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  • Find the centre of mass of a uniform a) half-disc b) quarter-disc

Find the centre of mass of a uniform

a) half-disc

b) quarter-disc

Answers (1)

Let the mass of half-disc be M

Then

Area=\frac{\pi R^{2}}{2}

mass per unit area m 

=\frac{2M}{\pi R^{2}}

Let us assume that the half-disc is divided into many semi-circular strips with radius from 0 to R

Surface area of a semicircular strip =\frac{\pi }{2}[(r+dr)^{2}-r^{2}]

Area  =\frac{\pi }{2}[r^{2}+dr^{2}+2rdr^{2}-r^{2}]

Mass of strip (dm) =

\left ( \frac{2M}{\pi R^{2}} \right )\pi r\; dr=\left ( \frac{2M}{R^{2}} \right )r\; dr

Let (x,y) be the co-ordinates of c.m. of this strip (x,y)

=\left ( 0,\frac{2r}{\pi } \right )

x=x_{cm}=\frac{1}{M}\int_{0}^{R}x\; dm=\int_{0}^{R}\; 0\; dm=0

y_{cm}=\frac{1}{M}\int_{0}^{R}y\; dm=\frac{1}{M}\int_{0}^{R}\; \frac{2r}{\pi }\; \frac{2M}{R^{2}}\; r\; dr

=\frac{1}{M}\; \frac{4M}{\pi R^{2}}\; \int_{0}^{R}\; r^{2}dr=\frac{4}{\pi R^{2}}\; \frac{[R^{3}-0^{3}]}{3}=\frac{4R}{3\pi }

So the centre of mass of the circular half-disc

=\left ( 0,\frac{4R}{3\pi } \right )

Mass per unit area of quarter disc centre of mass of a uniform quarter disc=\frac{M}{\frac{\pi R^{2}}{4}}=\frac{4M}{\pi R^{2}}

For a half disc along y-axis, m will be at y =\frac{4R}{3\pi} (Using symmetry)

So, centre of mass of quarter disc = \left ( \frac{4R}{3\pi},\frac{4R}{3\pi} \right )



 

Posted by

infoexpert23

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