The end B of the rod AB which makes angle `theta` with the floor is being pulled with a constant velocity `v_(v)` as shown. The length of the rod is `l`.

Options

A. (a)At `theta=37^(@)` velocity of end `A` is `(4)/(3)v_(0)` downwards
B. (b)At `theta=37^(@)` angular velocity of rod is `(5v_(0))/(3l)`
C. (c)Angular velocity of rod is constant
D. (d)velocity of end A is constant.

Explanation

Velocity component along `AB=0`
`thereforev_(0)costheta=vsintheta`
or `v=v_(0)cottheta=f_(1)(theta)`
at `theta=37^(@),v=(4)/(3)v_(0)`
`omega=("relative velocity "bot "to " AB)/(l)`
`=(v_(0)sintheta+vcostheta)/(l)=f_(2)(theta)`
`=((v_(0))(3//5)+(4//3v_(0))(4//5))/(l)=(5v_(0))/(3l)`