Q. 23 Find the center of mass of a uniform (a) half-disc, (b) quarter-disc.
Let M and R be the mass and radius of the half-disc, mass per unit area of the half-disc
m=M12πR2=2MπR2m=M12πR2=2MπR2
Step 2: Find the position of center of mass of small element.
(a) The half-disc can be supposed to be consists of a large number of semicircular rings of mass dm and thickness d, and radii ranging from r = 0 to r = R.
The surface area of semicircular ring of radius r and of thickness dr=122πr×dr=πrdrdr=122πr×dr=πrdr
∴∴ Mass of this elementary ring, dm=πrdr×2MπR2dm=πrdr×2MπR2
dm=2MR2rdrdm=2MR2rdr
If (x, y) are coordinates of the center of mass of this element, then,
(x,y)=(0,2rπ)∴ x=0 and y=2rπ
Let xCM and yCM be the coordinates of the center of mass of the semicircular disc.
Then
Step 3: Find the position of center of mass for half-disc.
xCM=1M∫R0xdm=1M∫R00dm=0yCM=1M∫R0ydm=1M∫R02rπ×(2MR2rdr) =4πR2∫R0r2dr=4πR2[r33]R0 =4πR2×(R33−0)=4R3π
∴ Centre of mass of the semicircular disc =(0,4R3π)
(b)
Step 4: Find centre of mass of a uniform quarter disc.
using symmetry
For a half-disc along the y-axis center of mass be at x=4R3π
For a half-disc along the x-axis center of mass will be at x=4R3π
Hence, for the quarter disc center of mass =(4R3π,4R3π)
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