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## the same gunpower has different energy?

 Quote by lywcy68526 After second shoot, the disk Angular velocity is 2ω
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.

 Kinetic energy increases from Iωω/2 to I2ω2ω/2, the growth ΔE2=3Iωω/2: Why :
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.

Blog Entries: 6
 Quote by lywcy68526 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?……
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.
 Don't you think so ? If not 2ω ,how much it should be in the opinion of people A sitting on the disk?

Blog Entries: 6
 Quote by lywcy68526 Don't you think so ? If not 2ω ,how much it should be in the opinion of people A sitting on the disk?
Not sure what you are asking here. What don't I think?

From my last post the speed of the disc according to a non rotating observer off the disc will be ω after the first shot and 2.0010005ω after the second shot.

To the observer on the disc the disc speed increases by ω after the first shot and by a further 1.0010005ω after the second shot. The speed of the each successive bullet fired will be a bit slower than the last according to the observer on the disc. This is because the disc is getting lighter and progressively more energy goes into the recoil than the bullet after each shot.

My calculations exactly back up what DH said qualitatively in post #18.

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