A rotating door is made from four rectangular glass panes, as shown in the drawing. The mass of each pane is 85 kg. A person pushes on the outer edge of one pane with a force of F = 63 N that is directed perpendicular to the pane. Determine the magnitude of the door's angular acceleration in rad/s2.

The distance from the drawing is 1.2m.

* Pls don't say u'll critique my work cos i've tried and failed. I need a solution. It's not homework, but i need to understand this for a test coming up.

Again, model the door as a beam rotating about one end, four of them attached at the center.

Idoor= (4)(1/3 mL^2) where 4 means four doors connected at center pivot, ant 1/3ml^2 is the inertia for each door swinging on the end.

F*l= Idoor*alpha

alpha= Force*L /(4/3 ml^2)

First, we need to calculate the moment of inertia (Idoor) for the four glass panes. We are given the mass (m) and the length (L) of each pane, so we can use the formula:

Idoor = 4 * (1/3) * m * L^2

Plugging the given values:

Idoor = 4 * (1/3) * 85 kg * (1.2 m)^2
Idoor = 4 * (1/3) * 85 kg * 1.44 m^2
Idoor = 4 * 40.8 kg * m^2
Idoor = 163.2 kg * m^2

Next, we need to calculate the angular acceleration (alpha) using the formula given:

alpha = Force * L / (4/3 * m * L^2)

Plugging in the given values:

alpha = 63 N * 1.2 m / (4/3 * 85 kg * 1.44 m^2)
alpha = 75.6 N * m / (4/3 * 122.4 kg * m^2)
alpha = 75.6 N * m / (163.2 kg * m^2)

Now, simplifying the expression:

alpha = 75.6 / 163.2 rad/s²
alpha ≈ 0.46 rad/s²

So, the magnitude of the door's angular acceleration is approximately 0.46 rad/s².

To find the magnitude of the door's angular acceleration, we can use the formula:

α = (F * L) / Idoor

Here, F is the force applied on the outer edge of one pane (63 N), L is the distance from the drawing (1.2 m), and Idoor is the moment of inertia of the door.

From the given information, the mass of each pane is 85 kg. We can calculate the moment of inertia of the door using the formula:

Idoor = (4 * (1/3) * m * L^2)

where m is the mass of each pane (85 kg) and L is the distance from the drawing (1.2 m).

Substituting the values into the equation:

Idoor = (4 * (1/3) * 85 kg * (1.2 m)^2)

Idoor = 163.2 kg·m^2

Now we can substitute this value along with the force and distance into the formula for angular acceleration:

α = (63 N * 1.2 m) / 163.2 kg·m^2

Calculating this:

α ≈ 0.461 rad/s^2

Therefore, the magnitude of the door's angular acceleration is approximately 0.461 rad/s^2.

To determine the magnitude of the door's angular acceleration, we can use the formula:

α = (F * L) / Idoor,

where α is the angular acceleration, F is the force applied to the door, L is the distance from the pivot point where the force is applied, and Idoor is the moment of inertia of the door.

From the information given in the problem, we know that the force applied to the door is F = 63 N, and the distance from the drawing is L = 1.2 m.

To calculate the moment of inertia of the door, we need to consider that the door is made up of four rectangular glass panes. Each pane has a mass of 85 kg.

The formula for the moment of inertia of a rectangular pane rotating about one end (as a beam) is:

Ipane = (1/3) * m * L^2,

where m is the mass of the pane, and L is the length of the pane.

Since there are four panes attached at the center, the moment of inertia of the door can be calculated as:

Idoor = (4) * Ipane.

Substituting the values, we have:

Idoor = (4) * ((1/3) * (85 kg) * (1.2 m)^2).

Now, with the values of F, L, and Idoor, we can calculate the angular acceleration (α) using the formula mentioned earlier:

α = (F * L) / Idoor.

Substituting the values, we have:

α = (63 N * 1.2 m) / ((4) * ((1/3) * (85 kg) * (1.2 m)^2)).

Simplifying this expression will give us the magnitude of the door's angular acceleration in rad/s².