Question

A bird (mass m = 500 gram) is flying horizontally at a speed of v0 = 4.00 m/s, when it suddenly hits a stationary vertical bar at 50 cm below the top end of the bar. The bar is uniform with a mass M = 1.5 kg and a length l = 1.00 m. The bar is hinged at the bottom end and is free to rotate about this pivot point. The collision stuns the bird, which then drops vertically to the ground.
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?

Answer 1

Answer:

a) W = 2rad/s

b) $W_f =5.78rad/s$

Explanation:

part a. (the angular velocity of the bar just after it hit by the bird)

using the conservationof the angular momentum:

$L_i = L_f$

so:

$MVd= IW$

where M is the mass of the bird, V the velocity of the bird, d the lever arm, I the moment of inerta of the bar and W the angular velocity of the bar just after the collition.

Now we have to find the moment of inertia of the bar as:

I = $\frac{1}{3}M_bL^2$

Where M is the mass of the bar and L the length of the bar. So, I is equal to:

I = $\frac{1}{3}(1.5kg)(1m)^2$

I = 0.5 kg*m^2

Then, replacing values on the first equation, we get:

$(0.5kg)(4m/s)(0.5m)= (0.5kg*m^2)W$

Solving for W:

W = 2rad/s

part b. (the angular velocity of the bar just as it reaches the ground)

we will use the conservation of energy:

$E_i = E_f$

so:

$M_bgh+\frac{1}{2}IW^2 = \frac{1}{2}IW_f^2$

Where $M_b$ is the mass of the bar, g is the gravity, h the altitude of the center of mass and $W_f$ the angular velocity of the bar just as it reaches the ground.

Then, replacing values, we get:

$(1.5kg)(9.8)(0.5m)+\frac{1}{2}(0.5)(2)^2 = \frac{1}{2}(0.5)W_f^2$

Finally, solving for $W_f$:

$W_f =5.78rad/s$