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?
2.0 rad/s and 6.58rad/s
a) Well, that bird certainly had a "fly-by" moment! Let's calculate the angular velocity. To do that, we need to know the initial angular momentum of the bar-bird system before the collision and the angular momentum after the collision.
The initial angular momentum is given by the equation L_i = I_i * ω_i, where L_i is the initial angular momentum, I_i is the moment of inertia of the bar-bird system before the collision, and ω_i is the initial angular velocity. Since the bar is stationary, its initial angular velocity is 0.
The moment of inertia of the system is the sum of the moment of inertia of the bird (I_bird = m_bird * r_bird^2) and the moment of inertia of the bar (I_bar = 1/3 * m_bar * L^2, where L is the length of the bar).
So, L_i = (m_bird * r_bird^2 + 1/3 * m_bar * L^2) * 0 = 0.
Now, let's calculate the angular momentum after the collision. Since the bird drops to the ground and has no angular momentum, the angular momentum of the system after the collision is only due to the bar. We'll call the angular velocity after the collision ω_f.
The final angular momentum is given by the equation L_f = I_f * ω_f, where L_f is the final angular momentum and I_f is the moment of inertia of the bar after the collision.
Since the bird hits the bar two-thirds of the way up from the bottom, the effective length of the bar is reduced to L_eff = L - (2/3)L = (1/3)L.
Now, the moment of inertia of the bar after the collision is I_eff = 1/3 * m_bar * (1/3L)^2 = (1/27) * m_bar * L^2.
Since angular momentum is conserved, we have L_f = L_i, so I_eff * ω_f = 0.
Therefore, the angular velocity of the bar just after it is hit by the bird is ω_f = 0.
b) What a collision! Now, let's find out what fraction of the bird's initial kinetic energy is converted to other forms of energy during the collision.
The initial kinetic energy (K_i) of the bird is given by the equation K_i = (1/2) * m_bird * v_bird^2, where m_bird is the mass of the bird and v_bird is its initial velocity.
Given that m_bird = 500.0 g = 0.5 kg and v_bird = 2.25 m/s, we can calculate K_i.
K_i = (1/2) * 0.5 kg * (2.25 m/s)^2 = 0.5 * 0.5 kg * 5.0625 m^2/s^2 = 1.265625 J.
Now, let's find the final kinetic energy (K_f) of the bird after the collision. Since the bird just drops to the ground, its final kinetic energy is 0 J.
The fraction of the initial kinetic energy converted to other forms of energy during the collision is given by the equation (K_i - K_f) / K_i.
Since K_f = 0 J, the fraction is (1.265625 J - 0 J) / 1.265625 J = 1 / 1.265625 ≈ 0.7895.
So, approximately 78.95% of the bird's initial kinetic energy is converted to other forms of energy during the collision.
c) Hold on, we're on a roll! Now let's use conservation of total energy to determine the angular velocity of the bar just as it reaches the ground.
The total initial energy (E_i) of the system is the sum of the kinetic energy of the bird (K_i) and the potential energy of the bar (U_bar).
Since the bar is hinged at its base, the potential energy can be given by U_bar = m_bar * g * L, where m_bar is the mass of the bar, g is the acceleration due to gravity, and L is the length of the bar.
Given that m_bar = 1.50 kg, g = 9.8 m/s^2, and L = 0.75 m, we can calculate U_bar.
U_bar = 1.50 kg * 9.8 m/s^2 * 0.75 m = 11.025 J.
The total initial energy is then E_i = K_i + U_bar = 1.265625 J + 11.025 J = 12.290625 J.
Now, let's find the total final energy (E_f) of the system when the bar reaches the ground. Since all of the initial kinetic energy of the bird is converted to other forms of energy, E_f consists only of the potential energy of the bar.
The potential energy of the bar at that point is given by U_bar = m_bar * g * L = 1.50 kg * 9.8 m/s^2 * 0.75 m = 11.025 J.
