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Suppose my system involves:

1) A mounted wheel with some outward flap

2) A bullet already in motion

Initially the net angular momentum is 0 and the net kinetic energy is just that of the speeding bullet.

The bullet hits the flap, causing the wheel to turn, and continues on (slightly slower).

Now the net angular momentum of the system is > 0 and the net kinetic energy is lower.

1) Is energy being converted into angular momentum here (so net energy is conserved)?

2) How is the net angular momentum of this system being conserved with the net amount before/after has changed?

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What does "some outward flap" mean? –  joshphysics May 2 at 23:52
    
Just something rigid the bullet can hit to make the wheel spin. –  Brendan Hill May 3 at 0:03

3 Answers 3

I'll address one underlying issue.

It's important to remember that objects moving in straight lines can have angular momentum. Your bullet can, for example.

The definition of angular momentum $\vec L$ for some point object is:

$$\vec L \equiv \vec r \times \vec p.$$

In that definition, $\vec r$ is the position vector of your object and $\vec p$ is the momentum of the object. So as long as the cross product of the position and momentum vectors is non-zero, something moving in a straight line can have angular momentum.

Now, there are other expressions for angular momentum. You may have seen $\vec L = I \vec\omega,$ which is quite useful for spinning objects. This is actually a special case that can be derived from the definition above.

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So in the system I described, would the bullet continue on with some angular momentum of its own in the opposite direction of the wheel? –  Brendan Hill May 3 at 0:07
    
It could end up either embedded in part of the wheel, punch a hole through and continue on (perhaps somewhat deflected) or ricochet off and head in some other direction. Either way, the final angular momentum (wheel + bullet) is equal to the initial angular momentum of the bullet (and wheel, but you're probably thinking of an axis about which the wheel initially has 0 angular momentum). –  Kyle May 3 at 2:28

Initially the net angular momentum is by no means 0, as you can easily convince yourself by drawing the bullet's momentum vector and it's position vector starting at the wheel's axis onto a sheet of paper. Once they comprise an angle with respect to each other the absolute value of their cross product is nonzero (remember how angular momentum is actually defined $\vec{L} = \vec{r} \times \vec{p}$). Working in the framework of notions that have been the essence of hundreds of years of physics research, it is without any doubt that energy, momentum and angular momentum are always conserved (corresponding to the fact that in classical mechanics there is no point on the time axis, nor a point in 3 dim-space , nor a direction in 3 dimensional space preferable against all others. This makes 7 conserved quantities: energy, 3 components of ang momentum, 3 components of momentum).

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So at the moment that the bullet hits the outward flap, the bullet will decelerate and the wheel will accelerate (at the moment of impact there is conservation of energy and momentum).

Now if the wheel was free floading (for example a bullet hit a wheel in space) the weel would start moving (and probably turning) in the direction of the bullet and the conservation of energy and momtum would be (somewhat) obvious.

But my guess is that you imagine a wheel on earth, that is mounted onto some (frictionless) turning mechanism. In that case we still have conservation of energy, but NO CONSERVATION OF ANGULAR MOMENTUM. The fact that the mount forces the weel to stay in place (and hence the flap to start turning) changes the angular momentum (the mount exerts an external force onto the system).

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