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Radius of Gyration

Find the posi...

Question

Find the position of center of mass of a disc of radius $R$ from which a hole of radius r is cut out. The center of the hole is at a distance $R / 2$ from the center of the disc.

\frac{R r^{2}}{2 (R^{2} - r^{2})}

towards right of

O

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\frac{R r^{2}}{2 (R^{2} - r^{2})}

towards left of

O

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\frac{2 R r^{2}}{(R^{2} + r^{2})}

towards right of

O

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\frac{2 R r^{2}}{(R^{2} + r^{2})}

towards left of

O

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Solution

The correct option is B $\frac{R r^{2}}{2 (R^{2} - r^{2})}$ towards left of $O$
Center of mass of bigger disc $= O (0, 0)$ Let $O$ be the origin.
Center of mass of smaller disc $= C (R / 2, 0)$
Let $x c m = x_{1}$
$x c m = R / 2 = x_{2}$
Now, $m_{1} + m_{2} = m$
$x c m = 0 = x$
By using the formula $\frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} = x$
We get center of mass $x_{1} = \frac{R r^{2}}{2 (R^{2} - r^{2})}$ towards left of $O$ .

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Find the position of center of mass of a disc of radius R from which a hole of radius r is cut out. The center of the hole is at a distance R/2 from the center of the disc.

The correct option is B Rr22(R2−r2)towards left of OCenter of mass of bigger disc =O(0,0) Let O be the origin.Center of mass of smaller disc=C(R/2,0)Let x cm=x1x cm=R/2=x2Now, m1+m2=mx cm=0=xBy using the formula m1x1+m2x2m1+m2=xWe get center of mass x1=Rr22(R2−r2) towards left of O.

Find the position of center of mass of a disc of radius $R$ from which a hole of radius r is cut out. The center of the hole is at a distance $R / 2$ from the center of the disc.