# What is the angular velocity of the door just after impact?

What is the angular velocity of the door just after impact?

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?

“bo finn”‘s analysis is not correct. What bo finn failed to take into account is that from the point of view of linear momentum, this is not a closed system! Specifically, there is an external force acting on the system (namely, the force of the hinge acting on the door), and because of this, the principle of conservation of linear momentum does not apply.

Therefore, we need to resort to the conservation of ANGULAR momentum (which gives us a very different answer than bo finn’s). The force exerted by the hinge passes through the axis of rotation, and therefore does not contribute to the torque; therefore there are no external torques, so this conservation principle applies.

The angular momentum before impact is contained entirely in the bullet. Even things moving in a straight line have angular momentum if they’re referenced relative to some rotation axis.

Angular momentum before:

(M_bullet)×(V_bullet)×(perpendicular distance to axis)

= (M_bullet)(V_bullet)(R) (where “R” = width of door)

The angular momentum after impact consists of rotating objects, so it’s convenient to express it in terms of angular speed “ω” and moment of inertia “I”:

Angular momentum after:

ωI

= ω(I_door + I_bullet)

For a swinging uniform plank (like a door), I = (1/3)MR²

For a point mass (like a bullet), I = MR²

So:

Angular momentum after = ω((1/3)(M_door)R² + (M_bullet)R²)

By conservation of angular momentum:

(M_bullet)(V_bullet)(R) = ω((1/3)(M_door)R² + (M_bullet)R²)

So:

ω = (M_bullet)(V_bullet)(R) / ((1/3)(M_door)R² + (M_bullet)R²)

= (M_bullet)(V_bullet) / (R(M_door/3 + M_bullet))

= (10g)(400m/s) / (1.0m(10kg/3 + 10g))