Relation between linear and angular momentum

  • #1
Krushnaraj Pandya
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Homework Statement


Assertion- If linear momentum of particle is constant, then its angular momentum about any axis will also remain constant
Reason-Linear momentum remains constant when net force is 0, angular momentum remains constant when net torque is zero
which of these statements is/are true? is the reason the correct explanation of the assertion or not?

Homework Equations


dP/dt=F(net)
dL/dt=torque(net)

The Attempt at a Solution


I took the example of a rod of length l, when two equal and opposite forces act on its ends the net force is zero but net torque about center is Fl. Therefore linear momentum is conserved but angular momentum is not but my textbook says both assertion and reason are true but reason is not correct explanation of assertion- I can't figure out how I'm wrong
 

Answers and Replies

  • #2
Nathanael
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Homework Statement


Assertion- If linear momentum of particle is constant, then its angular momentum about any axis will also remain constant
Reason-Linear momentum remains constant when net force is 0, angular momentum remains constant when net torque is zero
which of these statements is/are true? is the reason the correct explanation of the assertion or not?
I would say that’s a correct explanation. The way that comes to mind for me is to differentiate the definition of angular momentum:
dL/dt = d(rxp)/dt = rx(dp/dt) + (dr/dt)xp
(x is the vector cross product)
The first term is zero because dp/dt is zero by assertion. The second term is zero because it’s the cross product of parallel vectors. Thus, dL/dt = 0 or L = constant, as desired.
 
  • #3
Krushnaraj Pandya
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697
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I would say that’s a correct explanation. The way that comes to mind for me is to differentiate the definition of angular momentum:
dL/dt = d(rxp)/dt = rx(dp/dt) + (dr/dt)xp
(x is the vector cross product)
The first term is zero because dp/dt is zero by assertion. The second term is zero because it’s the cross product of parallel vectors. Thus, dL/dt = 0 or L = constant, as desired.
ah! that's a beautiful proof for the assertion to be correct, but what's wrong with my rod example? and in your proof you didn't use the 'reason' statement anywhere so that means both are correct but the 'reason' statement is not an explanation for the assertion?
 
  • #4
Nathanael
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ah! that's a beautiful proof for the assertion to be correct, but what's wrong with my rod example? and in your proof you didn't use the 'reason' statement anywhere so that means both are correct but the 'reason' statement is not an explanation for the assertion?
Ah yes, the rod, I meant to mention that ... rods are not particles!

The “reason” (F = 0) is correct but it’s only half the answer. The other thing to mention is that the torque τ (rate of change of L) is equal to r x F and is therefore also zero. It’s the same proof as mine I just start from L instead of τ
 
  • #5
Krushnaraj Pandya
Gold Member
697
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Ah yes, the rod, I meant to mention that ... rods are not particles!

The “reason” (F = 0) is correct but it’s only half the answer. The other thing to mention is that the torque τ (rate of change of L) is equal to r x F and is therefore also zero. It’s the same proof as mine I just start from L instead of τ
Oh yes, I overlooked that it isn't a particle, and I got your beautiful proof too, in your last message do you mean to say that for a particle if net F is 0, net torque is 0 too...that seems very intuitive for a point mass but the "reason" never directly mentioned the connection...which is probably why the textbook says it isn't a correct explanation of the assertion
 
  • #6
Nathanael
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in your last message do you mean to say that for a particle if net F is 0, net torque is 0 too...that seems very intuitive for a point mass but the "reason" never directly mentioned the connection...which is probably why the textbook says it isn't a correct explanation of the assertion
Yes, it seems the only thing left out of their reasoning to make it correct is the connection τ = r x (∑F) which gives the implication [∑F=0]⇒[τ=0]

Notice if it was an extended body or system of particles we would have τ = ∑(r x F) where the r cannot be factored out because the mass is not all at one point, and so the implication fails.
 
  • #7
Krushnaraj Pandya
Gold Member
697
73
Yes, it seems the only thing left out of their reasoning to make it correct is the connection τ = r x (∑F) which gives the implication [∑F=0]⇒[τ=0]

Notice if it was an extended body or system of particles we would have τ = ∑(r x F) where the r cannot be factored out because the mass is not all at one point, and so the implication fails.
Excellent! All concepts for the day made beautifully clear...no teacher in my school could do so such explicitly and for such an extended time. I love this forum
 

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