A rotating door is made from four rectangular glass panes. The mass of each pane is 90kg. A person pushes on the outer edge of one pane with a force of F=70N that is directed perpendicular to the pane. Determine the magnitude of the door's angular acceleration. in rad/s2
I will be happy to critique your work on this. The key is modeling the door so that you can get the moment of inertia.
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35rad/s²
543
To determine the magnitude of the door's angular acceleration, we need to calculate the moment of inertia of the door and then apply Newton's second law for rotation.
The moment of inertia (I) represents the resistance of an object to changes in its rotational motion and is calculated based on the object's mass distribution.
Given that the rotating door is made from four rectangular glass panes, each with a mass of 90kg, we can assume that the panes are of equal size and distribute the mass evenly. Therefore, the mass of each pane is 90kg / 4 = 22.5kg.
Now, let's calculate the moment of inertia for a single glass pane.
The moment of inertia for a rectangular pane about its center of mass is given by the formula:
I = (1/12) * m * (h^2 + w^2)
Where:
m = mass of the pane
h = height of the pane
w = width of the pane
Since the pane is rectangular, we can assume that the height and width are equal. Let's denote the width and height of the pane as b.
The given information does not provide the dimensions of the pane, so we'll need to make an assumption for the dimensions. Let's assume that the width (b) of the pane is 0.8 meters.
Substituting the values into the moment of inertia formula:
I = (1/12) * (22.5kg) * (0.8m^2 + 0.8m^2)
I = (1/12) * (22.5kg) * (1.6m^2)
I = 0.3 kg * m^2
Now, we can calculate the torque (τ) applied to the door by the person pushing on one pane. The torque is given by the formula:
τ = F * r
Where:
F = applied force (perpendicular to the pane)
r = distance from the rotation axis to the point of application of force
In this case, the force applied is 70N, and the distance from the rotation axis to the outer edge of the pane is the width of the pane (b) divided by 2:
r = b/2 = 0.8m / 2 = 0.4m
Substituting the values into the torque formula:
τ = (70N) * (0.4m)
τ = 28 N*m
Finally, we can use Newton's second law for rotation to relate the torque (τ) to the angular acceleration (α):
τ = I * α
Rearranging the equation to solve for α:
α = τ / I
Substituting the values:
α = (28 N*m) / (0.3 kg * m^2)
α ≈ 93.33 rad/s^2
Therefore, the magnitude of the door's angular acceleration is approximately 93.33 rad/s^2.