tan⁡θslide=μ=0.3\tan\theta_{\text{slide}}=\mu=0.3

    It slides when the incline reaches the angle of friction.

  4. Take the smaller of the two critical angles

    θ=min⁡(arctan⁡0.24, arctan⁡0.3)=13.5∘\theta=\min\left(\arctan0.24,\ \arctan0.3\right)=13.5^{\circ}

    Whichever critical angle is smaller is reached first, so that motion begins there.

  5. Recall the model of a uniform solid

    ρ=constant\rho=\text{constant}

    A uniform solid has constant density, so its centre of mass depends only on its shape.

  6. Recall the centre-of-mass formula for a solid of revolution about the xx-axis

    xˉ=∫xy2 dx∫y2 dx\bar{x}=\frac{\int x y^{2}\,dx}{\int y^{2}\,dx}

    The factor πρ\pi\rho appears in the moment and the mass, so it cancels.

  7. Recall the centre-of-mass formula for a solid of revolution about the yy-axis

    yˉ=∫yx2 dy∫x2 dy\bar{y}=\frac{\int y x^{2}\,dy}{\int x^{2}\,dy}

    By symmetry the centre of mass of such a solid lies on the axis of rotation.

  8. Note that the centre of mass lies on the axis of symmetry

    yˉ=0\bar{y}=0

    Every plane through the axis is a plane of symmetry, so the centre of mass is on the axis.

  9. Recall the volume of a solid of revolution

    V=π∫y2 dxV=\pi\int y^{2}\,dx

    This is the mass, divided by the density, up to the constant π\pi.

  10. Recall the standard result for a uniform solid cone

    xˉ=34h from the vertex\bar{x}=\tfrac{3}{4}h\ \text{from the vertex}

    A solid cone has its centre of mass three quarters of the way from the vertex to the base.

  11. Recall the standard result for a uniform solid hemisphere

    xˉ=38r from the centre\bar{x}=\tfrac{3}{8}r\ \text{from the centre}

    A solid hemisphere has its centre of mass three eighths of the radius from the flat face.

  12. Recall the standard result for a uniform hemispherical shell

    xˉ=12r from the centre\bar{x}=\tfrac{1}{2}r\ \text{from the centre}

    A hollow hemisphere has its centre of mass at the midpoint of the radius to the pole.

  13. Recall the volume of a cylinder

    V=πr2lV=\pi r^{2} l

    This is needed to weight the cylinder in the moment equation.

  14. Recall the volume of a cone

    V=13πr2hV=\tfrac{1}{3}\pi r^{2} h

    A cone has one third of the volume of the cylinder on the same base.

  15. Recall the volume of a hemisphere

    V=23πr3V=\tfrac{2}{3}\pi r^{3}

    A hemisphere has half the volume of the sphere of the same radius.

  16. Take moments about the reference plane

    xˉ∑Vi=∑Vixi\bar{x}\sum V_i=\sum V_i x_i

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

  17. Note that a removed part contributes a negative volume

    Vsolid=Vwhole−VholeV_{\text{solid}}=V_{\text{whole}}-V_{\text{hole}}

    Drilling a hole subtracts both volume and moment.

  18. State the required value

    θ=13.5∘\theta=13.5^{\circ}

    This is the critical angle (or its tangent) that the question asks for.

Answer
θ=13.5∘\theta=13.5^{\circ}
Question 3
9 markschallenging
A uniform solid formed by joining a cylinder of radius 5 cm5\text{ cm} and height 12 cm12\text{ cm} to a hemisphere of radius 5 cm5\text{ cm} on one plane face is freely suspended from a point on the rim of the free plane face of the cylinder. Find the angle, in degrees to 3 significant figures, that the axis of symmetry makes with the vertical. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Locate the centre of mass on the axis of symmetry

    xˉ=7.712 cm from the suspension face\bar{x}=7.712\ \text{cm from the suspension face}

    The centre of mass of the token lies on its axis of symmetry.

  2. Hang the body: the centre of mass is vertically below the pivot

    line pivot→centre of mass is vertical\text{line pivot} \to \text{centre of mass is vertical}

    A freely suspended body rests with its centre of mass beneath the point of suspension.

  3. Resolve the offset perpendicular and parallel to the axis

    tan⁡θ=rxˉ\tan\theta=\frac{r}{\bar{x}}

    The perpendicular offset is the radius 55 cm; the axial offset is the distance of the centre of mass from the rim.

  4. Take the inverse tangent

    θ=arctan⁡(0.6483)=33.0∘\theta=\arctan\left(0.6483\right)=33.0^{\circ}

    This is the angle between the axis and the vertical.

  5. Recall the model of a uniform solid

    ρ=constant\rho=\text{constant}

    A uniform solid has constant density, so its centre of mass depends only on its shape.

  6. Recall the centre-of-mass formula for a solid of revolution about the xx-axis

    xˉ=∫xy2 dx∫y2 dx\bar{x}=\frac{\int x y^{2}\,dx}{\int y^{2}\,dx}

    The factor πρ\pi\rho appears in the moment and the mass, so it cancels.

