A uniform 5.50-kg square solid wooden gate 2.00 m on each side hangs vertically from a frictionless pivot at its upper edge. A 1.00-kg raven flying horizontally at 5.00 m/s flies into this door at its center and bounces back at 2.00 m/s in the opposite direction. What is the angular speed of the gate just after it is struck by the unfortunate raven?

ω = 0.9545 rad/s

Explanation:

In this question, angular momentum is conserved.

Thus;

Initial angular momentum = final angular momentum.

Thus, L_1 = L_2

So,

m•v1•l = -m•v2•l + I_gate•ω

So;

m•v1•l + m•v2•l = I_gate•ω

l = l/2

So,we now have;

(ml/2)•(v1 + v2) = I_gate•ω

Now, I_gate is expressed as Ml²/3

Where M is mass of gate = 5.5 kg.

So,

(ml/2)•(v1 + v2) = (Ml²/3)•ω

ω = [3ml(v1 + v2)]/2Ml²

This will reduce further to;

ω = [3m(v1 + v2)]/2Ml

Where m is mass of raven = 1 kg

v1 = 5 m/s

v2 = 2 m/s

l = 2m

So;

ω = [3(1)(5 + 2)]/(2×5.5×2)

ω = 3×7/22

ω = 0.9545 rad/s