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?
iniitial angular momentum about hinge = m v r = .010 *400 * 1 = 4 kg m^2/s
or
omega = v/r = 400/1
I = m r^2 = .010 (1)^2)
so I omega = .01 * 400 = 4 agains
now final angular momentum is the same
I = .010 *1^2 + (1/3)(10)1^2
= .01 + 10/3
= 3.34 kg m^2
so
I omega = 4 = 3.34 omega
omage = 1.2 radians/sec
To determine the angular velocity of the door immediately after impact, we can use the principle of conservation of angular momentum. The angular momentum of an object is given by the formula:
L = Iω
Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
In this case, the dart sticks into the door and causes it to rotate. Therefore, the initial angular momentum of the system is only due to the dart, as the door was initially at rest. The moment of inertia of the door can be calculated using the formula:
I = 1/3 * m * r^2
Where m is the mass of the door and r is the distance from the hinge to the point where the dart sticks.
Given that the mass of the door is 10 kg, the width of the door is 1 m, and the dart sticks at the edge opposite the hinge, the distance from the hinge to the point where the dart sticks is also 1 m.
Now let's calculate the initial angular momentum of the system:
L_initial = I_dart * ω_initial
The mass of the dart is 10 g, which is 0.01 kg. The initial angular momentum can be calculated as follows:
L_initial = (0.01 kg) * (400 m/s) * (1 m)
L_initial = 4 kg·m^2/s
Now, assuming there are no external torques acting on the system, the conservation of angular momentum states that the initial angular momentum is equal to the final angular momentum.
L_initial = L_final
Since the dart is stuck in the door, the moment of inertia of the system increases. Therefore, we need to include the moment of inertia of the dart and the door in the calculation of the final angular momentum.
The moment of inertia of a point mass (dart) is given by the formula:
I_dart = m_dart * r^2
Where m_dart is the mass of the dart and r is the distance from the hinge to the point where the dart sticks.
In this case, the moment of inertia of the door can be calculated using the formula:
I_door = 1/3 * m_door * r^2
Where m_door is the mass of the door and r is the distance from the hinge to the center of mass of the door.
Now let's calculate the final moment of inertia of the system:
I_final = I_dart + I_door
Substituting the values:
I_final = (0.01 kg) * (1 m)^2 + (1/3) * (10 kg) * (1 m)^2
I_final = 0.01 kg·m^2 + 3.33 kg·m^2
I_final = 3.34 kg·m^2
Now we can rearrange the equation for angular momentum to solve for the final angular velocity:
L_final = I_final * ω_final
ω_final = L_final / I_final
Substituting the values:
ω_final = (4 kg·m^2/s) / (3.34 kg·m^2)
ω_final ≈ 1.20 rad/s
Therefore, the angular velocity of the door immediately after impact is approximately 1.20 rad/s.