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A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide The centre of mass of the new disc is $\alpha R$ from the centre of the bigger disc . The value of $\alpha$ is

Option 1)
$1/4$
Option 2)
$1/3$

Option 3)
$1/2$
Option 4)
$1/6$

Answers (1)

best_answer

As we learnt in

Position of centre of mass when some mass is removed -

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

- wherein

m₁ is value of whole mass and $\vec{r_{1}}$ is position of centre of mass for whole mass. m₂& $\vec{r_{2}}$ are values for mass which has been removed.

$(M+m)=M=\pi (2R)^{2}.\sigma$

$where\; \sigma =mass\; per\; unit \; area$

$m=\sigma R^{2}.\sigma\; ,M=3\pi R^{2}\sigma$

$\frac{3\pi R^{2}\sigma .x+\pi R^{2}\sigma .R}{M}=0$

Because for the full disc, the centre of mass is at the centre O

$\Rightarrow \; \; \; x=-\frac{R}{3}=\alpha R.\; \; \; \therefore \; \; \; \left | \alpha \right |=\left | \frac{-1}{3} \right |.$

The centre of mass is at R/3 to the left on the diameter of the original disc.

The question should be at a distance $\alpha R$ and not $\alpha/ R$ .

Option 1)

$1/4$

Option 2)

$1/3$

Option 3)

$1/2$

Option 4)

$1/6$

Posted by

Aadil

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