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

The acceleration of end `B` of the rod is

Options

A. `-(6g)/7hati-(12g)/7hatj`
B. `-(5g)/7hati-(6g)/7hatj`
C. `-(18g)/7hati-(5g)/7hatj`
D. `-(18g)/7hati-(9g)/7hatj`

Answer

D

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`