  7. Recall the centre-of-mass formula for a solid of revolution about the yy-axis

    yˉ=∫yx2 dy∫x2 dy\bar{y}=\frac{\int y x^{2}\,dy}{\int x^{2}\,dy}

    By symmetry the centre of mass of such a solid lies on the axis of rotation.

  8. Note that the centre of mass lies on the axis of symmetry

    yˉ=0\bar{y}=0

    Every plane through the axis is a plane of symmetry, so the centre of mass is on the axis.

  9. Recall the volume of a solid of revolution

    V=π∫y2 dxV=\pi\int y^{2}\,dx

    This is the mass, divided by the density, up to the constant π\pi.

  10. Recall the standard result for a uniform solid cone

    xˉ=34h from the vertex\bar{x}=\tfrac{3}{4}h\ \text{from the vertex}

    A solid cone has its centre of mass three quarters of the way from the vertex to the base.

  11. Recall the standard result for a uniform solid hemisphere

    xˉ=38r from the centre\bar{x}=\tfrac{3}{8}r\ \text{from the centre}

    A solid hemisphere has its centre of mass three eighths of the radius from the flat face.

  12. Recall the standard result for a uniform hemispherical shell

    xˉ=12r from the centre\bar{x}=\tfrac{1}{2}r\ \text{from the centre}

    A hollow hemisphere has its centre of mass at the midpoint of the radius to the pole.

  13. Recall the volume of a cylinder

    V=πr2lV=\pi r^{2} l

    This is needed to weight the cylinder in the moment equation.

  14. Recall the volume of a cone

    V=13πr2hV=\tfrac{1}{3}\pi r^{2} h

    A cone has one third of the volume of the cylinder on the same base.

  15. Recall the volume of a hemisphere

    V=23πr3V=\tfrac{2}{3}\pi r^{3}

    A hemisphere has half the volume of the sphere of the same radius.

  16. Take moments about the reference plane

    xˉ∑Vi=∑Vixi\bar{x}\sum V_i=\sum V_i x_i

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

  17. State the required value

    θ=arctan⁡(0.6483)=33.0∘\theta=\arctan\left(0.6483\right)=33.0^{\circ}

    This fixes the resting position of the suspended solid.

Answer
θ=arctan⁡(0.6483)=33.0∘\theta=\arctan\left(0.6483\right)=33.0^{\circ}
Question 4
9 markschallenging
A uniform solid cylinder has radius 6 cm6\text{ cm} and height 14 cm14\text{ cm}. A solid right circular cone of base radius 6 cm6\text{ cm} and depth 8 cm8\text{ cm} is drilled out from the centre of one plane face, the axis of the cone lying along the axis of the cylinder. The point OO is the centre of the other (intact) plane face. Find the distance of the centre of mass of the solid from the point OO, giving your answer to 3 significant figures. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Model the solid as a set of standard components

    xˉ=∑Vixi∑Vi\bar{x}=\frac{\sum V_i x_i}{\sum V_i}

    Each part contributes a volume and a moment about the reference plane.

  2. List the volume and centre of mass of each part

    V=504, xˉ=7;V=−96, xˉ=12V=504,\ \bar{x}=7; V=-96,\ \bar{x}=12

    The factor π\pi is common to every volume and cancels.

  3. Find the total volume (a removed part is negative)

    ∑Vi=504+−96=408\sum V_i=504 + -96=408

    This is proportional to the mass of the assembled solid.

  4. Find the total moment about the reference plane

    ∑Vixi=504×7+−96×12=2376\sum V_i x_i=504\times 7 + -96\times 12=2376

    Positive parts add moment; a drilled part subtracts it.

  5. Divide the total moment by the total volume

    xˉ=2376408=5.824\bar{x}=\frac{2376}{408}=5.824

    This locates the centre of mass along the axis.

  6. Recall the model of a uniform solid

    ρ=constant\rho=\text{constant}

    A uniform solid has constant density, so its centre of mass depends only on its shape.

  7. Recall the centre-of-mass formula for a solid of revolution about the xx-axis

    xˉ=∫xy2 dx∫y2 dx\bar{x}=\frac{\int x y^{2}\,dx}{\int y^{2}\,dx}

    The factor πρ\pi\rho appears in the moment and the mass, so it cancels.

  8. Recall the centre-of-mass formula for a solid of revolution about the yy-axis

    yˉ=∫yx2 dy∫x2 dy\bar{y}=\frac{\int y x^{2}\,dy}{\int x^{2}\,dy}

    By symmetry the centre of mass of such a solid lies on the axis of rotation.

  9. Note that the centre of mass lies on the axis of symmetry

    yˉ=0\bar{y}=0

    Every plane through the axis is a plane of symmetry, so the centre of mass is on the axis.

  10. Recall the volume of a solid of revolution

    V=π∫y2 dxV=\pi\int y^{2}\,dx

    This is the mass, divided by the density, up to the constant π\pi.

