A bullet hits a rod attached to a pivot at one of its ends....

[isabelle3]

Homework Statement

A bullet having mass $m$ and initial velocity $v$ hits and stick in a resting rod attached to the pivot. Find the angular velocity of the rod, if it is known, that the length of the rod is $L$, its mass is $M$, and the bullet hole is at a distance $l$ from the pivot.

Relevant Equations

L = cross(r, p) = Iw
E = const
I_rod = 1/3 M L^2
I_bullet = m l^2

My initial thought was to use the conservation of energy law since there're no external forces acting on the system bullet + rod. The rod is in rest, the bullet is moving. Then after the collision, the bullet and the rod are rotating around the pivot together, so the kinetic energy of the bullet transformed into the rotational kinetic energy of the system. So, I had this equation:
$$\frac12 mv^2 = \frac12 I_\text{system}w^2$$
$$mv^2 = (1/3 ML^2 + ml^2)w^2$$
$$ w = \sqrt{\frac{3m}{ML^2 + 3ml^2}} v$$

But if I use the law of conservation of angular momentum, the result is different.
$$L^\text{in} = r \times p = m||r|| ||v|| \sin \frac\pi 2= mlv$$
$$L^\text{fin} = Iw = (\frac13ML^2 + ml^2) w$$

Hence,
$$L^\text{in} = L^\text{fin}$$
$$mlv = (\frac13ML^2 + ml^2) w$$
$$w = \frac {3ml}{ML^2+3ml^2}v$$

As far as I understand, the second answer is correct (I've found a book that had a similar problem solved the second way). But that makes me wonder why the conservation of energy didn't work. Maybe I missed an additional term somewhere? Or maybe I can't just equate kinetic linear and angular energies like that? If that's the case, could you please provide some guidelines for when I can apply such a way of solving a task.

 

[Doc Al]

My initial thought was to use the conservation of energy law since there're no external forces acting on the system bullet + rod.

Careful! When objects collide, kinetic energy is not necessarily conserved. Especially if they stick together. (A totally inelastic collision.)

 

[haruspex]

since there're no external forces acting on the system

That's the condition for conservation of linear momentum, not energy.

 

[isabelle3]

Ahh! Understand.

My confusion was that I thought that in a perfectly inelastic collision the only scalar quantity that changes is the total linear kinetic energy. After additionally consulting wikipedia, here's my mistake that I made: in the conservation of energy law it should be
$$\frac12 mv^2 = \frac12 Iw^2 + \Delta E_\text{th}\\
\text{where } \Delta E_\text{th} \text{ is the thermal energy emitted after the collision.}$$

I don't think that I can calculate it with the given amount of information in the task.

Thank you very much.

 

[isabelle3]

kinetic energy is not necessarily conserved

I think I quoted it, but I didn't...

 

[isabelle3]

That's the condition for conservation of linear momentum, not energy.

Yes, you're right, that's another thing that I mixed up. The conservation of energy always holds for closed systems.

 

[jbriggs444]

Yes, you're right, that's another thing that I mixed up. The conservation of energy always holds for closed systems.

Energy is conserved in closed systems. But mechanical energy (e.g. kinetic and potential energy) is not. It can be converted to heat without leaving the system.

 

[PeroK]

Ahh! Understand.

My confusion was that I thought that in a perfectly inelastic collision the only scalar quantity that changes is the total linear kinetic energy. After additionally consulting wikipedia, here's my mistake that I made: in the conservation of energy law it should be
$$\frac12 mv^2 = \frac12 Iw^2 + \Delta E_\text{th}\\
\text{where } \Delta E_\text{th} \text{ is the thermal energy emitted after the collision.}$$

I don't think that I can calculate it with the given amount of information in the task.

Thank you very much.

From conservation of momentum you can generally calculate how much mechanical energy has been lost. In this case you could if you wanted to.

 

[isabelle3]

From conservation of momentum you can generally calculate how much mechanical energy has been lost. In this case you could if you wanted to.

I could? Do I need to know $w$ to do this? Because I tried to use them, but ended up replacing $Iw$ by $I\frac{v}{CM}$, where $CM$ is the distance from the pivot to the center of the mass of the bullet + rod. I got scary and, I think, incorrect value
$$ \Delta E_\text{th} = \frac{1}{2} v^2 \left( m - \frac{\frac23M^2L^2+2Mml^2 + \frac23 MmL^2+2m^2l^2}{ML+2ml}\right)$$

I think you meant a different way...

 

[PeroK]

I could? Do I need to know $w$ to do this? Because I tried to use them, but ended up replacing $Iw$ by $I\frac{v}{CM}$, where $CM$ is the distance from the pivot to the center of the mass of the bullet + rod. I got scary and, I think, incorrect value
$$ \Delta E_\text{th} = \frac{1}{2} v^2 \left( m - \frac{\frac23M^2L^2+2Mml^2 + \frac23 MmL^2+2m^2l^2}{ML+2ml}\right)$$

I think you meant a different way...

In general the KE (of a particle or rotating object) is determined by its momentum. Assuming you know the mass or moment of inertia.

I haven't looked at the details of this problem. Unless you are asked to calculate the energy loss, its not worth it.

 

[haruspex]

since there're no external forces acting on the system bullet + rod
...
I could? Do I need to know $w$ to do this? Because I tried to use them, but ended up replacing $Iw$ by $I\frac{v}{CM}$, where $CM$ is the distance from the pivot to the center of the mass of the bullet + rod. I got scary and, I think, incorrect value
$$ \Delta E_\text{th} = \frac{1}{2} v^2 \left( m - \frac{\frac23M^2L^2+2Mml^2 + \frac23 MmL^2+2m^2l^2}{ML+2ml}\right)$$

I think you meant a different way...

There is an external force: "rod attached to the pivot", so what conservation law can you use?

Please show your working for how you got the angular velocity of the rod.