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)

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. m\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

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