Angular Speed of Door at Horizontal Position

[vincisonfire]

Homework Statement

A truck has acceleration A. Its door has a width w and a heigth h. We want to find the angular speed of the door when it reaches the horizontal. (I attached the problem as an image)

Homework Equations

See solution attempt.

The Attempt at a Solution

There is a fictitious force acting on the center of mass of the door. It is situated at w/2.
This force is F = MA
Let T be the symbol for torque and y the angle from the horizontal.
T = r x F = w/2 * MA sin(pi/2 - y) = w/2 * MA cos(y)
Let L be the angular momentum and I the moment of inertia
T = dL/dt = d(I y')/dt
Chain rule gives us
T = I/2 d(y'^2)/dy
T = I/2 d(y'^2)/dy = w/2 * MA cos(y)
T = d(y'^2) = w/I * MA cos(y) dy
y'^2 = w/I * MA sin(y)
I = 1/2*m*w^2
y'^2 = 2*A*sin(y)/w
y' = sqrt(2*A*sin(y)/w)
PS : can't we use latex on this forum?

 

[anathema]

Hello,

Thank you for sharing your solution attempt. Your use of torque and angular momentum equations seems to be correct. However, I would like to point out a potential error in your calculation of the moment of inertia (I). The moment of inertia for a rectangular object like a door is not simply 1/2 * m * w^2, but rather it is dependent on the distribution of mass within the object. In this case, the door's moment of inertia would be closer to 1/12 * m * (w^2 + h^2).

Additionally, I would suggest using radians instead of degrees in your calculations to avoid any confusion with the trigonometric functions.

To answer your question, yes, you can use LaTeX on this forum by enclosing your equations between two dollar signs ($).

 

[thomate1]

I would approach this problem by first identifying the known variables and any relevant equations that can be used to solve for the angular speed of the door. From the given information, we know the acceleration of the truck (A), the width and height of the door (w and h), and that we are trying to find the angular speed of the door when it reaches the horizontal position.

One relevant equation that can be used is the equation for torque (T), which is equal to the product of the force (F) and the distance from the pivot point (r). In this case, the pivot point would be at the center of mass of the door, which is located at w/2. This gives us T = w/2 * F.

We also know that torque can be written as the derivative of angular momentum (L) with respect to time (t). This gives us the equation T = dL/dt. Additionally, we can use the equation for moment of inertia (I) which is equal to 1/2 * mass * radius^2.

Using these equations and the known values, we can write the equation T = I/2 * d(y'^2)/dy, where y is the angle from the horizontal and y' is the angular speed. Simplifying this equation, we get T = w/2 * F = I/2 * d(y'^2)/dy = w/2 * MA cos(y).

Now, we can solve for the angular speed by integrating the equation with respect to y. This gives us y'^2 = 2*A*sin(y)/w. Substituting in the value for the moment of inertia (I = 1/2 * m * w^2), we get y' = sqrt(2*A*sin(y)/w).

In conclusion, the angular speed of the door when it reaches the horizontal position can be calculated using the equation y' = sqrt(2*A*sin(y)/w), where A is the acceleration of the truck and w is the width of the door.

As for using LaTeX on this forum, unfortunately it is not supported. However, you can use symbols and equations by using the formatting options provided in the toolbar above the text box.