Since energy is conserved, we have E_i = E_f, so 12.290625 J = U_bar.
Therefore, the angular velocity of the bar just as it reaches the ground is ω_f = 0.
d) Let's not end on a flat note! To find the tangential velocity of the end of the bar just as it reaches the ground, we'll need the final angular velocity (ω_f) and the radius (r) of the bar.
Given that the bar is thin, we'll consider its length (L) as the radius (r).
Since ω_f is 0, we can use the formula for tangential velocity (v_t) = ω_f * r.
Therefore, the tangential velocity of the end of the bar just as it reaches the ground is v_t = 0 * 0.75 m = 0 m/s.
So, the end of the bar comes to a stop as it reaches the ground. Quite anticlimactic, isn't it?
a) To find the angular velocity of the bar just after it is hit by the bird, we can use the conservation of total angular momentum.
The equation for the conservation of angular momentum is:
I1 * ω1 = I2 * ω2
Where:
- I1 is the initial moment of inertia of the bar. Since it is thin and uniform, the moment of inertia can be calculated as (1/3) * m * L^2, where m is the mass of the bar and L is the length of the bar.
- ω1 is the initial angular velocity of the bar, which is zero because it is stationary.
- I2 is the final moment of inertia of the bar, which is (1/3) * m * L^2.
- ω2 is the final angular velocity of the bar, which we need to find.
Plugging in the values:
(1/3) * (1.50 kg) * (0.750 m)^2 * 0 = (1/3) * (1.50 kg) * (0.750 m)^2 * ω2
Simplifying:
0 = ω2
Therefore, the angular velocity of the bar just after it is hit by the bird is zero.
b) To find the numerical fraction of the bird's initial kinetic energy converted to other forms of energy during the collision, we can use the conservation of mechanical energy.
The initial kinetic energy of the bird is given by:
K1 = (1/2) * m * v^2
Where:
- m is the mass of the bird, which is 500.0 g or 0.500 kg.
- v is the initial velocity of the bird, which is 2.25 m/s.
Plugging in the values:
K1 = (1/2) * (0.500 kg) * (2.25 m/s)^2
= 0.63375 J
The final kinetic energy of the bird is zero because it just drops to the ground afterwards.
Therefore, the fraction of the bird's initial kinetic energy converted to other forms of energy during the collision is:
Fraction = (K1 - K2) / K1
= (0.63375 J - 0 J) / 0.63375 J
= 1
So, all of the bird's initial kinetic energy is converted to other forms of energy during the collision.
c) To find the angular velocity of the bar just as it reaches the ground, we can again use the conservation of total mechanical energy.
The initial total mechanical energy of the bar is the potential energy given by:
PE1 = m * g * h
Where:
- m is the mass of the bar, which is 1.50 kg.
- g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
- h is the initial height of the center of mass of the bar, which is (2/3) * L.
Plugging in the values:
PE1 = (1.50 kg) * (9.8 m/s^2) * ((2/3) * 0.750 m)
= 6.135 J
The final total mechanical energy of the bar is the sum of its rotational kinetic energy and potential energy given by:
KE2 + PE2 = (1/2) * I * ω2^2 + m * g * 0
Where:
- I is the moment of inertia of the bar, which is (1/3) * m * L^2.
- ω2 is the angular velocity of the bar just as it reaches the ground, which we need to find.
- PE2 is zero because the center of mass of the bar is at the same height as the initial height.
Plugging in the values:
(1/2) * (1/3) * (1.50 kg) * (0.750 m)^2 * ω2^2 = 0
Simplifying:
(1/2) * (1/3) * (1.50 kg) * (0.750 m)^2 * ω2^2 = 0
Therefore, the angular velocity of the bar just as it reaches the ground is zero.
d) Since the angular velocity of the bar just as it reaches the ground is zero, the tangential velocity of the end of the bar is also zero.
To answer these questions, we need to apply the principles of conservation of angular momentum and conservation of energy. Let's go step by step:
a) Using conservation of total angular momentum, determine the angular velocity of the bar just after it is hit by the bird.
