How Does a Pendulum Swing Conserve Momentum?

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A pendulum swing exhibits both linear and angular momentum, as the bob has a linear momentum vector during its motion. At the maximum height, the linear momentum is zero, resulting in zero angular momentum as well. When the pendulum reaches the bottom of its swing, it possesses maximum linear momentum, but the angular momentum is calculated differently. The discussion concludes that neither linear nor angular momentum is conserved throughout the swing, as their values differ at the top and bottom positions. The key takeaway is that while the pendulum demonstrates both types of momentum, they do not remain constant during the swing.
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Does a pendulum swing demonstrate angular momentum, linear, both, or neither?
 
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What do you think it is and why?
 
I believe its both because during the swing the bob has a linear momentum vector with the swing. This creates an angular momentum vector out of the plane of the pendulum swing. Once it reaches its max height during the swing the force of gravity causes it to fall back the other way creating a torque vector perpendicular to plane of swing.

I'm not sure if this is completely right...I ask for confirmation
 
Kaxa2000 said:
I believe its both because during the swing the bob has a linear momentum vector with the swing. This creates an angular momentum vector out of the plane of the pendulum swing. Once it reaches its max height during the swing the force of gravity causes it to fall back the other way creating a torque vector perpendicular to plane of swing.

I'm not sure if this is completely right...I ask for confirmation

then yes, it has both linear and angular momentum.
 
Is that the same thing as saying a pendulum swing conserves angular and linear momentum??
 
Let's see. At the top of the pendulum's swing (when v=0), what's the linear & angular momentum of the bob? At the bottom of the swing, what's the linear & angular momentum of the bob? Are the two the same?
 
No

Angular momentum : L = r x p

At top of swing when v = 0

p = 0
&
L = rAt bottom of swing
p = mv

L = r x p

So they are not conserved??
 
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Well, angular momentum at the top of the swing is actually 0 because p=0 and L=r x p. The conclusion is correct, however: neither is conserved.
 
Kaxa2000 said:
No

Angular momentum : L = r x p

At top of swing when v = 0

p = 0
&
L = r


At bottom of swing
p = mv

L = r x p

So they are not conserved??

Your question asked whether it has angular momentum and/or linear momentum, not whether it was conserved or not.
 
  • #10
x is the cross product in this case.

Is r cross product 0 = 0?

Sorry I kind of forgot cross product ...so I'm not surerockfreak, sorry maybe I shouldve restated it in the actual post but the topic title is "pendulum swing conserves what?"
 
  • #11
How does it show that neither are conserved? Is it because the result is not the same at the top of the swing and the bottom of the swing in both cases?
 
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