# Moment of inertia car problem

1. Aug 19, 2008

### i<3math

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$$\Sigma$$M = Ig$$\alpha$$ + $$\Sigma$$ r x mag ... (1)

$$\omega$$ = $$\omega$$0 + 2$$\alpha$$$$\theta$$

a = R$$\alpha$$

3. The attempt at a solution

$$\Sigma$$M = Ig$$\alpha$$ + $$\Sigma$$ r x mag

::: $$\Sigma$$M = Wg 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?

::: Ig$$\alpha$$ + $$\Sigma$$ r x mag

= 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 im suppose to find $$\alpha$$ and then use the kinematic equation:

$$\omega$$ = $$\omega$$0 + 2$$\alpha$$$$\theta$$
a = R$$\alpha$$

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

i feel like im missing an important part of the equation because im not getting the right answer...and i don't quite understand it. Im 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

2. Aug 19, 2008

### LowlyPion

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.