|Jun11-12, 11:51 AM||#18|
Blog Entries: 1
Force and point of application
|Jun11-12, 11:55 AM||#19|
|Jun11-12, 12:05 PM||#20|
Edit: I will put it here:
If that professor is right, than I can create more overall movement with less movement. I intuitevely thought this wasn't possible! But it appears to be. For instances, I thought that conservation of linear momentum wouldn't consider the signal of the velocity, but it does!
So imagine the following collision:
Body A: mass=1kg v=9m/s
Body B: mass=2kg v=0
What I call overall movement does not consider if velocity is positive or negative. In the inicial conditions I have 1kg of mass with 9m/s, this is what I call overall movement.
After collision, the real result is:
Body A: v=-3m/s
Linear momentum is -3*1 + 6*2= 9, like before. But wait! I would intuitevely say that this couldn't be true because I should work with modules | | because I don't care if velocity is positive or negative.
What I see in this result is:
1kg with 3m/s and 2 kg with 6m/s. This for me is on average 3kg of 5m/s or 1kg of 15m/s: this is more than the initial 9m/s in 1 kg!!! (I see mass as a number of something, like atoms). This intuitive idea is wrong.
I thought that if body A has lost 6m/s (9-3 and not 9-(-3)), it could only give 3m/s to the body B, because this body doubles the mass. But this is not what happens.
So I can conclude that In a closed system I can create more of this visual "overall movement", I don't know if this gives potential to anything since energy remains the same.
I don't know if I'm making myself clear, all this can be very silly :9
|Jun11-12, 02:19 PM||#21|
However the difference between an impulse and a steady force (such as gravity) is that you wanted to discuss energy.
Now energy = Force x distance when the force is a steady force such as gravity.
By steady I mean that it is always acting so if a body of mass m moves downward a distance d gravity is acting the whole time and the energy involved is mgd.
However if I extend a spring a distance d by applying a force F the energy involved is only 1/2Fd since the force 'builds up from zero'
For an impulsive force the situation is even lower because not only does the force start at zero but it ends at zero. So the energy is ∫Fδd and you need an expression of F in terms of d which is difficult. That is why we use momentum.
Another point is that F usually represents a steady force such as gravity or a push rod that is constrained to act along the line shown.
Such a the force will tend to brush the object out of its way unless applied at the centre of mass. But it does not remain tangential to the rotation.
An impulse of course does not have this problem.
|Jun11-12, 04:38 PM||#22|
The total kinetic energy of a rigid body can be decomposed into the kinetic energy associated with the translation of the center of mass (CM), and the kinetic energy associated with rotation of the rigid body about the CM.
If a force adds a specific amount of kinetic energy to the rigid body, then the energy has to be divided between the rotation about the CM and the translation of the CM. If the force were directly applied at the CM, the change in rotational kinetic about the CM won't change.
|Jun12-12, 12:08 PM||#23|
I think now I have a different idea of this.
Thanks everybody for all the replies
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