Question

roblem: A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter 42cm is removed from one edge of the plate. Find the position of the center of mass of the remaining portion.

Answer 1

See diagram.

Thickness of plate = t        Total mass of circular plate = M
Density of plate = M / (π r₁² t)           ,  r₁ = 28 cm
    r₂ = radius of cut out part = 21 cm

Mass of  small circular portion cut out = volume * density = π r₂² t * (M /π r₁² t)
              m₂ = M r₂² / r₁² 

Mass of the remaining part = m3 = M - M  r₂² / r₁² = M (r₁² - r₂²) / r₁²

Let the center of mass of remaining portion be at a distance d from the center of original full plate. From symmetry the center of mass of remaining portion lies along the line joining the center of original plate and the center of removed part.

 Let  Center of mass of total full plate = 0 
      0   =  1/M ( d * mass of remaining part + C₁C₂ * mass of removed part)
      0 = d * M (r₁² - r₂²) / r₁² +  7 * M r₂² / r₁²
       0 = d * (r₁² - r₂²) + 7 * r₂² 
           d = - 7 * r₂² / (r₁² - r₂²) = - 7 * 21² / (28² - 21²) = - 7 * 21 * 21 / 49*7 = - 9 cm

Center gravity of the remaining portion is 9 cm to the left of the original center of full plate on the line of symmetry.  Or, it is 19 cm from the edge of remaining plate along the line of symmetry.

Answer 2

See the following attachment

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