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the position vector of the centre of ...

the position vector of the centre of mass lt
`to`
r cm of an asymmetric uniform bar of negligible area of cross - section as shown in figure is :

A

`vecr_(cm)=(13)/(8) Lhatx+(5)/(8)Lhaty`

B

`vecr_(cm)=(5)/(8) Lhatx+(13)/(8)Lhaty`

C

`vecr_(cm)=(3)/(8) Lhatx+(11)/(8)Lhaty`

D

`vecr_(cm)=(11)/(8) Lhatx+(3)/(8)Lhaty`

Text Solution

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