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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 

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Asked by rushabh1234 21st October 2019, 9:49 PM
Answered by 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) ]
---------------------------------------------------------------------
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 Expert 22nd October 2019, 12:27 PM
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