## A mechanics problem

Nova
Posts: 7
Joined: Thu Sep 29, 2005 2:00 am

### A mechanics problem

I came across this problem in an undergrad text. Can someone help me with this (with explanations)?

A door 1.0 m wide and of 15 kg mass is hinged at one side so that it can rotate without friction about a vertical axis. It is unlatched. A police fires a bullet of 1 g at speed 400m/s perpendicular to the plane of the door. Find the angular veloctiy of the door just after the bullet imbeds itself in the door.

It is a ballastic pendulum kind of problem. Momentum conservation is the key here. How can one relate the linear momentum of the bullet prior to collision with total angular momentum of the bullet+door system after the inellastic collision?

eanzenberg
Posts: 10
Joined: Sat Nov 05, 2005 8:28 pm
L_b = L_a right...

So, L_b is still Iw, where I=mR^2 for the bullet. w is the tricky one, won't it be just = v/R? They don't give the radius of the bullet with respect to the hinge, so either it drops out, or I'm doing it wrong =x. L_a also deals with R so it should drop

Grant
Posts: 191
Joined: Tue May 11, 2004 7:55 pm
I believe the problem needs to specify where the bullet hits the door (i.e. the distance away from the axis of the hinge). If you had this essential information then you could solve the problem.

If this information was provided then the problem can be solved by conservation of angular momentum about the hinge axis.

Let “W” be the width of the door. Let “D” be the perpendicular distance between where the bullet hits the door and the hinge axis. Let “m” be the mass of the bullet. Let “v” be the initial speed of the bullet. Let “M” be the mass of the door. Let “w” be the final angular velocity of the door.

The initial angular momentum about the hinge is just the angular momentum of the bullet about the hinge:
L_i = m*v*D

The final angular momentum is the angular momentum of the door rotating about the hinge plus a small component of the angular momentum of the bullet that is embedded in the door.
L_f = (1/3*M*W^2)*w + (m*D^2)*w
(Note: Recall that I_door_about_hinge = 1/3*M*W^2)

Equating these two (L_i = L_f) and solving for the angular velocity “w” you get:
w = m*v*D/(1/3*M*W^2 + m*D^2)

Note: You can see that the distance “D” is essential to solving this problem. If D=0 then there is no final angular velocity of the door (i.e. a bullet fired directly at the hinge does not rotate the door).

sin_B
Posts: 1
Joined: Mon May 01, 2006 7:17 am

### Possible solution

About the problem involving a bullet being fired at a door and having to find the final angular velocity of the door. I think you should just use conservation of energy by assuming that the collision is elastic as follows.

(0.5 x 1 x 10^ -3) x 400^ 2 = 0.5(15 + (1/1000)) x (omega(1)) ^ 2
which should give you a value of omega as 3.27 rad/s to 3 sig. fig.

And since the question does not specify where the bullet collides with the door, perhaps you must assume that the door is a point particle that will subsequently follow an arc of radius equal to the door length?

Sota
Posts: 2
Joined: Mon Aug 07, 2006 12:34 pm
In general, collisions which involve things sticking to one another are not considered elastic.

Grant has the answer, though I'd neglect the bullet's contribution to moment of inertia, as 15 kg >> 1 g. The bullet has angular momentum about the hinge axis at the time of impact. (And before, but you don't care about that.)

Regarding the missing distance, it's no surprise that your textbook forget to mention exactly where the bullet hit. Just write "assuming the bullet hits the edge of the door..." in front of your answer.