I dont get how ice skaters angular momentum is conserved when arms are drawn in?

  • #1
in linear momentum, considering the collision of 2 particles to deudce that the momentum in a closed system is concerved
(W_1 + F_12)t = m.v_f1 - m.v_01 ....(i)
where
W_1=external force during impact
F_12=force acting on body 1 by body 2 in collision
t=impacting time
m.v_f1= momentum of body 1 after collision
m.v_01=momentum of body 1 before collision
similiarly
(W_2 + F_21)t = m.v_f2 - m.v_02 ....(ii)
adding 1 and 2
since F_12 = -F_21 by newton's third law
therefore (W_1+W_2)t = (m.v_f1 +m.v_f2) - (m.v_01 + m.v_02)
if the particles is isolated system, W_1+W_2=0 and the linear momentum of the system is conserved

similary result can be deueced by adding more of these equations of same form of (i) and (ii)

with calculus
F=dp/dt=d(mv)/dt= m. dv/dt= ma
with no external force a=o and linear momentum is conseved

I get how this works similiarly with rotational dynamics only in COLLISION

in the first method, conservation of angular momentum closed system durning collsion can be dudeuced by the 1st method by replacing
W_1 by torque by exteral force
F_12 by T_12= torque created on body(or disc) 1 by body 2
m.v_f1 by I.w_f1= angular momentum of body after collision
m.v_01 by I.w_01= angular momentum of body before collision

and by calculus
T= dL/dt = d(Iw)/dt =I. dw/dt = Ia

this is what i dont get: in the case of a ice skater, pulling arms in during rotation, I(moment of inertia) is changed, so how is conservation of momentum explained with calculus method when I of the system changes and i dont get at all how the non calculus method explains the conservation in this case (i only get it if it is during impact in rotational but in linear if totally understand how this method explains the conservation whether it impact of no impact)
 
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Answers and Replies

  • #2
olgranpappy
Homework Helper
1,271
3
this is what i dont get: in the case of a ice skater, pulling arms in during rotation, I(moment of inertia) is changed, so how is conservation of momentum explained with calculus method when I of the system changes and i dont get at all how the non calculus method explains the conservation in this case (i only get it if it is during impact in rotational but in linear if totally understand how this method explains the conservation whether it impact of no impact)
The angular momentum will be conserved when the sum of all the torques is zero, just like the linear momentum is conserved when the sum of all the forces is zero. The sum of torques on the skater is not exactly zero, but quite small, so her angular momentum is conserved.

Angular momentum is a bit more difficult because there is more vector calculus involved in most cases (exceptions being the usual simple textbook cases that reduce to rotation about one "easy" axis). But, consider this simple example:

A single mass point has angular momentum about some origin
[tex]
\vec L = m\vec r \times \vec v
[/tex]

thus
[tex]
\frac{d\vec L}{dt}=m \vec r \times \vec a + m \vec v \times \vec v
[/tex]

but [tex]\vec v \times \vec v=0[/tex] thus
[tex]
\frac{d\vec L}{dt}=m\vec r \times \vec a=\vec r \times \vec F_{tot}
[/tex]

the far RHS of the above equation is by definition the sum of all the torques, and so if it is zero the angular momentum is conserved.
 
  • #3
actually i dont get v.v=0
is one of the v the tangential velocity and the second the velocity perpendicuar to that( going towards
the axis)?
 
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  • #4
also can u explain this case with the non calculus method?
is it because when the arms are drawn in, the body exerts a torque to keep the arms in a lower orbit and the arms exert an equal and opposite torque, and so even the I of the whole system changes, it is conserved due to no external torque?
 
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  • #5
i still dont quiet understand as the ice skater draws arms in (assume friction is 0), how come momentum is conserved even though his/her moment of inertia is changed ?
 
  • #6
olgranpappy
Homework Helper
1,271
3
Her inertia is changed (decreases), but her angular speed is also changed (increases)! In exactly such a way that

[tex]
L=I\omega
[/tex]

is constant
 
  • #7
I THINK i get it now !
Thx 4 ur help!
 
  • #8
my original question is acutally how L = Iw is explained by conservation of momentum coz i did not get how the skates momentum is actaully conserved!
 
  • #9
but i think i got it now!
 

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