Re: [題目] 角動量守恆的問題
※ 引述《allen750920 (So Fun)》之銘言：
: [題目]當某質點被繩繫住作圓周運動，今天增加繩子張力，半徑固定，角動
: 量是否守恆?
: [瓶頸]
: 1.角動量守恆端看有無受力矩作用，今天繩子張力通過軸心，力矩為零，
: 故角動量守恆。
: 2.當繩子張力增加，質點切線速度增加，由於半徑固定，所以L=mvXr
: 之值改變，角動量不守恆?
: 從1的角度來看確實角動量守恆，但從2的角度來看卻無法解釋為什麼角動
: 量不守恒。想請問各位，我有什麼地方沒有考慮到嗎?
Initial condition, the object undergoes circular motion
with x=rcosb ; y=rsinb (b is the parameter of the angle)
In this case, there is no potential energy given (assuming U=0)
therefore, L=T-U=T (L= lagrangian; T=kinetic energy; U=potential)
L=T=1/2 m(x'^2+y'^2) (where x'=-rb'sinbt; y'=rb'sinbt; b'=angualr velocity)
=1/2 mr^2 b'^2
so, pb=dL/db' (where pb is the angular momentum)
pb=d/dt(1/2mr^2 b'^2)=mb'r^2
1st case, if b'=constant (which is the initial condition)
then pb=constant (angular momentum conserves)
2nd case, if you increase tension consistently (non-stop)
b" is a constant(angular acceleration), but b' is not a constant
then pb is not a constant (angular momentum "NOT" conserves)
3rd case, if you increase tension for a small period and stay the same after,
during the time you increased the tension b' is not constant→pb not conserve
after a while when you stay with the same tension b'=constant (new constant)
pb→conserves but with a larger value.
Hope this can help you something, if not or you think there is something
you dont agree with, feel free to point out and hopefully we can dicuss more
--
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◆ From: 99.245.10.248
※ 編輯: lewisweilin 來自: 99.245.10.248 (03/27 13:48)
※ 編輯: lewisweilin 來自: 99.245.10.248 (03/27 13:54)
推
03/27 20:46, , 1F
03/27 20:46, 1F
→
03/27 20:46, , 2F
03/27 20:46, 2F
Well, from the lagrangian formulation, it is a little bit too hard for a
biginner. But, the physics is the same. There are some pretty straight
forward phenomenun as we can see here. The angular momentum of course is
related to its angular velocity, if the angular velocity remain the same, then
the quantity is conserved. If not (the tension force is changing), clearly,
there is a change in angular velocity which means b" is not zero. Therefore,
the angular momentum is not conserved. Or in other words, we can say that
F=mv^2/r, if the force increases gradually (with r fixed) that means we will
have a larger tangential velocity. From here, we can use the relationship,
L=mvXr, to see that an inconsistent v caused by the tension force will result
the angular momentum varies (not consistent→not conserved)
※ 編輯: lewisweilin 來自: 99.245.10.248 (03/28 07:08)
推
03/28 20:48, , 3F
03/28 20:48, 3F
→
03/29 08:02, , 4F
03/29 08:02, 4F
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