Question

A uniform rod of mass m and length `l` can rotate in vertical plane about a smooth horizontal axis hinged at point `H`. Find angular acceleration `alpha` of the rod just after it is released from initial position making an angle of `37^@` with horizontal from rest?

Answer 1

Correct Answer - `(mgsqrt(10))/5`
Torque about hing `=tauH=Ialpha`
`Mg cos37^@ l/2=(ml^(2))/3 alpha`
`alpha 6g//5l`

`a_(t)=alpha l/2=(3g)/5`
`Mgcos37^@-N-1=ma_(1)`
`N_(1)=(mg)/5`
Angular velocity of rod is zero so `N-2=mgsin37^@=3mg//5`
`N=sqrt(N-1^(2)+N_(2)^(2))=sqrt(((mg)/5)^(2)+((3mg)/5)^(2))=(mgsqrt(10))/5`