# The same gunpower has different energy?

1. Jun 12, 2012

### lywcy68526

the same gunpowder has different energy?

A gun fixed on the disk which can rotate freely,
after first shoot, the disk Angular velocity is ω, Kinetic energy is ΔE1=Iωω/2,
after second shoot, the disk Angular velocity is 2ω, Kinetic energy increases from Iωω/2 to
I2ω2ω/2, the growth ΔE2=3Iωω/2:
Why :
different bullet gunpowder does do different works?
and how about the tenth bullet?……

Last edited: Jun 13, 2012
2. Jun 13, 2012

### willem2

After the first shot, the gun will be moving backwards, so the second bullet will move slower than the first, so there is more kinetic energy left for the gun and the disk.

3. Jun 13, 2012

### lywcy68526

different people will make different conclusion：
1，the people A sit on the disk will find the bullet has the same velocity;
2, only the people B standing by side will agree with you;
3, but another problem appears : after analysis people A will statement that
environment's tempreture is lower than people B's calculation.
E0=Chemical energy of gunpowder
E1=bullet's Kinetic energy
E2=disk's rotating Kinetic energy
E3=heat energy absorbed by environment
E0=E1+E2+E3 every bullet has the same E0
CAN WE CONCLUDE THAT:
in different frame, the tempreture is relative?

Last edited: Jun 13, 2012
4. Jun 13, 2012

### lywcy68526

Is anything wrong?
Equivalence principle?
In any energy conversion process we should put the saddle on the right horse.
Another example:
Two ship A and B move towards the moon, now ship A slow down by jetting gas forward .
In A's opinion the ship's kinetic energy decreases and converts into other kind of energy,so does the gas's internal energy;
But in B's opinion ship A is speeding up but not slow down! Ship A's kinetic energy is increasing up. Ship A need absorb energy !
Would you like to tell me :who is the loser and who is the winner or receiver?

5. Jun 14, 2012

### willem2

The disk isn't an inertial frame, so you can't expect that energy is conserverd. For the people on the disk, the first bullet would suddenly appear to get more energy when you fire the second bullet.

6. Jun 14, 2012

### lywcy68526

You are brave enough.
But the first bullet has gone with the wind. Can energy be transferred through remote sensing? Like Telekinesis ?
And every moving body can entrust different porters carrying different energy share —— even uncertain value ——corresponding to different observers in different frames just in order to match their relative velocity?
My God!

7. Jun 14, 2012

### Staff: Mentor

Energy can change by changing your coordinate system - which observers on the disk do (even if it is so large that the rotational motion is nearly linear for the timescales involved).

Ship B will see a high energy of the ejected gas, as it is now quicker than ship A.

Each reference system will observe different energies for different parts of the same system, and even a different total energy. That is fine. Every system will see a conservation of this energy in its own frame.

8. Jun 14, 2012

### lywcy68526

Would you like to tell me in different frame, the temperature is relative?

9. Jun 14, 2012

### Staff: Mentor

Temperature is usually defined in a reference frame where the object does not move. Therefore, it is pointless to discuss "temperature in a different reference frame".

In addition, I do not see the relevance here. You can look at the energy content from heat - this is the same in all reference frames (at least without special relativity).

10. Jun 14, 2012

### lywcy68526

E0=Chemical energy of gunpowder
E1=bullet's Kinetic energy
E2=disk's rotating Kinetic energy
E3=heat energy absorbed by environment

E0=E1+E2+E3 every bullet has the same E0
The problem is the linear velocity is relative and angular velocity is absolute.
If people A find E1(a) is bigger than people B's conclusion about E1(b), He can only judge that E3(a) gets less share from E0.

11. Jun 14, 2012

### lywcy68526

After n bullets were shot the angular velocity indreased from (n-1)ω to nω——according to the law of momentum conservation;
And the rotating kinetic energy increased from I(n-1)ω(n-1)ω/2 to Inωnω/2: the growth was nIω2-Iω2/2.
The gun can be controlled to fire toward 12,3,6 or 9 o'clock direction in turn!

