Centres of mass: solids and calculus Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Centres of mass: solids and calculus questions. See exactly how to solve problems on centres-of-mass, standard-solids, solid-of-revolution, integration.

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.

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 right circular cone of base radius 5 cm5\text{ cm} and height 12 cm12\text{ cm} is used as a paperweight. Find the distance of its centre of mass from its vertex.

Worked solution

  1. Model the paperweight as a uniform solid

    ρ=constant\rho=\text{constant}

    The centre of mass of a uniform body lies on its axis of symmetry.

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

    xˉ=∫0hx⋅x2 dx∫0hx2 dx=34h from the vertex\bar{x}=\frac{\int_{0}^{h} x\cdot x^{2}\,dx}{\int_{0}^{h} x^{2}\,dx}=\tfrac{3}{4}h\ \text{from the vertex}

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

  3. Substitute the given dimension

    xˉ=3h/4=9\bar{x}=3h/4=9

    Putting in the stated length gives the numerical position.

  4. State the distance of the centre of mass

    xˉ=9 cm\bar{x}=9\ \text{cm}

    This is the required distance from the stated point.

Answer
xˉ=9 cm\bar{x}=9\ \text{cm}
Question 3
2 markseasy
A uniform solid right circular cone of base radius 6 cm6\text{ cm} and height 16 cm16\text{ cm} is used as a plug. Find the distance of its centre of mass from the centre of its base.

Worked solution

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

    xˉ=∫0hx⋅x2 dx∫0hx2 dx=34h from the vertex\bar{x}=\frac{\int_{0}^{h} x\cdot x^{2}\,dx}{\int_{0}^{h} x^{2}\,dx}=\tfrac{3}{4}h\ \text{from the vertex}

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

  2. Substitute the given dimension

    xˉ=h/4=4\bar{x}=h/4=4

    Putting in the stated length gives the numerical position.

  3. State the distance of the centre of mass

    xˉ=4 cm\bar{x}=4\ \text{cm}

    This is the required distance from the stated point.

Answer
xˉ=4 cm\bar{x}=4\ \text{cm}
Question 4
2 markseasy
A uniform hollow hemispherical shell of radius 10 cm10\text{ cm} is used as a casting. Find the distance of its centre of mass from the centre OO of its plane rim.

Worked solution

  1. Model the casting as a uniform solid

    ρ=constant\rho=\text{constant}

    The centre of mass of a uniform body lies on its axis of symmetry.

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

    xˉ=∫0rx r dx∫0rr dx=12r\bar{x}=\frac{\int_{0}^{r} x\,r\,dx}{\int_{0}^{r} r\,dx}=\frac{1}{2}r

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

  3. Substitute the given dimension

    xˉ=r/2=5\bar{x}=r/2=5

    Putting in the stated length gives the numerical position.

  4. State the distance of the centre of mass

    xˉ=5 cm\bar{x}=5\ \text{cm}

    This is the required distance from the stated point.

Answer
xˉ=5 cm\bar{x}=5\ \text{cm}
Question 5
2 markseasy
A uniform solid uniform pyramid of height 20 cm20\text{ cm} on a square base is used as a billet. Find the distance of its centre of mass from the centre of its base.

Worked solution

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

    xˉ=∫0hx⋅(area∝x2) dx∫0h(∝x2) dx=14h from the base\bar{x}=\frac{\int_{0}^{h} x\cdot(\text{area}\propto x^{2})\,dx}{\int_{0}^{h}(\propto x^{2})\,dx}=\tfrac{1}{4}h\ \text{from the base}

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

  2. Substitute the given dimension

    xˉ=h/4=5\bar{x}=h/4=5

    Putting in the stated length gives the numerical position.

  3. State the distance of the centre of mass

    xˉ=5 cm\bar{x}=5\ \text{cm}

    This is the required distance from the stated point.

Answer
xˉ=5 cm\bar{x}=5\ \text{cm}

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