How Does a Car Door's Angular Velocity Change as It Slams Shut?

[i<3math]

Homework Statement

A driver starts his car with the door on the passenger’s side
wide open (\omega = 0) the 36-kg door has a centroidal radius of gyration
k= 250 mm, and its mass center is located at a distance r = 440 mm from
its vertical axis of rotation. Knowing that the driver maintains a constant
acceleration of 2 m/s², determine the angular velocity of the door as it
slams shut (\omega = 90°).

R = 440mm = .44m
k = 250mm = .25m
a = 2 m/s
m = 36kg

Homework Equations

\SigmaM = I_g\alpha + \Sigma r x ma_g ... (1)

\omega = \omega₀ + 2\alpha\theta

a = R\alpha

The Attempt at a Solution

\SigmaM = I_g\alpha + \Sigma r x ma_g

::: \SigmaM = W_g R cos(\theta)
* even though i did this...what do i use for theta? (90? degrees)
* also, i think the only thing that contributes to the moment is the weight of the door?

::: I_g\alpha + \Sigma r x ma_g

= I\alpha + maR
= mk² \alpha + mR²\alpha

after i complete (1) i get this when i solve for alpha:

\alpha = Rg/ (k² + R²cos(\theta)


and then i get stuck here. i didn't know what to use for the theta, but i think I am suppose to find \alpha and then use the kinematic equation:

\omega = \omega₀ + 2\alpha\theta
a = R\alpha

to find omega, and i know omega initial = 0 ??

i feel like I am missing an important part of the equation because I am not getting the right answer...and i don't quite understand it. I am a little confused on the cosine part, am i suppose to use that for the acceleration, and not the moment of the weight? if someone can please help?!? i would appreciate it! Thanks! :D :D

 

[LowlyPion]

Homework Statement

A driver starts his car with the door on the passenger’s side
wide open (\omega = 0) the 36-kg door has a centroidal radius of gyration k= 250 mm, and its mass center is located at a distance r = 440 mm from its vertical axis of rotation. Knowing that the driver maintains a constant acceleration of 2 m/s², determine the angular velocity of the door as it slams shut (\omega = 90°).

R = 440mm = .44m
k = 250mm = .25m
a = 2 m/s
m = 36kg

Have you thought about approaching this problem from the frame of reference of the accelerating car?

In that case wouldn't the change in potential energy - the center of mass moving in the accelerating frame from 0 to the radius of the center of mass below the door post (viewed from the top) - translate into the kinetic energy of the door moving at 1/2*Iw^2 ?

Looks like you can calculate I from the radius of gyration and its mass. Rg^2 = I/m.

Not sure however if this is the way they might want you to solve it for the course you are taking.