# conservation of angular momentum/conservation of energy?

Perhaps a dumb question to ask, but I was given the following problem to solve:

A 10 g bullet traveling at 400 m/s strikes a 10 kg, 1.0-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open. What is the angular velocity of the door just after impact?

When I saw this problem my first attempt to solve this was to use conservation of energy: $m_b$=mass of bullet

$m_d$=mass of door

$L$=length of the door

$I=I_{bullet}+I_{door}= \frac 13m_dL^2+ m_bL^2$

$\frac 12m_bv^2=\frac 12Iw^2$

$w=\sqrt{\frac {m_bv^2}I}$

which is far off from the actual answer. Now, I know that the correct method of solving this type of problem is to use conservation of angular momentum, which will give:

$m_bvL=Iw$

$w=\frac {m_bvL}{I}$

Why is the first method wrong? Thanks in advance.

• Because it's angular momentum rather than angular energy? – John Duffield Jun 5 '15 at 6:44
• Yes I know angular momentum is conserved in this scenario and it is probably obvious to use conservation of angular momentum to solve the problem. But I was wondering why conservation of energy does not work in this case. – whoami Jun 5 '15 at 6:47
• where did you get the moments of inertia of the bullet and the door? – Joaquin Liniado Jun 5 '15 at 6:47
• I used moment of inertia of a plane $I=\frac 13Ma^2$ for the door and moment of inertia of a point mass $I=mr^2$ for the bullet – whoami Jun 5 '15 at 6:51
• Angular momentum is only conserved when net torque is zero, but in this case, why is it zero? I think that the collision with the door changes the momentum of the system during the short period of time the collision takes place, which means that there is a force perpendicular to the axis of rotation during that period of time. I this reasoning correct? Its just I don´t understand why angular momentum is conserved. – nonameable Nov 18 '15 at 16:23