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.
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)
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