Conservation of Linear Momentum of Rigid Body

In summary, the conversation discusses two different methods for solving a problem involving the conservation of linear momentum and the calculation of angular speed. The first method uses the velocities of each particle before and after impact, while the second method uses the conservation of linear momentum of the center of gravity of the rigid body. However, the second method may not be entirely accurate due to the external, impulsive force of the support during the collision. The correct answer may involve assuming that particle C has a negligible moment of inertia and friction with the stop bracket.
  • #1
Uchida
23
6
Homework Statement
Consider a determined structure of three particles, each with mass m, interconnected by rigid bars of negligible mass forming an equilateral triangle on the side a.
The structure falls from a height h, as shown in the figure and the particle C hooks on the stop.

The angular velocity of the structure immediately after the impact is:
Relevant Equations
Conservation of Linear Momentum
Imagem 135.jpg


I solved it by two methods:

-----------------------------------------------------

First, by conservation of linear momentum, using the vector velocities of each particle:

In the imminence of the impact, the velocity of all the three particles are the same, [itex] \vec v_0 = - \sqrt{2gh} \hat j [/itex]
After que impact, C will have zero velocity, and particles A and B will have velocities with the same magnitude [itex] v_f [/itex], in the perpendicular direction of the bars connected to them. Therefore:

[itex] \vec v_A = - v_fsin(30º) \hat i - v_fcos(30º) \hat j [/itex]
[itex] \vec v_B = v_fsin(30º) \hat i - v_fcos(30º) \hat j [/itex]

By conservation of linear momentum:

[itex] m_A\vec v_0 + m_B\vec v_0 + m_C\vec v_0 = m_A\vec v_A + m_B\vec v_B + m_B\vec v_C[/itex]

Considering [itex] m_A = m_B = m_C = m[/itex] and [itex] \vec v_C = 0[/itex], we have:

[itex] 3m\vec v_0 = 2m(\vec v_A + \vec v_B)[/itex]

[itex] -3\sqrt{2gh} \hat j = -2v_fcos(30º) \hat j[/itex]

Which give us: [itex] v_f = ||v_A|| = ||v_B|| = \sqrt{6gh} [/itex]

The angular speed of the rigid body can be calculated as [itex] ω = \frac{v_t}{r}[/itex]. where [itex] v_t = v_f[/itex] and [itex] r = a [/itex].

Finally:
[tex] ω = \frac{\sqrt{6gh}}{a}[/tex]-----------------------------------------------------

The second method, by conservation of linear momentum of the center of gravity of the rigid body.
This method is more straight forward, and the calculations give the velocity of CG after the impact equal to the velocity in the imminence of the impact. [itex] v_{f_{CG}} = v_{0_{CG}} = \sqrt{2gh} [/itex]

The distance of CG from the particle C is [itex] r_{CG} = a\frac{2}{3}{cos(30º)} = \frac{a \sqrt{3}}{3} [/itex]

The angular speed of the rigid body can be calculated as [itex] ω = \frac{v_{f_{CG}}}{r_{CG}}[/itex].

Finally:
[tex] ω = \frac{\sqrt{6gh}}{a}[/tex]
The problem is, that the correct answer to the question is:
onde.png


The "correct" answer has a 2 dividing the result I found, and there is no answer key to the result I got.

Can someone please help me find where I am wrong, or validate my methods (which would mean that the answer key to the question must be wrong).
 
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  • #2
The total linear momentum of the 3-particle system is not conserved when particle C collides with the support. The support exerts an external, impulsive force on the system that changes the linear momentum.
 
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  • #3
TSny deals with the first method and the conservation of momentum issue of the second.
Another problem with the second approach is, I think, that you take ##v_{CG}## as the velocity of all the masses immediately prior to impact.
On impact, C stops and A & B do not increase their vertical speed (the only possible forces on them are upwards and sideways.) The vertical speed of the CG must be between that of C (zero) and that of A & B, hence less than ##\sqrt{2gh}##, though that speed may not be relevant if momentum is changed.

The only approach that got me the official answer, was to assume C is small, having negligible moment of inertia and negligible friction with the stop bracket.
 
  • #4
The angular momentum relative to point C just before colisión Is ##L=2m\sqrt(2gh)a\sin(60)## and it's conserved. Upon collision the sistem rotates around C
##2m\sqrt(2gh)a\sin(60)=2ma^2\omega##
##\omega=\frac{\sqrt(6gh)}{2a}##
 

1. What is the definition of conservation of linear momentum of rigid body?

The conservation of linear momentum of a rigid body is a fundamental principle in physics that states that the total linear momentum of a system remains constant as long as there are no external forces acting on the system.

2. How is the conservation of linear momentum of rigid body related to Newton's laws of motion?

The conservation of linear momentum of a rigid body is closely related to Newton's laws of motion. Specifically, it is a result of Newton's third law, which states that for every action, there is an equal and opposite reaction. This means that when two objects interact, the total momentum of the system remains constant.

3. What are the conditions for conservation of linear momentum of rigid body?

The conservation of linear momentum of a rigid body holds true under two conditions: first, there must be no external forces acting on the system, and second, the system must be a closed system, meaning that there are no external objects interacting with the system.

4. How is the conservation of linear momentum of rigid body applied in real-life situations?

The conservation of linear momentum of a rigid body is applied in many real-life situations, such as collisions between objects, rocket propulsion, and even sports like billiards. It helps us understand and predict the motion of objects in these situations.

5. What are some common misconceptions about conservation of linear momentum of rigid body?

One common misconception is that the total momentum of a system must always be zero for conservation of linear momentum to hold true. However, this is not the case. The total momentum can be any value as long as it remains constant. Another misconception is that the conservation of linear momentum only applies to objects moving in a straight line. In reality, it applies to all types of motion, including rotational motion.

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