A circular plate of uniform thickness as a diameter of 56 cm .And a circular portion of diameter 42 cm is removed from edge of the plate .Then find the shift in centre of mass?
Answers
Answer:
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
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