Mass center of system (airplane and helicopter crash)

In summary, a small airplane and a helicopter collide above a tower in a wooded area. The resulting wreckage is split into two pieces, H1 and H2, and the coordinates of these pieces are given. Using equations of motion, the final position vector of the mass center is calculated to be (-10933 m)i + (4000 m)j + (1200 m)k. The motion of the pieces after the collision is in free fall and undergoes acceleration due to gravity. However, it is unclear how to use this information to calculate the final position vector of the airplane wreckage.
  • #1
JJBladester
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Homework Statement



A small airplane of mass 1500 kg and a helicopter of mass 3000 kg flying at an altitude of 1200 m are observed to collide directly above a tower located at O in a wooded area. Four minutes earlier the helicopter had been sighted 8.4km due west of the tower and the airplane 16 km west and 12 km north of the tower. As a result of the collision the helicopter was split into two pieces, H1 and H2, of mass m1 = 1000 kg and m2 = 2000 kg, respectively; the airplane remained in one piece as it fell to the ground. Knowing that the two fragments of the helicopter were located at points H1 (500 m, -100 m) and H2 (600 m, -500 m), respectively, and assuming that all pieces hit the ground at the same time, determine the coordinates of the point A where the wreckage of the airplane will be found.

Answer:
(1004 m)i - (48.7 m)j

airplane%20helicopter.jpg


Homework Equations



[tex]m\mathbf{\overline{r}}=\sum_{i=1}^{n}m_i\mathbf{r_i}[/tex]

[itex]\mathbf{\overline{r}}[/itex] = position vector of mass center of system

"Problems of this type can be solved by writing the equation of motion of the mass center of the system in vectorial form and expressing its position vector before and after the collision in terms of the position vectors of the relevant bodies. You can then rewrite the vector equation as two or three scalar equations and solve these equations for an equivalent number of unknowns."

The Attempt at a Solution



[tex]m_A=1500kg[/tex]

[tex]m_H=3000kg[/tex]

[tex]t=4\, min=240\, s[/tex]

[tex]\mathbf{r_A}=-16000\mathbf{i}+12000\mathbf{j}+1200\mathbf{k}[/tex]

[tex]\mathbf{r_H}=-8400\mathbf{i}+0\mathbf{j}+1200\mathbf{k}[/tex]

[tex]\mathbf{\overline{r}}=\frac{m_A\mathbf{r_A}+m_H\mathbf{r_H}}{m_A+m_H}[/tex]

[tex]\mathbf{\overline{r}}=\frac{(1500kg)(-16000\mathbf{i}+12000\mathbf{j}+1200\mathbf{k})+(3000kg)(-8400\mathbf{i}+0\mathbf{j}+1200\mathbf{k})}{1500kg+3000kg}[/tex]

[tex]\mathbf{\overline{r}}=-10933\mathbf{i}+4000\mathbf{j}+1200\mathbf{k}[/tex]

Airplane motion before collision (not in freefall):
[tex]\mathbf{v_0}=\frac{16000m}{240s}\mathbf{i}+\frac{12000m}{240s}\mathbf{j}+0\mathbf{k}=66.7\mathbf{i}+50\mathbf{j}+0\mathbf{k}[/tex]

Helicopter motion before collision (not in freefall):
[tex]\mathbf{v_0}=\frac{8400m}{240s}\mathbf{i}+0\mathbf{j}+0\mathbf{k}=35\mathbf{i}+0\mathbf{j}+0\mathbf{k}[/tex]

Airplane motion after collision (in freefall):
Not sure... But I know the velocities in the x- and y-directions do not undergo acceleration. The velocity in the k-direction undergoes acceleration equal to gravity (g).

Helicopter motion after collision (in freefall):
Not sure... But I know the velocities in the x- and y-directions do not undergo acceleration. The velocity in the k-direction undergoes acceleration equal to gravity (g).

Final center of mass position vector:
[tex]\mathbf{\overline{r}}=\frac{m_A\mathbf{r_{A}}+m_1\mathbf{r_{H_1}}+m_2\mathbf{r_{H_2}}}{m_A+m_1+m_2}[/tex]

I know the goal is to find the final i and j coordinates of the airplane (we know the k coordinate is zero (all pieces fall on the ground [height = 0]). I know the theory behind this problem is that even though the collision occurs and we end up with pieces everywhere, the pieces should follow the trajectory of the mass center.

I know that at the moment of the collision, the plane and helicopter pieces undergo projectile motion in free fall and are accelerated downward at the rate of 9.81m/s2.

However, I am finding it hard to put all of the pieces together.
 
Last edited:
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  • #2
I am not sure what the velocities of the airplane and helicopter pieces are after the collision, in order to calculate the final position vector. I know that the velocity of the mass center is just the sum of the individual velocities. But I am not sure how to use this information to get the final position vector.Any help would be greatly appreciated. Thank you!
 

Related to Mass center of system (airplane and helicopter crash)

1. What is the mass center of a system?

The mass center of a system refers to the point where the entire mass of the system can be considered to be concentrated. It is also known as the center of mass or center of gravity.

2. How is the mass center of a system determined?

The mass center of a system can be determined by taking into account the mass of each individual component in the system and their respective positions. The mass center can be calculated using the formula: xcm = (m1x1 + m2x2 + ... + mnxn)/mtotal, where x is the position and m is the mass of each component.

3. How does the mass center affect the stability of a system?

The mass center plays a crucial role in determining the stability of a system. If the mass center is located above the base of support, the system will be stable. However, if the mass center is located outside the base of support, the system will be unstable and may topple over.

4. What happens to the mass center in an airplane and helicopter crash?

In an airplane and helicopter crash, the mass center of the system changes as the two objects collide. The mass center will shift towards the direction of the heavier object, and depending on the angle and force of impact, it may also change in position.

5. How does the mass center affect the trajectory of a system?

The mass center of a system also affects its trajectory. If the mass center is not aligned with the direction of motion, it will cause the system to rotate. This is why airplanes and helicopters have their mass centers carefully balanced to ensure stable flight.

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