I of bar around base of bar = (1/3) m L^2
= (1/3)(1.5)(.75)^2 = .281 kg m^2
initial angular momentum of bird around base of bar = m (2L/3)^2 [V/(2L/3)]
= m (2L/3)V = .5 (.5)(2.25) = .5625
If the bird just stopped, then that is the angular momentum of the bar
.5625 = I omega = .281 omega
so
(a) omega = 2 radians/second
(b) do (1/2) m v^2 of bird before and (1/2) I omega^2 of bar after
(c) change of PE of falling bar = m g h
= m g (L/2)
(1/2) I omega^2 at ground = (1/2)I omega^2 at collision + m g (L/2)
(d) omega from (c) * L
A 500.0 g bird is flying horizontally at 2.25 m/s, not paying much attention, when it suddenly flies into a stationary vertical bar, hitting it
two-thirds of the way up from the bottom. The bar is uniform and very
thin, is 0.750 m long, has a mass of 1.50 kg, and is hinged at its base.
The collision stuns the bird so that it just drops to the ground afterwards.
a) Using conservation of total angular momentum, determine the angular velocity of the bar just
after it is hit by the bird.
b) What numerical fraction of the bird’s initial kinetic energy is converted to other forms of energy
during the collision?
c) Using conservation of total energy, determine the angular velocity of the bar just as it reaches
the ground.
d) What is the tangential velocity of the end of the bar just as it reaches the ground?
2 answers
2.0 rad/s and 6.58rad/s