A 10 g dart traveling at 400 m/s sticks into a 10 kg, 1 m wide door at the edge opposite the hinge causing the door to swing open. What is the angular velocity of the door immediately after impact?
To find the angular velocity of the door immediately after impact, we can use the principle of conservation of angular momentum. Angular momentum is defined as the product of the moment of inertia and angular velocity.
The moment of inertia of the door can be calculated using the formula:
I = (1/3) * M * L^2
Where:
I is the moment of inertia
M is the mass of the door
L is the length of the door
In this case, the length of the door is given as 1 m, and the mass of the door is given as 10 kg. Substituting these values into the formula:
I = (1/3) * 10 kg * (1 m)^2
I = (1/3) * 10 kg * 1 m^2
I = 10/3 kg*m^2
The initial angular momentum of the dart before impact is given as:
L_initial = m * v
L_initial = 10 g * 400 m/s
Converting the mass of the dart to kilograms:
L_initial = 10 g * (1 kg / 1000 g) * 400 m/s
L_initial = 0.01 kg * 400 m/s
L_initial = 4 kg*m/s
According to the principle of conservation of angular momentum, the initial angular momentum of the dart is equal to the final angular momentum of the dart and the door. So:
L_initial = L_final
The final angular momentum of the dart and the door can be calculated as:
L_final = (I + m * r^2) * ω
Where:
L_final is the final angular momentum
I is the moment of inertia of the door
m is the mass of the dart
r is the distance of the dart from the hinge (assumed to be half the width of the door)
ω is the angular velocity of the door
Rearranging the formula:
ω = (L_final - m * r^2) / (I)
Substituting the known values:
ω = (4 kg*m/s - (10 g * r^2)) / (10/3 kg*m^2)
Now, substitute r = 0.5 m (since the dart hits the door at the edge opposite the hinge):
ω = (4 kg*m/s - (10 g * (0.5 m)^2)) / (10/3 kg*m^2)
ω = (4 kg*m/s - (10 g * 0.25 m^2)) / (10/3 kg*m^2)
Perform the calculations:
ω = (4 kg*m/s - 0.0025 kg*m/s) / (10/3 kg*m^2)
ω = (3.9975 kg*m/s) / (10/3 kg*m^2)
ω = 3.9975 * 3/10 m/s
ω ≈ 1.199 m/s
Therefore, the angular velocity of the door immediately after impact is approximately 1.199 m/s.