A 550.0 g bird is flying horizontally at 2.50 m/s, not paying much attention, wh

Question

A 550.0 g bird is flying horizontally at 2.50 m/s, not paying much attention, when it suddenly flies into a stationary vertical bar, hitting it 25.0 cm below the top (the figure ). The bar is uniform, 0.800 m long, has a mass of 1.30 kg, and is hinged at its base. The collision stuns the bird so that it just drops to the ground afterward (but soon recovers to fly happily away).

What is the angular velocity of the bar just after it is hit by the bird, and what is the angular velocity of the bar just as it reaches the ground?

Explanation / Answer

Mass of the rod M = 1.30 Kg

Mass of the bird m = 550 g = 0.55 Kg

Speed of the bird before collision v = 2.50 m/s

Length of the bar L = 0.8 m

Moment of inertia I of the rod = 1/3*M*L^2

                                                 = 1/3*1.30*0.8^2

                                                 = 0.27733 kg-m^2


Initial angular momentum before (due to the bird) I* = angular momentum after due to the bar

So m*r^2*v/r = 1/3*M*L^2*


Therefore = m*r*v/(1/3*M*L^2)

                    = 0.55*(0.8 - 0.250)*2.5/(1/3*1.3*0.8^2)

                    = 2.726 rad/s

B) Now use conservation of energy

So (U + K)1 = K2                   (Potential energy U is based on the height of the center of mass)

M*g*L/2 + 1/2*I*1^2 = 1/2*I*2^2

1.30*9.8*0.8/2 + 1/2*0.27733*2.726^2 = 1/2*0.27733*2^2

or 0.1386*2^2 = 6.1264

So = sqrt(6.1264/0.1386)

         = 6.647 rad/s or 6.65 rad/s

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