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"Reflect, review and renew constantly." ...Wisdom

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 Author Topic: Dynamics With Variable Mass  (Read 12378 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
j142
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 « Embed this message on: May 28, 2009, 03:18:59 pm » posted from:Surat,Gujarat,India

What will be the dynamics of the system if its mass is varying?

What will be the effect on dynamic system if mass is added or removed?

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Fu-Kwun Hwang
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 « Embed this message Reply #1 on: May 28, 2009, 05:13:16 pm » posted from:Taipei,T\'ai-pei,Taiwan

Mass is related to the inertia of the system.
The dynamic will be different if the mass is changed in diffent ways.
You need to describe what is the system and the way mass will be  changed.
Increase/decrease the mass with what rate?
How is it being increased/decreased? (It could introduce interaction force when mass is increased or decreased).
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j142
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 « Embed this message Reply #2 on: May 28, 2009, 05:32:27 pm » posted from:Vadodara,Gujarat,India

Can we have generalized idea or behaviour of the system with change of mass?

Suppose i say that mass is varying by function F(x), then what will be impact of this on the system? let us assume that system is 1 DOF, like pendulum or spring mass system.
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Fu-Kwun Hwang
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 « Embed this message Reply #3 on: May 29, 2009, 12:05:02 am » posted from:Taipei,T\'ai-pei,Taiwan

The condition which cause the mass change is very important, and can not be ignored in general.
Because when you add mass to a system, you are some interaction occured. (for example: it could be in-elastic collision or collision occurs)
And the dynamic of the system is determined by the external force (i.e. the way interaction occurs).

I will assume the momentum and angular momentum will be conserved, but energy is not conserved for most of the case (unless only elastic collision occurs).
So if the mass was added to the system and the process did not increase momentum to the system, then the velocity and angular velocity will become smaller.
However, if the momentum was also increase, then it is a different story.

There are too many possible outcome become you did not specify any condition.
If we really want to discuss a problem, we should make the problem clear.
It will be easier if it is a well defined problem.
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j142
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 « Embed this message Reply #4 on: May 31, 2009, 08:07:05 pm » posted from:Ahmadabad,Gujarat,India

Thanks you for your beautiful answer, to continue our discussions ahead, i define three different cases as per your advise.

Case 1:-Let us assume that one pendulum made up of the hollow ball and water is filled inside that. There is a hole at the bottom of the ball and water is continuously coming out from that as we start its oscillations.

Case 2:-A spring mass system is oscillating and its mass becomes double when it reaches to extreme position.

Case 3:- A spring mass system is oscillating and its mass becomes halved when it is in equilibrium condition.
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Fu-Kwun Hwang
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 « Embed this message Reply #5 on: May 31, 2009, 09:46:22 pm » posted from:Taipei,T\'ai-pei,Taiwan

1. For the pendulum system, the equation of motion is

$\frac{d^2 \theta}{dt^2}= -\frac{\ell}{g} \sin\theta$
The motion of the pendumum will be the same if the water always leak out in the radial direction
(Did not change momentum in tangential direction).
However, the pendumum motion will be changed if the water leak out has momentum in the tangential direction of the pendulum motion.
2. Does the mass always doubled every time it reach maximum length or just doubled once?

You can guess what will happen with the following analysis:
The potential energy for a spring is $U(x)=\frac{1}{2}k x^2$, where x is the displacement.
All the potential energy will convert to kinetic energy when it reach the equilibrium position.
i.e. $\frac{1}{2}k x^2 =\frac{1}{2} m v^2$.
If the mass is doubled, then the velocity will become smaller $v'=v/\sqrt{2}$,
and the oscillation frequency will be smaller too. $\omega= \sqrt{k/m}$
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"Reflect, review and renew constantly." ...Wisdom