Angular momentum: why can't I answer this problem using linear momentum rules?

In summary, a 1.0 g bullet is fired into a 0.5 kg block attached to the end of a 0.6 m nonuniform rod of mass 0.5 kg. The block-rod-bullet system then rotates in the plane of the figure about a fixed axis at A (the figure shows a vertical rod labeled A at its top. A block is attached to its bottom end. The bullet flies into the block). The rotational inertia of the rod alone about the axis at A is 0.060 (kg)m^2. Treat the block as a particle. If the angular speed (ω) of the system about A just after impact is 4.5 rad/s, what is the bullet's
  • #1
mm2424
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Homework Statement



A 1.0 g bullet is fired into a 0.5 kg block attached to the end of a 0.6 m nonuniform rod of mass 0.5 kg. The block-rod-bullet system then rotates in the plane of the figure about a fixed axis at A (the figure shows a vertical rod labeled A at its top. A block is attached to its bottom end. The bullet flies into the block). The rotational inertia of the rod alone about the axis at A is 0.060 (kg)m^2. Treat the block as a particle. If the angular speed (ω) of the system about A just after impact is 4.5 rad/s, what is the bullet's speed just before impact.

Homework Equations



I have the worked out solution and see that we need to use L = Iω and L = (linear momentum) x r.

The Attempt at a Solution



When I first tried this problem, I tried to do it as a linear momentum problem. I set it up as (mass of the bullet)(initial velocity of bullet) = (mass of the block, bullet and rod system)(velocity of the block, bullet, rod system). I then found the velocity of the block, bullet, rod system using V=ωR.

I understand that this is an angular momentum problem and that I should have used angular momentum equations to solve it. However, I tried to turn it into a linear momentum problem. My strategy was to use the w given to find v of the block/rod/bullet system after impact (using V=ωR). Since my strategy didn't yield the correct answer, I now see that my reasoning was flawed :).

One thing in particular that confuses me about all of this is that the actual solution requires that we take the linear momentum of the bullet and multiply it by the length of the rod...which seems like taking linear momentum and turning it into angular momentum. I therefore thought/hoped/prayed there was a way to turn angular momentum into linear momentum (by finding V from the w given)...in part because the L = r x p and L = Iω equations seem like magic whereas linear momentum equations seem to make sense.

Can someone please explain the error in my logic? While I'm sure that a completely adequate explanation requires an explanation of the cross product, I think I'm clinically incapable of understanding what it is...so if there's a conceptual way of explaining why I can't tackle this problem with linear momentum...and why the actual method of multiplying the bullet's linear momentum by R and equating it to Iω works, I'd be super grateful!

Thanks!
 
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  • #2
Welcome to PF .

One end of the rod is fixed, so the center of mass of the rod has only 1/2 of the velocity of the rest of the system.
 
  • #3
Oh, I see. I have a quick follow up, if that's ok.

I think what confuses me about this problem is the fact that angular momentum can be written as both r x p and Iω.

The solution to this problem seems to require that we write the angular momentum of the bullet before it strikes as r x p while writing the angular momentum of the system after the strike as Iω. What's conceptually hard for me to reconcile is that it doesn't seem possible to write the angular momentum of the bullet in the Iω form before the strike, since it has no rotational velocity. It seems like ω for the bullet prior to the strike is 0, which would make the angular momentum before the strike 0...but this conflicts with the fact that r x p for the bullet is not 0.

Right now, the decision of whether to express angular momentum as r x p or Iω seems arbitrary, which makes the concept seem a bit mysterious. Is there some way of thinking about these problems that I'm missing? Thanks a million for any advice!
 

1. What is angular momentum and how does it differ from linear momentum?

Angular momentum is a property of rotating objects that describes the quantity of rotational motion. It is a vector quantity that takes into account both the mass and the distribution of mass around an axis. Linear momentum, on the other hand, is a property of moving objects in a straight line and only takes into account the mass and velocity of an object.

2. Why can't I use linear momentum rules to solve problems involving angular momentum?

Linear momentum rules only apply to objects moving in a straight line, while angular momentum rules apply to objects that are rotating. This means that the equations and principles used to solve for linear momentum cannot be applied to angular momentum problems.

3. Can I convert angular momentum into linear momentum?

No, angular momentum and linear momentum are two separate and distinct quantities. While they both involve the concept of motion, they are not interchangeable and cannot be converted into one another.

4. How does the angular momentum of a system change when there is no external torque acting on it?

According to the law of conservation of angular momentum, the total angular momentum of a system remains constant when there is no external torque acting on it. This means that the angular momentum of a system cannot be created or destroyed, only transferred or transformed within the system.

5. What is the relationship between angular momentum and moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in its rotational motion, while angular momentum is a measure of the quantity of rotational motion. The two are related through the equation L = Iω, where L is angular momentum, I is moment of inertia, and ω is angular velocity.

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