A disc of radius R is cut out from a larger disc of radius 2R in such a way that the edge of the hole touches the edge of the disc. Locate the center of mass of the residual disc.
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ANSWER::
See figure for better understanding.
Lets assume the centre of the bigger disc to be the origin.
The smaller disc is cut out from the bigger disc.
Radius of bigger disc = 2R
Radius of smaller disc = R
m₁=πR² x T x ρ x₁=R y₁=0
m₂=π(2R)² x T x ρ x₂=0 y₂=0
where as ,T = Thickness of both discs
ρ = Density of both discs
Position of centre of mass =
[(m₁x₁+m₂x₂)/(m₁+m₂) , (m₁y₁+m₂y₂)/(m₁+m₂)]
= [(-πR²TρR + 0) / (-πR²Tρ + π(2R)²TρR ) , 0 +0 / (m₁+m₂)]
= [(-πR²TρR) / (3πR²Tρ) , 0 ]
= (-R/3 , 0)
Centre of mass is at R/3 from the centre of bigger disc which is away from the centre of the hole.
Hope it helps!
ANSWER::
See figure for better understanding.
Lets assume the centre of the bigger disc to be the origin.
The smaller disc is cut out from the bigger disc.
Radius of bigger disc = 2R
Radius of smaller disc = R
m₁=πR² x T x ρ x₁=R y₁=0
m₂=π(2R)² x T x ρ x₂=0 y₂=0
where as ,T = Thickness of both discs
ρ = Density of both discs
Position of centre of mass =
[(m₁x₁+m₂x₂)/(m₁+m₂) , (m₁y₁+m₂y₂)/(m₁+m₂)]
= [(-πR²TρR + 0) / (-πR²Tρ + π(2R)²TρR ) , 0 +0 / (m₁+m₂)]
= [(-πR²TρR) / (3πR²Tρ) , 0 ]
= (-R/3 , 0)
Centre of mass is at R/3 from the centre of bigger disc which is away from the centre of the hole.
Hope it helps!
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