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Byju's Answer
Standard XII
Physics
Center of Mass as an Average Point
Five homogene...
Question
Five homogeneous bricks, each of length
L
, are arranged as shown in figure. Each brick is displaced with respect to the one in contact by
L
/
5
. Find the x-coordinate of the centre of mass relative to the origin
O
shown.
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Solution
→
X
c
m
=
m
(
L
2
+
L
2
+
L
5
+
L
2
+
2
L
5
+
L
2
+
L
5
+
L
2
)
5
m
where,
L
2
- Ist block
L
2
+
L
5
- 2nd block
L
2
+
2
L
5
- 3rd block
L
2
+
L
5
- 4th block
L
2
- 5th block
COM
=
33
L
50
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Figure