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प्रश्न
Find the centre of mass of a uniform (a) half-disc, (b) quarter-disc.
उत्तर
Let M and R be the mass and radius of the half-disc, mass per unit area of the half-disc
(a) The half-disc can be supposed to be consists of a large number of semicircular rings of mass om and thickness and radii ranging from r = 0 to r = R.
The surface area of the semicircular ring of radius r and of thickness dr = `1/2 2πr xx dr = πrdr`
∴ Mass of this elementary ring, dm = `πrdr xx (2M)/(rR^2)`
dm = `(2M)/R^2 rdr`
If (x, y) are coordinates of the centre of mass of this element,
Then, `(x, y) = (0, (2r)/π)`
Therefore, x = 0 and y = `(2r)/π`
Let xCM and yCM be the coordinates of the centre of mass of the semicircular disc.
Then xCM = `1/M int_0^R xdm = 1/M int_0^R 0 dm = 0`
yCM = `1/M int_0^R ydm = 1/M int_0^R (2r)/π xx ((2M)/R^2 rdr)`
= `4/(πR^2) int_0^R r^2dr`
= `4/(πR^2) [r^3/3]_0^R`
= `4/(πR^2) xx (R^3/3 - 0)`
= `(4R)/(3π)`
∴ Centre of mass of the semicircular disc = `(0, (4R)/(3π))`
(b) Centre of mass of a uniform quarter disc.
Mass per unit area of the quarter disc = `M/((πR^2)/4) = (4M)/(πR^2)`
Using symmetry
For a half-disc along the y-axis centre of mass will be at `x = (4R)/(3π)`
For a half-disc along the x-axis centre of mass will be at `x = (4R)/(3π)`
Hence, for the quarter disc centre of mass = `((4R)/(3π), (4R)/(3π))`
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