End `A` of a rod `AB` is being pulled on the floor with a constant velocity `v_(0)` as shown. Taking the length of the rod as `l`, at an instant when the rod makes an angle `37^@` with the horizontal, calculate

the velocity of the `CM` of the rod

Options

A. `5/7v_(0)` at `tan^(-1)4/3` below horizontal
B. `5/7v_(0)` at `tan^(-1)3/4` below horizontal
C. `5/6v_(0)` at `tan^(-1)3/4` below horizontal
D. `5/6v_(0)` at `tan^(-1) 4/3` below horizontal

Answer

D

Explanation

`x_(CM)=x/2`
`impliesV_(CM)=1/2(dx)/(dt)0=(v_(0))/2(rarr)`
`implies y_(CM)=y/2`
`y_(CM)=1/2 (dy)/(dt)=(v_(B))/2=2/3v_(0)darr`
`V_(CM)=sqrt((v_(0)^(2))/4+(4v_(0)^(2))/9)=5/6v_(0)` at `tan^(-1)(4/3)` below horizontal