Question

a uniform disc of radius r is put over another uniform disc of radius 2r of same thickness and density.the peripheries of 2 discs touch each other.the position of there centre of mass is

Answer 1

Let the center of bigger disc be O. Let the center of smaller disc be P.
OP = r.
mass of the smaller disc = m
Mass of the bigger disc = 4 m  , as area is 4 times.

Center of mass from O:   (4m * 0 + m * r)/(4m+m) = r/5

Answer 2

The position of their center of mass is at R/5 from the center of the bigger disc towards the center of the smaller disc.

Explanation:

The radius of smaller disc = r

The radius of bigger disc = 2r

$m_{1}$ = π$r^{2}$ * T * р

$m_{2}$ = π$(2R)^{2}$ I T * р

The thickness of the two discs = T

The density of the two discs = р

∵ The position of the center of the mass

{$\frac{m_{1} X_{1} + m_{2} X_{2}}{m_{1} + m_{2} },$ $\frac{m_{1} y_{1} + m_{2} y_{2} }{m_{1} + m_{2} }$}

As we know,

x1 = r  y1 = 0

x2 = 0 y2 = 0

= (4m * 0 + m * r)/(4m+m) = r/5

Learn more: center of mass

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