Centres of mass: laminas and frameworks Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Centres of mass: laminas and frameworks questions. See exactly how to solve problems on centre-of-mass, system-of-particles, standard-lamina, triangle.

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.

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
Three particles of mass 1 kg1\text{ kg}, 2 kg2\text{ kg} and 3 kg3\text{ kg} are placed at the points (0, 0)(0,\ 0), (6, 0)(6,\ 0) and (0, 4)(0,\ 4) respectively, referred to a fixed origin OO. Find the coordinates of the centre of mass of the system.

Worked solution

  1. Write down the total mass

    M=1+2+3=6M=1+2+3=6

    The denominator of each coordinate is the total mass of the system.

  2. Take moments about the yy-axis

    xˉ=1×0+2×6+3×06=2\bar{x}=\frac{1\times 0+2\times 6+3\times 0}{6}=2

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

  3. Take moments about the xx-axis

    yˉ=1×0+2×0+3×46=2\bar{y}=\frac{1\times 0+2\times 0+3\times 4}{6}=2

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

  4. State the centre of mass

    G=(2, 2)G=\left(2,\ 2\right)

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

Answer
(2, 2)\left(2,\ 2\right)
Question 3
2 markseasy
Three particles of mass 4 kg4\text{ kg}, 1 kg1\text{ kg} and 3 kg3\text{ kg} are placed at the points (2, 1)(2,\ 1), (5, 3)(5,\ 3) and (−1, 4)(-1,\ 4) respectively, referred to a fixed origin OO. Find the xx-coordinate of the centre of mass of the system.

Worked solution

  1. Write down the total mass

    M=4+1+3=8M=4+1+3=8

    The denominator of each coordinate is the total mass of the system.

  2. Take moments about the yy-axis

    xˉ=4×2+1×5+3×−18=54\bar{x}=\frac{4\times 2+1\times 5+3\times -1}{8}=\frac{5}{4}

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

  3. State the xx-coordinate

    xˉ=54\bar{x}=\frac{5}{4}

    This is the xx-coordinate of the centre of mass.

Answer
54\frac{5}{4}
Question 4
2 markseasy
Three particles of mass 2 kg2\text{ kg}, 2 kg2\text{ kg} and 4 kg4\text{ kg} are placed at the points (3, 5)(3,\ 5), (7, 1)(7,\ 1) and (1, −2)(1,\ -2) respectively, referred to a fixed origin OO. Find the yy-coordinate of the centre of mass of the system.

Worked solution

  1. Write down the total mass

    M=2+2+4=8M=2+2+4=8

    The denominator of each coordinate is the total mass of the system.

  2. Take moments about the xx-axis

    yˉ=2×5+2×1+4×−28=12\bar{y}=\frac{2\times 5+2\times 1+4\times -2}{8}=\frac{1}{2}

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

  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. State the yy-coordinate

    yˉ=12\bar{y}=\frac{1}{2}

    This is the yy-coordinate of the centre of mass.

Answer
12\frac{1}{2}
Question 5
2 markseasy
A uniform triangular lamina has vertices at (0, 0)(0,\ 0), (6, 0)(6,\ 0) and (3, 9)(3,\ 9). Find the coordinates of the centre of mass of the lamina.

Worked solution

  1. Average the xx-coordinates

    xˉ=0+6+33=3\bar{x}=\frac{0+6+3}{3}=3

    Add the three xx-coordinates and divide by three.

  2. Average the yy-coordinates

    yˉ=0+0+93=3\bar{y}=\frac{0+0+9}{3}=3

    Add the three yy-coordinates and divide by three.

  3. State the centre of mass

    G=(3, 3)G=\left(3,\ 3\right)

    This is the centroid of the triangular lamina.

Answer
(3, 3)\left(3,\ 3\right)

Unlock 65 more Centres of mass: laminas and frameworks questions

Create a free account to work through every Further Maths Centres of mass: laminas and frameworks question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Centres of mass: laminas and frameworks practice

Hard QuestionsPractice Questions

Related Mechanics topics

Elastic collisions in one dimensionElastic collisions in two dimensionsCentres of mass: solids and calculusElastic strings and springsFurther dynamicsHorizontal circular motion