Verified answer

The angular momentum of the bullet about the hinge just prior to impact is rp (really vector r cross vector p, but I assume r and p are at 90° at impact), where R is the radius (= W, distance from hinge) and p the bullet linear momentum = m_bullet v_bullet, so

L = W m_bullet v_bullet

Just after impact, all momentum has been transferred to door+bullet. The bullet can be neglected, it's mass is so much smaller than the door. Angular momentum of the rotating door is I_door ω_door, where I_door is the moment of inertia of the door, which is m_door W^2 / 3

L = m_door ω_door W^2 / 3

Conservation of angular momentum -> W m_bullet v_bullet = m_door ω_door W^2 / 3

So ω_door = 3 m_bullet v_bullet / (m_door W) = 3 (.0133 kg)(370 m/s) / ((13.5 kg)(1.06m)) = 1.03 rad/s

Note: moment of inertia of the door can be computed by treating the door as having a linear mass density m/W = 13.5kg / 1.06m :

I = ∭ρr^2 dV = ∫m/W r^2 dr (integrated from 0 to W)

Centripetal acceleration a = w^2r, the place w is the angular speed and r is the radius of rotation So, a = 1500 g = 1500 x 9.eighty one m/s^2 r = 15 cm = 0.15 m w = sqrt (a / r) = ... rads/s 313.2 rads/s ..