Rotation - Collision of Rotating Cylinders Cylinders

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Homework Help Overview

The discussion revolves around the collision of two rotating cylinders and the conservation of angular momentum. Participants are exploring the implications of torque and the correct application of moments of inertia in their calculations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the impulse-momentum theorem to relate angular momentum before and after the collision. They question the validity of their method, particularly regarding the calculation of angular momentum and the effects of torque.
  • Some participants inquire about the choice of reference point for calculating angular momentum and whether it was done correctly. They also discuss the implications of torque on the conservation of momentum.

Discussion Status

Participants are actively engaging in clarifying the original poster's approach and checking the assumptions made regarding the conservation of angular momentum. There is a focus on understanding the correct application of moments of inertia and the conditions under which momentum is conserved.

Contextual Notes

There is a noted ambiguity regarding the velocity of approach of the cylinders, which is not specified in the problem statement. This uncertainty may affect the interpretation of the collision dynamics.

cupid.callin
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Homework Statement



attachment.php?attachmentid=33231&stc=1&d=1300475161.jpg



The Attempt at a Solution


The book did it like taking impulse of forces f and writing eqn Impulse = Δ(Angular momentum)
or J = ΔL

How i tried:

attachment.php?attachmentid=33232&stc=1&d=1300475161.png


let the 2 cylinders meet at A

(they have not mentioned the velocity of approach but as it says that the 2 cylinders remain in contact and don't fly off after collision so i guessed that velocity is negligible)

Considering Torque at A
As no torque acts during whole process so L about A remains constant

[tex]L_{initial} = I_1w_1 \ - \ I_2w_2[/tex]

now let that after they meet and friction does it work they have angular speeds w1' and w2'

[tex]L_{final} = I_1w_1' - I_2w_2'[/tex]

and also

[tex]w_1'r_1 = w_2'r_2[/tex]
as the point of contact of 2 cylinders has same velocity

But this gives wrong answer!

Can someone tell me why this method is wrong?
 

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hi cupid.callin! :smile:

your method should work :confused:

can i check … when you said …
cupid.callin said:
As no torque acts during whole process so L about A remains constant

… did you calculate L using the moments of inertia about the centres (which is correct), or about A (which isn't) ?
 
tiny-tim said:
… did you calculate L using the moments of inertia about the centres (which is correct), or about A (which isn't) ?

moment if inertia of a body about any point,

[tex]\vec{L} = \vec{L}_{CM} + m\vec{r}X\vec{v}_o[/tex]

where vo is velocity of CM

And as i said that cylinders do not fly away after collision, vo ≈ 0

So

[tex]\vec{L} = \vec{L}_{CM}[/tex]
 
And if the torque about A is zero then shouldn't momentum be conserved about A and thus calculate moments of inertia about A ?

If i calculate moments of inertia along CM then the friction will have torque and then momentum is not conserved !
 
hi cupid.callin! :smile:
cupid.callin said:
[tex]\vec{L} = \vec{L}_{CM}[/tex]

yes, that's right …

LA = Lc.o.m.

but you said you used the formula Iω for LA, so i was just checking whether you used IAω or Ic.o.m.ω :wink:
 
What form is the answer in?

Can you show your solution?
 

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