Moment of inertia car problem

[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

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

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

a = R$$\alpha$$

The Attempt at a Solution

$$\Sigma$$M = I_g$$\alpha$$ + $$\Sigma$$ r x ma_g

::: $$\Sigma$$M = 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:

$$\omega$$ = $$\omega$$₀ + 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.

 

[baba89]

The moment of inertia of the door can be calculated by using the formula I = mk^2, where m is the mass of the door and k is the radius of gyration. In this case, the moment of inertia would be I = (36kg)(0.25m)^2 = 2.25 kg*m^2.

To find the angular acceleration, we can use the formula a = R\alpha, where a is the linear acceleration and R is the distance from the axis of rotation to the center of mass. In this case, a = 2 m/s^2 and R = 0.44 m, so the angular acceleration would be \alpha = a/R = 2 m/s^2 / 0.44 m = 4.55 rad/s^2.

Now, we can use the kinematic equation \omega = \omega_0 + 2\alpha\theta to find the final angular velocity. Since the door starts at rest (\omega_0 = 0) and rotates 90 degrees (\theta = \pi/2), we can plug in these values to get \omega = 0 + 2(4.55 rad/s^2)(\pi/2) = 14.13 rad/s.

Therefore, the angular velocity of the door as it slams shut is 14.13 rad/s. It is important to note that the cosine term in the moment equation is not needed in this case, as it only accounts for the direction of the force, which is not relevant in this problem. The only force acting on the door is its own weight, which can be represented by W = mg.

I hope this helps clarify the problem and how to approach it. Good luck with your homework!