IIT JEE , AIEEE Forum - Physics , Mechanics

srujana

The end B of the rod AB makes an angle @ with the floor. The end B of the rod is being pulled with constant velocity V₀ as shown. The length of the rod is l , prove that at the instant when @= 37^o angular velocity of the rod is 5/3 V₀/l

iitkgp_bipin

Let the base OB = x

From right-angled triangle : x = l.cos@

Now differentiate both sides with respect to time :

dx/dt = l.sin@.d@/dt

Velocity of point B = V₀ which is same as rate of change of the length OB.
So, dx/dt = V₀

Angular velocity = rate of change of angle, w = d@/dt

V₀ = l.sin@.w

At @ = 37 degrees, sin@ = 3/5

V₀ = l.(3/5).w

w = 5V₀/3l

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