Angular velocity of a loop after being struck by bullet

AI Thread Summary
The discussion centers on the calculation of angular velocity after a bullet strikes a circular hoop, emphasizing the importance of using the center of mass (CoM) for conservation of angular momentum. Participants clarify that while one can choose any reference point, using the CoM simplifies calculations as it avoids complications from external forces and torques. The initial angular momentum is calculated based on the bullet's impact, but confusion arises when trying to apply the same principles using the center of the loop as a reference point. The correct angular velocity is derived as ω = v/3R when using the CoM, highlighting the need for careful consideration of reference points in such problems. The conversation underscores the complexities of angular momentum conservation in systems with moving reference points.
  • #51
brochesspro said:
Could you please edit this message? It only shows the plain text, without the implementation of formatting.
No. I have obviously tried. It is reported to staff.
 
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  • #52
Orodruin said:
There is no dot product.
When we expand the vector triple product, we get a linear combination of the vectors inside the cross product whose coefficients are in the form of dot product.
Orodruin said:
The first term is very relevant.
I know, and that is the very thing I fail to understand.
 
  • #53
Orodruin said:
No. I have obviously tried. It is reported to staff.
Scrap that, I found the missing }
 
  • #54
brochesspro said:
I get a vector triple product in the 2nd term.
brochesspro said:
When we expand the vector triple product,
You don’t need to expand it. It is the definition of ##I_r(\vec \omega)##.

Edit: I mean, you can expand it to rewrite it as
$$
m\vec d \times [\vec \omega \times \vec d]
=
md^2 \vec \omega - \vec d (\omega\cdot \vec d) = md^2 \vec \omega
$$
for an expression more familiar in 2D.
 
Last edited:
  • #55
Orodruin said:
As for the com issue. Consider
$$
\sum_i m_i (\vec x_i - \vec x_r) \times \vec v_r
=
M \left[ \underbrace{\frac 1M \sum_i m_i \vec x_i}_{\equiv \vec x_{cm}} - \vec x_r \frac 1M \underbrace{\sum_i m_i}_{\equiv M} \right] \times \vec v_r
=
(M \vec x_{cm} - M \vec x_r)\times \vec v_r
$$
Why can we take the Xr out of the Σ? Is it not variable? Since Vr represents its rate of change.
Orodruin said:
You don’t need to expand it. It is the definition of ##I_r(\vec \omega)##.
Could you tell its definition?
 
  • #56
brochesspro said:
Why can we take the Xr out of the Σ? Is it not variable?
The reference point is fixed.

Edit: Even if it were variable it is independent of ##i##, which is what the sum is over…
brochesspro said:
Since Vr represents its rate of change.
No. It represents the velocity of the solid body at ##\vec x_r##.

brochesspro said:
Could you tell its definition?
$$
I_r(\vec \omega) = \sum_i m_i (\vec x_i - \vec x_r)\times [\vec \omega \times (\vec x_i - \vec x_r)]
$$
 
  • #57
Orodruin said:
No. It represents the velocity of the solid body at x→r.
Wait, ##\vec x_r## and ##\vec v_r## are not the position and the velocity of the reference point respectively? Then what do they mean?
 
  • #58
brochesspro said:
Wait, ##\vec x_r## and ##\vec v_r## are not the position and the velocity of the reference point respectively? Then what do they mean?
Exactly what I said they mean in my last post.
 
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