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The figure shows a uniform square plate from which four identical squares at the corners will be removed. (a) Where is the center of mass of the plate originally? Where is it after the removal of (b) square 1; (c) squares 1 and 2; (d) squares 1 and 3; (e) squares 1, 2, and 3; (f) all four squares?
Explain briefly
Asked by rushabh1234 | 21 Oct, 2019, 09:49: PM
Expert Answer
centre of mass of full square plate is at geometrical centre O,
Let us have x-y coordinate system with origin O at centre of mass of full square plate.
(1) Square-1 is removed
By taking x-moment of masses with respect to O, we have
- a2 ρ [ (l/2)-(a/2) ] + (l2 - a2) ρ x = 0 or x = (1/2) a2 /(l+a)
first term in above expression is x-moment of square-1 and second term is x-moment of [ full square - square-1 ]
ρ is mass per unit area and x is x-coordinate of centre of mass
Similarly by taking y-moment, we have
a2 ρ [ (l/2)-(a/2) ] + (l2 - a2) ρ y = 0 or y = -(1/2) a2 /(l+a)
Hence corrdinates of centre of mass from O :- [ (1/2) a2 /(l+a), -(1/2) a2 /(l+a) ]
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Square-1 and square-2 are removed
By taking x-moment of masses with respect to O, we have
- a2 ρ [ (l/2)-(a/2) ] + a2 ρ [ (l/2)-(a/2) ] + (l2 - 2a2) ρ x = 0 or x = 0
first term in above expression is x-moment of square-1. second term is x-moment of square-2 and
third term is x-moment of [ full square - square-1 - square-2]
Similarly by taking y-moment, we have
a2 ρ [ (l/2)-(a/2) ] + a2 ρ [ (l/2)-(a/2) ]+ (l2 - 2a2) ρ y = 0 or y = - a2(l-a) /(l2-2a2)
Hence corrdinates of centre of mass from O :- [ 0 , - a2(l-a) /(l2-2a2) ]
-------------------------------------------------------------------------------
Square-1 and square-3 are removed
By taking x-moment of masses with respect to O, we have
- a2 ρ [ (l/2)-(a/2) ] + a2 ρ [ (l/2)-(a/2) ] + (l2 - 2a2) ρ x = 0 or x = 0
first term in above expression is x-moment of square-1. second term is x-moment of square-3 and
third term is x-moment of [ full square - square-1 - square-3]
Similarly by taking y-moment, we have
a2 ρ [ (l/2)-(a/2) ] - a2 ρ [ (l/2)-(a/2) ]+ (l2 - 2a2) ρ y = 0 or y = 0
Hence corrdinates of centre of mass from O :- [ 0 , 0 ]
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User is advised to do the remaining part of this problem as explained above
Answered by Thiyagarajan K | 22 Oct, 2019, 12:27: PM
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