- #1
isabelle3
- 5
- 1
- 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.
$$\frac12 mv^2 = \frac12 I_\text{system}w^2$$
$$mv^2 = (1/3 ML^2 + ml^2)w^2$$
$$ w = \sqrt{\frac{3m}{ML^2 + 3ml^2}} v$$
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.