Question

A 10 g bullet traveling at 350 m/s strikes a 9.0 kg , 1.2-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?

Answer 1

YThe 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)(350m/s) / (1.2m(9kg/3 + 10g))
= 0.9695 rad/sec