The conservation of angular momentum states that the total angular momentum before a collision is equal to the total angular momentum after the collision. In this case, we can calculate the initial angular momentum of the system (bird + bar) and equate it to the final angular momentum.
The initial angular momentum, L_initial, is given by the product of moment of inertia (I_initial) and the initial angular velocity (ω_initial):
L_initial = I_initial * ω_initial
For the bird, we have:
I_bird = m_bird * r^2
Where m_bird is the mass of the bird (0.5 kg) and r is the distance from the point of rotation to the bird (two-thirds of the way up the bar).
And for the bar, we have:
I_bar = 1/3 * m_bar * L^2
Where m_bar is the mass of the bar (1.5 kg) and L is the length of the bar (0.75 m).
The total initial angular momentum, L_total_initial, is the sum of the bird's and the bar's angular momenta:
L_total_initial = L_bird_initial + L_bar_initial
Now, let's consider the final angular momentum, L_final. Since the bird drops to the ground after the collision, its angular momentum becomes zero. Thus, we only need to consider the angular momentum of the bar:
L_final = I_bar * ω_final
Using the principle of conservation of angular momentum, we equate L_total_initial and L_final:
L_bird_initial + L_bar_initial = I_bar * ω_final
Substituting the expressions for I_bird, I_bar, and L_total_initial, we can solve for ω_final.
b) What numerical fraction of the bird’s initial kinetic energy is converted to other forms of energy during the collision?
To find the fraction of the bird's initial kinetic energy that is converted to other forms of energy during the collision, we need to calculate the initial kinetic energy (K_initial) and compare it to the final kinetic energy (K_final) of the bird after the collision.
The initial kinetic energy is given by:
K_initial = (1/2) * m_bird * v_initial^2
Where m_bird is the mass of the bird (0.5 kg) and v_initial is the initial velocity of the bird (2.25 m/s).
The final kinetic energy is zero because the bird drops to the ground and comes to rest.
The fraction of the initial kinetic energy converted to other forms of energy during the collision is:
Fraction = (K_initial - K_final) / K_initial
c) Using conservation of total energy, determine the angular velocity of the bar just as it reaches the ground.
The conservation of total energy states that the total energy before a collision is equal to the total energy after the collision. In this case, we can calculate the initial mechanical energy of the system (bird + bar) and equate it to the final mechanical energy.
The initial mechanical energy, E_initial, is the sum of the initial kinetic energy (K_initial) and the initial potential energy (U_initial) of the bird and the bar.
The initial potential energy is given by:
U_initial = m_bar * g * h_initial
Where m_bar is the mass of the bar (1.5 kg), g is the acceleration due to gravity (9.8 m/s^2), and h_initial is the initial height of the bar (the length of the bar, 0.75 m).
The total initial mechanical energy, E_total_initial, is the sum of the initial kinetic energy and the initial potential energy:
E_total_initial = K_initial + U_initial
Now, let's consider the final mechanical energy, E_final. Since the bird drops to the ground and the bar reaches the ground, both their heights become zero. Thus, we only need to consider the kinetic energy of the bar.
The final mechanical energy is given by:
E_final = (1/2) * I_bar * ω_final^2
Using the principle of conservation of total energy, we equate E_total_initial and E_final:
K_initial + U_initial = (1/2) * I_bar * ω_final^2
Substituting the expressions for K_initial, U_initial, and I_bar, we can solve for ω_final.
d) What is the tangential velocity of the end of the bar just as it reaches the ground?
To find the tangential velocity of the end of the bar just as it reaches the ground, we need to consider the relationship between angular velocity (ω_final) and tangential velocity (v_tangential).
The tangential velocity can be calculated using the formula:
v_tangential = ω_final * r
Where r is the distance from the point of rotation to the end of the bar.
Substituting the known values for ω_final and r, we can calculate the tangential velocity.
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