Teach me the steps to solve questions using examples under centre of mass ànd centre of gravity in maths mechanics

Physics

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Answer

(2.4, 3.2)

(73,116)(\frac{7}{3}, \frac{11}{6})

(32−π16−π,32−π16−π)(\frac{32 - \pi}{16 - \pi}, \frac{32 - \pi}{16 - \pi})

Step-by-step solution

18 steps

The center of mass (CM) is the unique point where the weighted relative position of the distributed mass sums to zero. It's the average position of all the mass in a system. The center of gravity (CG) is the point where the entire weight of an object appears to act. For objects in a uniform gravitational field, the center of mass and center of gravity are at the same location.

Here are the steps and examples for finding the center of mass:

Formulas for Center of Mass

For a system of discrete particles with masses m1,m2,…,mnm_1, m_2, \dots, m_n located at positions (x1,y1),(x2,y2),…,(xn,yn)(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n):

xˉ=m1x1+m2x2+⋯+mnxnm1+m2+⋯+mn=∑i=1nmixi∑i=1nmi\bar{x} = \frac{m_1 x_1 + m_2 x_2 + \dots + m_n x_n}{m_1 + m_2 + \dots + m_n} = \frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i} yˉ=m1y1+m2y2+⋯+mnynm1+m2+⋯+mn=∑i=1nmiyi∑i=1nmi\bar{y} = \frac{m_1 y_1 + m_2 y_2 + \dots + m_n y_n}{m_1 + m_2 + \dots + m_n} = \frac{\sum_{i=1}^{n} m_i y_i}{\sum_{i=1}^{n} m_i}

For a composite lamina (a flat, thin object of uniform density), the mass is proportional to its area. So, we can replace mass (mm) with area (AA) in the formulas, where AiA_i is the area of each part and (xi,yi)(x_i, y_i) is the centroid of that part:

xˉ=∑i=1nAixi∑i=1nAi\bar{x} = \frac{\sum_{i=1}^{n} A_i x_i}{\sum_{i=1}^{n} A_i} yˉ=∑i=1nAiyi∑i=1nAi\bar{y} = \frac{\sum_{i=1}^{n} A_i y_i}{\sum_{i=1}^{n} A_i}

Example 1: Discrete Particles

Find the center of mass of three particles with masses 2 kg2\,\text{kg}, 3 kg3\,\text{kg}, and 5 kg5\,\text{kg} located at coordinates (1,2)(1, 2), (4,1)(4, 1), and (2,5)(2, 5) respectively.

  1. Identify the masses and their coordinates.

    • m1=2 kgm_1 = 2\,\text{kg} at (x1,y1)=(1,2)(x_1, y_1) = (1, 2)
    • m2=3 kgm_2 = 3\,\text{kg} at (x2,y2)=(4,1)(x_2, y_2) = (4, 1)
    • m3=5 kgm_3 = 5\,\text{kg} at (x3,y3)=(2,5)(x_3, y_3) = (2, 5)

  2. Calculate the sum of the masses.

    ∑mi=m1+m2+m3=2 kg+3 kg+5 kg=10 kg\sum m_i = m_1 + m_2 + m_3 = 2\,\text{kg} + 3\,\text{kg} + 5\,\text{kg} = 10\,\text{kg}

  3. Calculate the sum of the products of mass and x-coordinate (∑mixi\sum m_i x_i).

    ∑mixi=(2 kg×1)+(3 kg×4)+(5 kg×2)\sum m_i x_i = (2\,\text{kg} \times 1) + (3\,\text{kg} \times 4) + (5\,\text{kg} \times 2) ∑mixi=2+12+10=24 kg⋅units\sum m_i x_i = 2 + 12 + 10 = 24\,\text{kg}\cdot\text{units}

  4. Calculate the x-coordinate of the center of mass (xˉ\bar{x}).

    xˉ=∑mixi∑mi=24 kg⋅units10 kg=2.4 units\bar{x} = \frac{\sum m_i x_i}{\sum m_i} = \frac{24\,\text{kg}\cdot\text{units}}{10\,\text{kg}} = 2.4\,\text{units}

  5. Calculate the sum of the products of mass and y-coordinate (∑miyi\sum m_i y_i).

    ∑miyi=(2 kg×2)+(3 kg×1)+(5 kg×5)\sum m_i y_i = (2\,\text{kg} \times 2) + (3\,\text{kg} \times 1) + (5\,\text{kg} \times 5) ∑miyi=4+3+25=32 kg⋅units\sum m_i y_i = 4 + 3 + 25 = 32\,\text{kg}\cdot\text{units}

  6. Calculate the y-coordinate of the center of mass (yˉ\bar{y}).

    yˉ=∑miyi∑mi=32 kg⋅units10 kg=3.2 units\bar{y} = \frac{\sum m_i y_i}{\sum m_i} = \frac{32\,\text{kg}\cdot\text{units}}{10\,\text{kg}} = 3.2\,\text{units}

    The center of mass is at (2.4,3.2)\boxed{(2.4, 3.2)}.

    Example 2: Composite Lamina (L-shaped)

    Find the center of mass of a uniform L-shaped lamina formed by two rectangles:

    • Rectangle 1: Base from x=0x=0 to x=6x=6, height from y=0y=0 to y=2y=2.
    • Rectangle 2: Base from x=0x=0 to x=2x=2, height from y=2y=2 to y=5y=5.

