A square shaped hole of side l=\frac{a}{2} is carved out at a distance d=\frac{a}{2} from the centre 'O' of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from O is -\frac{a}{X},value of X (to the nearest integer)is .......
Option: 1 2
Option: 2 5
Option: 3 23
Option: 4 56

Answers (1)
  1. Position of centre of mass when some mass is removed

                        {\vec{r}_{cm}}=\frac{m_{1}\vec{r_{1}}-{m_{2}}\vec{r}_{2}}{m_{1}-m_{2}}

 

Where m_1is value of whole mass and \vec{r_{1}}is position of centre of mass for whole mass.Similarly m_2 & \vec{r_{2}} are values for mass  which has been removed.

From question

Let the density is \rho

So

 \mathrm{m}_{1}=\pi \mathrm{a}^{2} \sigma

And

m_2=\frac{a^{2}}{4} \cdot \sigma

And r_1=0 \ \ and \ \ r_2=\frac{a}{2}

So

X_{cm}= \frac{(\sigma \pi a^{2} \times 0)- (\sigma \frac{a^{2}}{4} \times \frac{a}{2})}{\sigma \pi a^{2}-\frac{\sigma a^{2}}{4}}

X_{cm}=\frac{-a^{3} / 8}{\left(\pi-\frac{1}{4}\right) a^{2}}=\frac{-a}{2(4 \pi-1)}=\frac{-a}{8 \pi-2}=-\frac{a}{23}

So X=23

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