Last edited: Jun 14, 2012
12. Jun 14, 2012

### lywcy68526

Let's suggest there is a wooden barrel containing the disk and sharing the same center axis.
According to the law of momentum conservation the barrel will gain the same size angular velocity with reverse direction.
What will happen? A paradox or a disaster?

Last edited: Jun 14, 2012
13. Jun 14, 2012

### lywcy68526

14. Jun 14, 2012

### lywcy68526

15. Jun 14, 2012

### lywcy68526

A gun fixed on the disk which can rotate freely,
After first shoot, the disk's angular velocity is ω, kinetic energy is ΔE1=Iωω/2,
After second shoot, the disk Angular velocity is 2ω, Kinetic energy increases from Iωω/2 to
I2ω2ω/2, the growth ΔE2=3Iωω/2:
Why :
Different bullet gunpowder does do different works?
And how about the tenth bullet?……

E0=Chemical energy of gunpowder
E1=bullet's Kinetic energy
E2=disk's rotating Kinetic energy
E3=heat energy absorbed by environment

E0=E1+E2+E3 every bullet has the same E0

Let's suggest there is a wooden barrel containing the disk and sharing the same center axis.
According to the law of momentum conservation the barrel will gain the same size angular velocity with reverse direction.

Is anything wrong? Equivalence principle?

After n bullets were shot the angular velocity indreased from (n-1)ω to nω——according to the law of momentum conservation;
And the rotating kinetic energy increased from I(n-1)ω(n-1)ω/2 to Inωnω/2: the growth was nIω2-Iω2/2.
The gun can be controlled to fire toward 12,3,6 or 9 o'clock direction in turn!

Gradually the kinetic energy of the disk and the barrel will increase more and more rapidly step by step ——what will happen: a paradox or a disaster?

16. Jun 14, 2012

### Staff: Mentor

Re: A paradox or a disaster?

How did you determine this?

17. Jun 14, 2012

### yuiop

This is an interesting question and there does seem to be a problem conserving both momentum and energy or I am missing something elementary. I think it will help to rephrase the question in terms of linear velocity which would be easier to analyse. For example use a gun on a straight frictionless track. After the first shot the velocity of the gun is v. After the second shot the velocity is 2v. This implies the kinetic energy increases from mv^2/2 to 2*mv^2 or a fourfold increase in the kinetic energy and only using twice the gunpowder. How is this resolved?

18. Jun 14, 2012

### D H

Staff Emeritus
Re: A paradox or a disaster?

After the second shot the angular velocity will be slightly more than 2ω.

What you have created is the rotational equivalent of a rocket. Your device will have all of the benefits of a rocket, but also all of the drawbacks that plague a rocket.

Suppose you take thousands and thousands and thousands of bullets with you to the top of this barrel. Those bullets are going to change the momentum of inertia of your barrel, by a whole lot. The Δω from that first first shot is going to be very small because that huge load of bullets makes the moment of inertia of the barrel+rifle+bullets system very large. On the next shot you'll get a slightly larger change in angular velocity because the moment of inertia has decreased slightly as that first bullet is no longer part of the system. And so on. What you'll end up with is something that looks very much like the ideal rocket equation.

That's right. Well, almost right. You didn't account for how firing first bullet changed the moment of inertia of the barrel+rifle+bullets system.

What you are forgetting about is the kinetic energy of the bullet. Suppose you fired the gun from a fixed platform and measure the exit velocity of the bullet. The shot changed the bullet's energy by $\frac 1 2 m v_b^2$, where m is the mass of the bullet and vb is the exit velocity.

On your rotating barrel, this energy will be split between the bullet and the barrel. The total change in energy will remain constant for each firing of the gun, but how it is apportioned between bullet and barrel will vary. On the second shot, the platform is already rotating, so an inertial observer will see the second bullet as moving slower than the first. More of the energy of the shot goes into the barrel. This continues up to the point where rω=vb. The bullet drops straight to the ground with this shot, which means all of the energy of the goes into the barrel's rotation. After this shot, the bullets come out with non-zero velocity, so the portion of energy going into the barrel's rotation decreases.

19. Jun 14, 2012

### yuiop

Re: the same gunpowder has different energy?

