Torque: Swinging Door; mass, I, r, t and Angular Displacement Given

[JFlash]

Hello. The question reads, "...This is the world's heaviest hinged door. The door has a mass of 44,000kg, a rotational inertia about an axis through its hinges of 8.7X10^4 kg*m^2, and a width of 2.4m. Neglecting friction, what steady force, applied at its outer edge and perpendicular to the plane of the door, can move it from rest through an angle of 90 degrees in 30s? Assume no friction acts on the hinges."

The work I did below followed 4 steps:
1. Use this equation: $$\Theta = (1/2)(\omega initial + \omega)t$$ to find the angular velocity at 30s.
2. Find the average angular acceleration over the 30s period of applied force.
3. Set the torque equal to the rotational inertia of the door times its angular acceleration.
4. Set torque equal to r X F, and solve for F.

My answer came to 40.27N, which is about a third of the actual answer: 130N. As I look over my work, I realize that I never used the mass of the door, which is probably a reason why my answer is wrong. Anyway, here's what I did:

$$\Theta = \pi/2 = (1/2)(0 + \omega)30s... \omega = \pi/30 rad/s$$
$$\alpha = (\pi/30 rad/s)/(30s-0s) = \pi/900 rad/s^2$$
$$\tau = 87000kgm^2 * \pi/900 rad/s^2 = 96.67$$
$$\tau = rF sin \phi = 2.4m * F * sin(90 degrees) F = 40.27N$$

Does anyone know where I went wrong?

 

[gtoguy97]

You need to use the mass of the door in order to calculate the angular acceleration. The equation for angular acceleration is: \alpha = a/r where a is the linear acceleration of the door, and r is the radius of the door (which is 2.4m in this case). You can calculate a by using the equation a = F/m, where F is the applied force and m is the mass of the door. Once you have \alpha, you can use it to calculate the torque as you did above.

 

[charng]

Hello, thank you for sharing your work and thought process. From what I can see, your calculations and approach seem correct. However, the mistake may lie in the assumption of neglecting friction. In reality, there will always be some friction present, especially in a door of this size and weight. This could explain the discrepancy between your calculated force and the actual force needed to move the door.

Additionally, the mass of the door is not necessary in this calculation as it is already taken into account in the calculation of the rotational inertia. The only other factor that could potentially affect the force needed would be the distribution of mass in the door, but without that information, it is difficult to determine the exact reason for the difference in the calculated and actual force.

In conclusion, it seems like your approach and calculations were correct, but the assumption of neglecting friction may have led to the discrepancy in your final answer. Keep in mind that in real-world situations, there are always external factors that can affect the outcome, and it's important to consider them when making calculations.