  7. Divide the composite shape into simpler geometric shapes and find their individual areas and centroids.

    • Rectangle 1 (R1):
      • Dimensions: width 6, height 2.
      • Area A1=6×2=12A_1 = 6 \times 2 = 12.
      • Centroid (x1,y1)(x_1, y_1): Midpoint of width is 6/2=36/2 = 3. Midpoint of height is 2/2=12/2 = 1. So, (x1,y1)=(3,1)(x_1, y_1) = (3, 1).
    • Rectangle 2 (R2):
      • Dimensions: width 2, height 3 (from y=2y=2 to y=5y=5).
      • Area A2=2×3=6A_2 = 2 \times 3 = 6.
      • Centroid (x2,y2)(x_2, y_2): Midpoint of width is 2/2=12/2 = 1. Midpoint of height is 2+(3/2)=2+1.5=3.52 + (3/2) = 2 + 1.5 = 3.5. So, (x2,y2)=(1,3.5)(x_2, y_2) = (1, 3.5).

  8. Calculate the total area of the lamina.

    ∑Ai=A1+A2=12+6=18\sum A_i = A_1 + A_2 = 12 + 6 = 18

  9. Calculate the sum of the products of area and x-coordinate (∑Aixi\sum A_i x_i).

    ∑Aixi=(A1×x1)+(A2×x2)\sum A_i x_i = (A_1 \times x_1) + (A_2 \times x_2) ∑Aixi=(12×3)+(6×1)=36+6=42\sum A_i x_i = (12 \times 3) + (6 \times 1) = 36 + 6 = 42

  10. Calculate the x-coordinate of the center of mass (xˉ\bar{x}).

    xˉ=∑Aixi∑Ai=4218=73≈2.33\bar{x} = \frac{\sum A_i x_i}{\sum A_i} = \frac{42}{18} = \frac{7}{3} \approx 2.33

  11. Calculate the sum of the products of area and y-coordinate (∑Aiyi\sum A_i y_i).

    ∑Aiyi=(A1×y1)+(A2×y2)\sum A_i y_i = (A_1 \times y_1) + (A_2 \times y_2) ∑Aiyi=(12×1)+(6×3.5)=12+21=33\sum A_i y_i = (12 \times 1) + (6 \times 3.5) = 12 + 21 = 33

  12. Calculate the y-coordinate of the center of mass (yˉ\bar{y}).

    yˉ=∑Aiyi∑Ai=3318=116≈1.83\bar{y} = \frac{\sum A_i y_i}{\sum A_i} = \frac{33}{18} = \frac{11}{6} \approx 1.83

    The center of mass is at (73,116)\boxed{(\frac{7}{3}, \frac{11}{6})}.

    Example 3: Lamina with a Hole

    Find the center of mass of a square lamina with side length 4 m4\,\text{m} (from x=0x=0 to x=4x=4, y=0y=0 to y=4y=4) with a circular hole of radius 1 m1\,\text{m} centered at (1,1)(1, 1).

  13. Treat the shape as a large square with a "negative mass" (or negative area) circular hole.

    • Square (S):
      • Dimensions: side length 4 m4\,\text{m}.
      • Area AS=4×4=16 m2A_S = 4 \times 4 = 16\,\text{m}^2.
      • Centroid (xS,yS)(x_S, y_S): Midpoint of width is 4/2=24/2 = 2. Midpoint of height is 4/2=24/2 = 2. So, (xS,yS)=(2,2)(x_S, y_S) = (2, 2).
    • Circular Hole (H):
      • Radius r=1 mr = 1\,\text{m}.
      • Area AH=πr2=π(1)2=π m2A_H = \pi r^2 = \pi (1)^2 = \pi\,\text{m}^2.
      • Centroid (xH,yH)(x_H, y_H): Given as (1,1)(1, 1).
      • For calculation, we use this area as negative: −AH=−π-A_H = -\pi.

  14. Calculate the total effective area.

    ∑Ai=AS−AH=16−π\sum A_i = A_S - A_H = 16 - \pi

  15. Calculate the sum of the products of area and x-coordinate (∑Aixi\sum A_i x_i).

    ∑Aixi=(AS×xS)+(−AH×xH)\sum A_i x_i = (A_S \times x_S) + (-A_H \times x_H) ∑Aixi=(16×2)+(−π×1)=32−π\sum A_i x_i = (16 \times 2) + (-\pi \times 1) = 32 - \pi

  16. Calculate the x-coordinate of the center of mass (xˉ\bar{x}).

    xˉ=∑Aixi∑Ai=32−π16−π\bar{x} = \frac{\sum A_i x_i}{\sum A_i} = \frac{32 - \pi}{16 - \pi} xˉ≈32−3.1415916−3.14159=28.8584112.85841≈2.244\bar{x} \approx \frac{32 - 3.14159}{16 - 3.14159} = \frac{28.85841}{12.85841} \approx 2.244

  17. Calculate the sum of the products of area and y-coordinate (∑Aiyi\sum A_i y_i).

    ∑Aiyi=(AS×yS)+(−AH×yH)\sum A_i y_i = (A_S \times y_S) + (-A_H \times y_H) ∑Aiyi=(16×2)+(−π×1)=32−π\sum A_i y_i = (16 \times 2) + (-\pi \times 1) = 32 - \pi

  18. Calculate the y-coordinate of the center of mass (yˉ\bar{y}).

    yˉ=∑Aiyi∑Ai=32−π16−π\bar{y} = \frac{\sum A_i y_i}{\sum A_i} = \frac{32 - \pi}{16 - \pi} yˉ≈2.244\bar{y} \approx 2.244

    The center of mass is at (32−π16−π,32−π16−π)\boxed{(\frac{32 - \pi}{16 - \pi}, \frac{32 - \pi}{16 - \pi})}.

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