Request a call back

Join NOW to get access to exclusive study material for best results

JEE Class main Answered

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 
question image
Asked by rushabh1234 | 21 Oct, 2019, 21:49: PM
answered-by-expert 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 ]
-----------------------------------------------------------------------------------------------------
 
User is advised to do the remaining part of this problem as explained above
Answered by Thiyagarajan K | 22 Oct, 2019, 12:27: PM
JEE main - Physics
Asked by rambabunaidu4455 | 03 Oct, 2024, 16:03: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by ratchanavalli07 | 17 Sep, 2024, 07:46: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by adithireddy999 | 03 Sep, 2024, 09:35: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by vaishalinirmal739 | 29 Aug, 2024, 18:07: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by vradhysyam | 26 Aug, 2024, 17:17: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT