Question

From a uniform disc of radius R, a circular hole of radius R / 2 is cut out. The centre of the hole is R / 2 from the centre of the original disc. The centre of mass of the resulting body is at from the centre of original disc opposite to the centre of the cut portion.

• A. R / 6
• B. R / 4
• C. R / 3
• D. R / 2

Answer 1

Answer: (A)

Mass of remaining body
$=\frac{\left[\pi R^{2}-\pi\left(\frac{R}{2}\right)^{2}\right]}{\pi R^{2}} \times M=\frac{3}{4} M$
$x_{\text {Disc }}=\frac{\left(\frac{M}{4}\right)\left(\frac{R}{2}\right)+\left(\frac{3 M}{4}\right) \cdot x_{\text {Remaining Body }}}{M}$
As $x_{\text {Disc }}=0$
$ \Rightarrow \left(\frac{3 M}{4}\right)\left(x_{\text {Remaining Body }}\right)=-\frac{M R}{8}$
$\Rightarrow x_{\text {Remaining Body }}=-\left(\frac{R}{6}\right)$