from a uniform circular disc of radius r, acircular disc of radius r/6 and having centre at a distance r/2 from the centre of the disc is removed . determine the centre of mass of remaining portion of the disc.
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See the figure ,
OO1 = R/2
Let mass/area of disc = m
So, mass of disc(M) = area of disc × mass per unit area
= πR²m
Mass of the portion removed form the disc ( m') = area of removing part × mass per unit area
= π(R/2)²m = mr²/4 = M/4
If the center of mass of remaining part from the centre O is at a distance D ,
D ={ M×0 - m'×(R/2)}/(M-m')
= {-M/4×R/2}/(M-M/4)
= -R/6
Hence, centre of mass of the remaining part at R/6 to the left of centre O.
OO1 = R/2
Let mass/area of disc = m
So, mass of disc(M) = area of disc × mass per unit area
= πR²m
Mass of the portion removed form the disc ( m') = area of removing part × mass per unit area
= π(R/2)²m = mr²/4 = M/4
If the center of mass of remaining part from the centre O is at a distance D ,
D ={ M×0 - m'×(R/2)}/(M-m')
= {-M/4×R/2}/(M-M/4)
= -R/6
Hence, centre of mass of the remaining part at R/6 to the left of centre O.
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The first step is to set the equations for gravitational potential energy and work equal to each other and solve for force. W = PE is F × d = m × g × h, so F = (m × g × h) ÷ d. The second and final step is to plug the values from the problem into the equation for force.
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