There is a door that is 1.0 m wide and can rotate about a vertical axis without friction. The door’s mass is 15 kg and is free to swing open. Someone fires a 10 g bullet at 400 m/s into the exact center of the door, perpendicular to it.
Find the angular velocity of the door just after the bullet sticks in the door.
Show whether kinetic energy is conserved.
To find the angular velocity of the door just after the bullet sticks in it, we can use the principle of conservation of angular momentum.
The equation for angular momentum is L = I * ω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
In this case, the bullet sticks in the door, so its moment of inertia doesn't change. The bullet's moment of inertia can be approximated as the mass of the bullet times the perpendicular distance between the axis of rotation and the line of motion.
First, let's calculate the moment of inertia of the bullet. The mass of the bullet is 10 g, which is equivalent to 0.01 kg. The perpendicular distance between the axis of rotation (center of the door) and the line of motion (the bullet's path) is half the width of the door, so it is 1.0 m / 2 = 0.5 m.
The moment of inertia of the bullet is therefore I_bullet = mass_bullet * distance^2 = 0.01 kg * (0.5 m)^2 = 0.0025 kg·m^2.
Since the bullet is fired perpendicular to the door, its initial angular momentum is zero, as it has no initial angular velocity.
When the bullet sticks in the door, its angular momentum becomes I_bullet * ω_door, where ω_door is the angular velocity of the door after the bullet sticks.
Conservation of angular momentum tells us that the total angular momentum before and after the collision is the same. So, we can write:
I_bullet * ω_door = 0.
Solving for ω_door:
ω_door = 0 / I_bullet = 0.
Therefore, the angular velocity of the door just after the bullet sticks in it is zero. The door comes to a complete stop.
Now let's determine whether kinetic energy is conserved.
The initial kinetic energy is the energy associated with the bullet's linear motion before it collides with the door. The kinetic energy can be calculated using the formula:
KE_initial = 1/2 * mass_bullet * velocity_bullet^2.
Plugging in the values:
KE_initial = 1/2 * 0.01 kg * (400 m/s)^2 = 800 J.
The final kinetic energy is the energy associated with the door's rotational motion after the bullet sticks in it. The kinetic energy can be calculated using the formula:
KE_final = 1/2 * I_door * ω_door^2,
where I_door is the moment of inertia of the door and ω_door is the angular velocity of the door just after the bullet sticks in it.
Since the angular velocity of the door is zero as we calculated earlier, ω_door = 0. Therefore, the final kinetic energy is KE_final = 0.
Comparing the initial and final kinetic energies, we see that KE_initial ≠ KE_final. Therefore, kinetic energy is not conserved in this scenario.