A circular plate of uniform thickness has diameter 56 cm circular motion of diameter 42 cm is removed from one direction the centre of mass of remaining plate
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Heya
here's your 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
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.
hope it helps...✌✌
here's your 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
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.
hope it helps...✌✌
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