Two Particles Connected by Massless Rod: Dynamics Analysis

In summary, a particle of mass M collides with a particle located at the origin with a speed 𝑣⃗0. After the collision, a particle of mass M bounces straight back to its direction of entry. The total angular momentum about the origin before the collision is LM(v0+v)/m-r'(M)Mv.
  • #1
Booze
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New user is reminded to show their work on schoolwork problems at PF
Homework Statement
The particle of mass M collapses with the body of two point-like particles of mass m.
Relevant Equations
Help needed
Two point-like particles of mass m. The particles are rigidly connected to each other with a mass-less rod of length L. The particles are initially at rest in such a way that one particle is at the origin and the other is at the point (0, L). A point-like particle of mass M collides with a particle located at the origin with a speed 𝑣⃗0.

a) After the collision, a particle of mass M bounces straight back to its direction of entry. Show that two m-the center of mass of the body formed by the particle must then move so that 𝑉𝑐𝑚 = 𝑀(𝑣0+𝑣)/2𝑚 .
b) Determine the orbital angular momentum 𝐿⃗⃗ 𝑡𝑟𝑎𝑛𝑠 and the rotation rate 𝐿⃗⃗ 𝑟𝑜𝑡 of the given
using quantities (relative to the origin) before and after the collision. What can you say about the total angular momentum value?
c) Show that after the collision, the body of two m-particles rotates counterclockwise with angular velocity 𝜔 = 𝑀(𝑣0+𝑣)/𝑚𝐿.
 

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  • #2
Hello @Booze ,

:welcome: ##\qquad ## !​

Please read the PF homework guidelines. We can only help if you post your attempt at solution.

What have you learned so far that can help you solve this one ?

Oh, and "Help needed" is NOT a relevant equation. :wink:

##\ ##
 
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  • #3
in question a) I don't get where does come from M(v0+v)!
i did the law of conservation of momentum before=after. i got that 2m is equel M, but i think it is wrong!
in b) i assumed before collision that Lrot is 0 and Ltrans= r(M) x p(M). and after collision i was thinking that Ltrans=r'(M) x p(M) + Rcm x p(body)=r'(M) x M*v0 +Rcm x Vcm*2m= Rcm*Vcm*2m - r'(M)*M*v=...=LM(v0+v)/m-r'(M)Mv
Vcm u know from a) and i calculate Rcm=mL/2m
i got c) right w=Vcm/Rcm= answer
 
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  • #4
Booze said:
in question a) I don't get where does come from M(v0+v)!
i did the law of conservation of momentum before=after. i got that 2m is equel M, but i think it is wrong!
in b) i assumed before collision that Lrot is 0 and Ltrans= r(M) x p(M). and after collision i was thinking that Ltrans=r'(M) x p(M) + Rcm x p(body)=r'(M) x M*v0 +Rcm x Vcm*2m= Rcm*Vcm*2m - r'(M)*M*v=...=LM(v0+v)/m-r'(M)Mv
Vcm u know from a) and i calculate Rcm=mL/2m
i got c) right w=Vcm/Rcm= answer
Please show how you got your result in part (a), specifically how you applied linear momentum conservation. In part (b) you need to consider angular momentum conservation. What is the total angular momentum about the origin before the collision? Hint: ##\mathbf{L}=\mathbf{r}\times\mathbf{p}##. Part (c) will sort itself out after you have obtained correct answers for parts (a) and (b).
 
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  • #5
Booze said:
I don't get where does come from M(v0+v)!
Neither do I unless you are told that M bounces back with speed v (i.e. a velocity -v). Did you leave that out or is it missing from the original?
 
  • #6
haruspex said:
Neither do I unless you are told that M bounces back with speed v (i.e. a velocity -v). Did you leave that out or is it missing from the original?
I had the same question about the meaning of v. Then I saw in the figure posted in #1 that M is shown bouncing back with speed v.
 

Related to Two Particles Connected by Massless Rod: Dynamics Analysis

1. What is the purpose of analyzing the dynamics of two particles connected by a massless rod?

The purpose of analyzing the dynamics of two particles connected by a massless rod is to understand the motion of the particles and the forces acting on them. This analysis can help in predicting the future motion of the particles and designing systems that involve such arrangements.

2. What are the assumptions made in the dynamics analysis of two particles connected by a massless rod?

The main assumptions made in this analysis are that the rod is massless, the particles are point masses, and there are no external forces acting on the system. Additionally, it is assumed that the rod is rigid and does not deform under the applied forces.

3. How is the motion of the particles affected by the length of the massless rod?

The length of the massless rod affects the motion of the particles by determining the distance between them. As the length of the rod changes, the particles will move closer or farther away from each other, altering their motion and the forces acting on them.

4. Can the dynamics analysis of two particles connected by a massless rod be applied to real-world situations?

Yes, the dynamics analysis of two particles connected by a massless rod can be applied to real-world situations. This type of system is commonly seen in mechanics and engineering, such as in the motion of pendulums and the design of linkages and levers.

5. How does the position of the center of mass affect the dynamics of two particles connected by a massless rod?

The position of the center of mass affects the dynamics of two particles connected by a massless rod by determining the point around which the system rotates. The motion of the particles will be influenced by the position of the center of mass, and it is important to consider this in the analysis.

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