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From the circular disc of radius 4R two ...

From the circular disc of radius `4R` two small discs of radius R are cut off. The centre of mass of the new structure will be at

A

`hat(i)(R)/(5)+hat(j)(R)/(5)`

B

`-hat(i)(R)/(5)+hat(j)(R)/(5)`

C

`-hat(i)(R)/(5)-hat(j)(R)/(5)`

D

`-(3R)/(14)(hat(i)+hat(j))`

Text Solution

Verified by Experts

The correct Answer is:
D
Centre of mass of circular disc of radius
`4R=(0,0)`
Centre of mass of upper disc `=(0,3R)`
Centre of mass lower disc `=(3R,0)`
Let `M` be mass of complete disc and then the mass of cut out disc are `(M)/(16)`
Hence, centre of mass of new structure is given by
`bar (x)=(m_(1)x_(1)-m_(2)x_(2)-m_(3)x_(2))/(m_(1)-m_(2)-m_(3))`
`=(M(0)-(M)/(16)(3r)-(M)/(16)(0))/(M-(M)/(16)-(M)/(16))=(-3R)/(14)`
Position vector of `C.M. =-(3R)/(14)(hat(i)+hat(j))`
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