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

In summary: However, in step 4, the equation should be \tau = rF, since the angle between r and F is 0 degrees. Solving for F will give you the correct answer of 130N. In summary, to find the steady force needed to move the world's heaviest hinged door through an angle of 90 degrees in 30 seconds, you need to use the mass of the door in your calculations and use the equation \alpha = a/r to calculate the angular acceleration. Then, use the equation \tau = rF to solve for F, which will give you the correct answer of 130N.
  • #1
JFlash
6
0
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: [tex]\Theta = (1/2)(\omega initial + \omega)t[/tex] 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:

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

Does anyone know where I went wrong?
 
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  • #2
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.
 
  • #3


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.
 

1. What is torque and how is it related to swinging doors?

Torque is a measure of the rotational force applied to an object. In the case of a swinging door, torque is the force that is applied to the door to make it rotate around its hinges.

2. How does the mass of a door affect its torque?

The mass of a door affects its torque by increasing the amount of force needed to make it rotate. The larger the mass, the greater the torque required to move the door.

3. What is the role of moment of inertia (I) in calculating torque for a swinging door?

The moment of inertia (I) represents the resistance of an object to changes in its rotational motion. In the case of a swinging door, the moment of inertia depends on the door's mass and how it is distributed around its axis of rotation.

4. How does the distance from the axis of rotation (r) affect the torque of a swinging door?

The distance (r) from the axis of rotation to the point where the force is applied plays a crucial role in determining the torque of a swinging door. The farther the distance, the greater the torque needed to move the door.

5. How do time (t) and angular displacement affect the torque of a swinging door?

The time (t) and angular displacement of a swinging door are directly related to its torque. The longer the door is allowed to rotate (t), or the larger the angle it rotates (angular displacement), the greater the torque required to move the door.

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