I now understand what it was that both you and me (in my last post) were overlooking. It is the fact that most of the energy is transferred to the bullet and the fact that the gun gains 4 times as much kinetic energy after the second shot is not a worry because it is a small percentage of the total energy and the small loss in velocity of the second bullet compensates for this. What DH said in his last post is correct, but I will quantify it with a numerical example.

Initial conditions after the first shot:

Mass of bullet Mb1 = 1 Kg
Velocity of bullet Vb1 = 1000 m/s
Momentum of bullet = Mb1*Vb1 = 1000 Kg m/s
Kinetic Energy of bullet = (1/2)*Mb1*Vb1^2 = 500000 joules

Mass of gun Mg1 = 1000 Kg
Velocity of gun Vg1 = 1 m/s
Momentum of gun = Mg1*Vg1 = 1000 Kg m/s
Kinetic Energy of gun = (1/2)*Mg1*Vg1^2 = 500 joules

Total Kinetic Energy (TKE) obtained from the gunpowder = 500500 joules.

Now assume that the total kinetic energy (TKE) obtained from the second gunpowder shot is the same for the first. We can then state in the new centre of momentum frame that:

TKE = (1/2)*Mb2*Vb2^2 + (1/2)*Mg2*Vg2^2

and conservation of momentum in this frame dictates that:

Vg2 = Mb2*Vb2/Mg2

Substituting this equation into the first gives:

TKE = (1/2)*Mb2*Vb2^2 + (1/2)*Mb2^2*Vb2^2/Mg2

Solve for the velocity of the bullet after the second shot:

Vb2 = sqrt(2*TKE/(Mb2+Mb2^2/Mg2))

The mass of all the bullets is 1kg so Mb2=Mb1 and the mass of the gun Mb2 is 999 kgs.

Vb2 = sqrt(2*500500/(1+1^2/999)) = 999.9995 m/s

(This equates to 998.9995 m/s in the original frame.)

A similar process gives the velocity of the gun after the second shot as:

Vg2 = sqrt(2*500500/(999+999^2/1)) = 1.0010005 m/s

(This equates to 2.0010005 m/s in the original frame.)

Final conditions after the second shot in the centre of momentum frame:

Mass of bullet Mb2 = 1 Kg
Velocity of bullet Vb2 = 999.9995 m/s
Momentum of bullet = Mb2*Vb2 = 999.9995 Kg m/s
Kinetic Energy of bullet = (1/2)*Mb2*Vb2^2 = 499999.5 joules

Mass of gun Mg2 = 999 Kg
Velocity of gun Vg2 = 1.0010005 m/s
Momentum of gun = Mg2*Vg2 = 999.9995 Kg m/s
Kinetic Energy of gun = (1/2)*Mg2*Vg2^2 = 500.5 joules

Total Kinetic Energy (TKE) obtained from the gunpowder = 500500 joules.

To an observer that remained at rest in the original reference frame before the gun was ever fired:

Kinetic Energy of 1st bullet = (1/2)*1*1000^2 = 500000 joules
Kinetic energy of 2nd bullet = (1/2)*1*998.9995^2 = 499000.0005 joules
Kinetic Energy of gun after second shot = (1/2)*999*2.0010005^2 = 1999.9995 joules

Total kinetic energy of both bullets and the gun = 1001000 joules which is exactly twice the energy obtained from one gunpowder shot. Note that the kinetic energy of the gun after the second shot is almost exactly 4 times the kinetic energy of the gun after the first shot, but this is exactly compensated for by the reduced kinetic energy of the second bullet compared to that of the first bullet.

A general equation for the muzzle velocity of the last bullet after N shots is:

VbN = sqrt(2*TKE/(Mb+Mb^2/(Mg-Mb*N)))

and the velocity increase of the gun on firing the N'th shot is:

VgN = sqrt(2*TKE/((Mg-Mb*N)+(Mg-Mb*N)^2/Mb))

All the above methods are easily applicable to the rotational case.

Last edited: Jun 14, 2012
20. Jun 15, 2012

### lywcy68526

Don't you think so ? If not 2ω ,how much it should be in the opinion of people A sitting on the disk?