Spring Projectile problem: center of mass

In summary, the conversation revolves around the problem of determining the position of two masses connected by a massless spring, given an initial velocity. The equations of motion are set up and solved, but it is noted that they are only valid under certain assumptions. The question of including a non-zero normal force is also brought up.
  • #1
Zenshin
5
0
Hello everyone. I'm burning my head at this one, maybe someone could offer some insight? Here's the deal: imagine two masses, m1 and m2, united by a massless spring, of length l (relaxed).

m1
|
| _ > spring, at length l (relaxed)
|
___m2____ ground

The system has this config at rest. Then, it'f given a velocity v , upward, to the mass m1, and the question is: what is the position of the masses in any given time? I've figured out that I've to break the problem in two parts: find out the center of mass motion and then determine the position of the masses relative to it. However, how can I do that? I've already found out the center of mass motion, but I can't find the position of the masses relative to it. I assume it's a oscillatory motion, but I can't find it's amplitude. The only thing I have is the initial velocity of the mass m1, nothing else. Please, any advice would be HIGHLY useful hehehe

Thanks in advance!
 
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  • #2
All right:
Let us set up the individual laws of motion here:
[tex]-m_{1}g-k(x_{1}(t)-x_{2}(t)-l)=m_{1}a_{1}[/tex]
[tex]-m_{2}g+k(x_{1}(t)-x_{2}(t)-l)=m_{2}a_{2}[/tex]
where [tex]x_{1},x_{2}[/tex] are the respective positions of the masses, with associated accelerations [tex]a_{1},a_{2}[/tex]
This is a second-order linear system with constant coefficients which is readily solvable.
 
  • #3
since the motion is vertical, we use y instead of x (y positive upward). We divide this soution into two parts. 0.5mv1^2 + mgy1 = 0.5mv2^2 + mgy2
0 + mgh = 0.5mv2^2 + 0
v2 = squ root2gh)
 
  • #4
klmdad said:
since the motion is vertical, we use y instead of x (y positive upward). We divide this soution into two parts. 0.5mv1^2 + mgy1 = 0.5mv2^2 + mgy2
0 + mgh = 0.5mv2^2 + 0
v2 = squ root2gh)
Whatever are you talking about?
Where, for example, have you included the potential energy of the system contained in the spring?

Zenshin:
While I believe that the equations given are the ones you were asked to find, nevertheless it is important to realize that these equations are only valid under the assumption NO NORMAL FORCE ACTING ON THE SYSTEM FROM THE GROUND. If you want to include the possibility of a non-zero normal force as well, you've got a much trickier situation.
 

1. What is the center of mass in a spring projectile problem?

The center of mass in a spring projectile problem refers to the point at which the entire mass of the object is concentrated. It is the average position of all the individual particles that make up the object.

2. How is the center of mass calculated in a spring projectile problem?

The center of mass can be calculated using the formula:
x = (m1x1 + m2x2 + m3x3 + ...)/ (m1 + m2 + m3 + ...)
where m is the mass of each particle and x is the position of each particle along the horizontal axis.

3. Why is the center of mass important in a spring projectile problem?

The center of mass is important because it helps us understand the overall motion of the object. It acts as a point through which the object can be treated as a single point mass, making calculations and predictions easier.

4. How does the center of mass affect the trajectory of a spring projectile?

The center of mass does not directly affect the trajectory of a spring projectile. However, it does play a role in determining the overall motion and stability of the object. If the center of mass is not within the base of support, the object will experience a torque and may topple over.

5. Can the center of mass of a spring projectile change during its trajectory?

Yes, the center of mass can change during the trajectory of a spring projectile. As the object moves, the position of each particle making up the object may change, thus altering the overall position of the center of mass. However, if the object is undergoing uniform motion, the center of mass will remain in the same position.

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