A uniform rod `AB` hinged about a fixed point `P` is initially vertical. The rod is released from vertical position. When rod is in horizontal position:

The acceleration of the centre of mass of the rod is

Options

A. `-(8mg)/7hati-(12mg)/7hatj`
B. `-(3mg)/7hati-(9mg)/7hatj`
C. `-(6mg)/7hati-(4mg)/7hatj`
D. `-(12mg)/7hati-(6mg)/7hatj`

Answer

C

Explanation

From conservation of energy we get
`(mgl)/4=1/2m[(l^(2))/12+(l^(2))/16]omega^(2),omega^(2)=(24g)/(7l)`

From Newton's law
`(mgl)/4=((ml^(2))/12+(ml^(2))/16)alpha`……….i
`mg-R_(y)=m(1/4alpha)` ………….ii
`R_(x)=(ml)/4omega^(2)`..........iii
`alpha=(12g)/(7l),veca_(CM)=-(omega^(2)l)/4hati-1/4alphahatj`
`=(6g)/7hati-(3g)/7hatj`
`veca_(B)=-omega^(2)((3l)/4)hati-((3l)/4)alphahatj`
`vecR_(x)=-(6mg)/7hati,vecR_(h)=+(4mg)/7hatj`