Lower end B of a thin rod AB of length 1 m is made to move with a constant velocity of 1 m/s as shown in figure. Initial angle of inclination the rod with horizontal is ${90}^0.$  Then magnitude of angular acceleration of the rod at t = 0.5 second is

detailed solution

Correct option is C

At time t, Let θ be the angle of inclination of the rod with horizontal and at that instant of time ‘e’ is the instantaneous centre of velocity of the rod. Therefore instantaneous angular velocity of the rod is given by ω=VBBC =  Vl sin θ⇒   α  =  dωdt =  −  Vl  cosec θ  cot θNow, cos θ  =  vtl  and  at  t=0.5  sec,  cos θ  =  1 × 0.51 =  12⇒   θ=600.∴      At   t=0.5  s,   α = − 1V  cosec 600.  cot 600  rad/sec2⇒   α = −23  .  13   rad/sec2  = −23  rad/sec2.