  11. Recall the standard result for a uniform solid cone

    xˉ=34h from the vertex\bar{x}=\tfrac{3}{4}h\ \text{from the vertex}

    A solid cone has its centre of mass three quarters of the way from the vertex to the base.

  12. Recall the standard result for a uniform solid hemisphere

    xˉ=38r from the centre\bar{x}=\tfrac{3}{8}r\ \text{from the centre}

    A solid hemisphere has its centre of mass three eighths of the radius from the flat face.

  13. Recall the standard result for a uniform hemispherical shell

    xˉ=12r from the centre\bar{x}=\tfrac{1}{2}r\ \text{from the centre}

    A hollow hemisphere has its centre of mass at the midpoint of the radius to the pole.

  14. Recall the volume of a cylinder

    V=πr2lV=\pi r^{2} l

    This is needed to weight the cylinder in the moment equation.

  15. Recall the volume of a cone

    V=13πr2hV=\tfrac{1}{3}\pi r^{2} h

    A cone has one third of the volume of the cylinder on the same base.

  16. State the distance of the centre of mass

    xˉ=5.82 cm\bar{x}=5.82\ \text{cm}

    This is measured from the stated reference point.

Answer
xˉ=5.82 cm\bar{x}=5.82\ \text{cm}
Question 5
9 markschallenging
A uniform toy is made by joining a solid hemisphere of radius 12 cm12\text{ cm} to a solid right circular cone of base radius 12 cm12\text{ cm} and height 30 cm30\text{ cm}, so that their plane circular faces coincide. The point OO is the centre of the common circular face. Find the distance of the centre of mass of the solid from the point OO, giving your answer to 3 significant figures. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Model the solid as a set of standard components

    xˉ=∑Vixi∑Vi\bar{x}=\frac{\sum V_i x_i}{\sum V_i}

    Each part contributes a volume and a moment about the reference plane.

  2. List the volume and centre of mass of each part

    V=1152, xˉ=−4.5;V=1440, xˉ=7.5V=1152,\ \bar{x}=-4.5; V=1440,\ \bar{x}=7.5

    The factor π\pi is common to every volume and cancels.

  3. Find the total volume (a removed part is negative)

    ∑Vi=1152+1440=2592\sum V_i=1152 + 1440=2592

    This is proportional to the mass of the assembled solid.

  4. Find the total moment about the reference plane

    ∑Vixi=1152×−4.5+1440×7.5=5616\sum V_i x_i=1152\times -4.5 + 1440\times 7.5=5616

    Positive parts add moment; a drilled part subtracts it.

  5. Divide the total moment by the total volume

    xˉ=56162592=2.167\bar{x}=\frac{5616}{2592}=2.167

    This locates the centre of mass along the axis.

  6. Recall the model of a uniform solid

    ρ=constant\rho=\text{constant}

    A uniform solid has constant density, so its centre of mass depends only on its shape.

  7. Recall the centre-of-mass formula for a solid of revolution about the xx-axis

    xˉ=∫xy2 dx∫y2 dx\bar{x}=\frac{\int x y^{2}\,dx}{\int y^{2}\,dx}

    The factor πρ\pi\rho appears in the moment and the mass, so it cancels.

  8. Recall the centre-of-mass formula for a solid of revolution about the yy-axis

    yˉ=∫yx2 dy∫x2 dy\bar{y}=\frac{\int y x^{2}\,dy}{\int x^{2}\,dy}

    By symmetry the centre of mass of such a solid lies on the axis of rotation.

  9. Note that the centre of mass lies on the axis of symmetry

    yˉ=0\bar{y}=0

    Every plane through the axis is a plane of symmetry, so the centre of mass is on the axis.

  10. Recall the volume of a solid of revolution

    V=π∫y2 dxV=\pi\int y^{2}\,dx

    This is the mass, divided by the density, up to the constant π\pi.

  11. Recall the standard result for a uniform solid cone

    xˉ=34h from the vertex\bar{x}=\tfrac{3}{4}h\ \text{from the vertex}

    A solid cone has its centre of mass three quarters of the way from the vertex to the base.

  12. Recall the standard result for a uniform solid hemisphere

    xˉ=38r from the centre\bar{x}=\tfrac{3}{8}r\ \text{from the centre}

    A solid hemisphere has its centre of mass three eighths of the radius from the flat face.

  13. Recall the standard result for a uniform hemispherical shell

    xˉ=12r from the centre\bar{x}=\tfrac{1}{2}r\ \text{from the centre}

    A hollow hemisphere has its centre of mass at the midpoint of the radius to the pole.

  14. Recall the volume of a cylinder

    V=πr2lV=\pi r^{2} l

    This is needed to weight the cylinder in the moment equation.

  15. State the distance of the centre of mass

    xˉ=2.17 cm\bar{x}=2.17\ \text{cm}

    This is measured from the stated reference point.

Answer
xˉ=2.17 cm\bar{x}=2.17\ \text{cm}

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