Further Maths Centres of mass: solids and calculus Practice Questions

Free Further Maths Centres of mass: solids and calculus practice questions with full step-by-step worked solutions. Covers centres-of-mass, standard-solids, solid-of-revolution, integration. Practise exam-style problems and check your method.

centres-of-massstandard-solidssolid-of-revolutionintegrationcomposite-solidsmoments
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A uniform solid hemisphere of radius 8 cm8\text{ cm} is used as a component. Find the distance of its centre of mass from the centre OO of its plane face.
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Worked solution

  1. Recall (or derive) the standard centre-of-mass result

    xˉ=∫0rx(r2−x2) dx∫0r(r2−x2) dx=38r\bar{x}=\frac{\int_{0}^{r} x (r^{2}-x^{2})\,dx}{\int_{0}^{r} (r^{2}-x^{2})\,dx}=\frac{3}{8}r

    This result comes from the centre-of-mass integral for the solid.

  2. Substitute the given dimension

    xˉ=3r/8=3\bar{x}=3r/8=3

    Putting in the stated length gives the numerical position.

  3. State the distance of the centre of mass

    xˉ=3 cm\bar{x}=3\ \text{cm}

    This is the required distance from the stated point.

Answer
xˉ=3 cm\bar{x}=3\ \text{cm}
Question 2
2 markseasy
A uniform solid paperweight is formed by rotating the region RR bounded by the curve y=xy=x, the xx-axis and the line x=12x=12 through 360∘360^{\circ} about the xx-axis, where the units are centimetres. Find the distance of the centre of mass of the paperweight from the yy-axis. Select the correct value from the options given.
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Worked solution

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

    xˉ=∫012xy2 dx∫012y2 dx\bar{x}=\frac{\int_{0}^{12} x y^{2}\,dx}{\int_{0}^{12} y^{2}\,dx}

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

  2. Evaluate the denominator (proportional to the volume)

    ∫012y2 dx=576\int_{0}^{12} y^{2}\,dx=576

    This integral is proportional to the mass of the solid.

  3. Divide the moment by the volume

    xˉ=5184576=9\bar{x}=\frac{5184}{576}=9

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

  4. State the required distance

    xˉ=9 cm\bar{x}=9\ \text{cm}

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

Answer
xˉ=9 cm\bar{x}=9\ \text{cm}
Question 3
4 marksintermediate
A uniform solid uniform cylinder of radius 4 cm4\text{ cm} and height 15 cm15\text{ cm} stands on one of its circular ends on a rough horizontal plane. The plane is slowly tilted about a horizontal axis until the component is on the point of moving. Find the exact value of tan⁡θ\tan\theta, where θ\theta is the angle of inclination at which the component is on the point of toppling. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Model the component 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=47.5=0.5333\tan\theta_{\text{topple}}=\frac{4}{7.5}=0.5333

    It topples when the vertical through the centre of mass reaches the edge of the base; the half-width is 44 cm and the centre of mass is 7.57.5 cm high.

  3. 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.

  4. 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.

  5. 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.

  6. 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.

  7. State the required value

    tan⁡θ=0.533\tan\theta=0.533

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

Answer
tan⁡θ=0.533\tan\theta=0.533
Question 4
6 markshard
A uniform solid right circular cone of base radius 4 cm4\text{ cm} and height 15 cm15\text{ cm} stands on its base on a rough horizontal plane. The plane is slowly tilted about a horizontal axis until the float is on the point of moving. Find the angle of inclination, in degrees to 3 significant figures, at which the float is on the point of toppling. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Model the float 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=43.75=1.067\tan\theta_{\text{topple}}=\frac{4}{3.75}=1.067

    It topples when the vertical through the centre of mass reaches the edge of the base; the half-width is 44 cm and the centre of mass is 3.753.75 cm high.

  3. Take the inverse tangent

    θ=arctan⁡(1.067)=46.8∘\theta=\arctan\left(1.067\right)=46.8^{\circ}

    This converts the tangent into the angle of inclination.

  4. 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.

  5. 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.

  6. 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.

  7. 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.

  8. 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.

  9. 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.

  10. 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.

  11. 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.

  12. Recall the volume of a cylinder

    V=πr2lV=\pi r^{2} l

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

  13. 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.

  14. State the required value

    θ=46.8∘\theta=46.8^{\circ}

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

Answer
θ=46.8∘\theta=46.8^{\circ}
Question 5
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}

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