A rotating door is made from four rectangular sections, as indicated in the drawing. The mass of each section is 86 kg. A person pushes on the outer edge of one section with a force of F = 61 N that is directed perpendicular to the section. Determine the magnitude of the door's angular acceleration.
The picture shows the door with four doors connecting in a circle, as a revoliving door looks like. and the radius is 1.20 m
For one section the moment of inertia about the axis of rotation is
Iₒ =ma²/3 = 86•(1.2)²/3 =41.28 kg•m²
where: m = mass of section, a= distance to outer edge
There are 4 sections so the combined inertia is
I= 4•Iₒ = 4•41.28 = 165.12 kg•m²
The Torque (M) applied to the door is
M = F•a =61 •1.2 =73.2 N•m.
The Newton’s 2 Law for rotation
M=I•ε,
ε = M/I = 73.2/165.12 = 0.44 rad/s.
To determine the magnitude of the door's angular acceleration, we can use the equation:
τ = I α
where τ is the torque applied, I is the moment of inertia, and α is the angular acceleration.
The moment of inertia of each rectangular section can be calculated using the formula for a rectangular plate rotating around its center of mass:
I = (1/12) m (h^2 + b^2)
where m is the mass of the section, h is the height, and b is the width.
Considering that each section has the same mass and dimensions, we can calculate the moment of inertia for one section and multiply it by 4 to account for all four sections:
I = 4 * (1/12) * m * (h^2 + b^2)
Substituting the given values, where m = 86 kg, h and b are the dimensions of the rectangular section, and r is the radius of the door (given as 1.20 m):
I = 4 * (1/12) * 86 kg * (h^2 + b^2)
Next, we can calculate the torque applied to one section using the formula:
τ = rF
where r is the radius of the door and F is the force applied at the outer edge of the section.
τ = 1.20 m * 61 N
Finally, we can rearrange the equation τ = I α to solve for α:
α = τ / I
Substituting the calculated values, we can calculate the magnitude of the door's angular acceleration.
To determine the magnitude of the door's angular acceleration, you can use the equations of rotational motion.
The equation that relates the torque, moment of inertia, and angular acceleration is:
τ = Iα
Where:
τ is the torque (force applied multiplied by the lever arm),
I is the moment of inertia, and
α is the angular acceleration.
To calculate the torque, you need to find the perpendicular distance from the point where the force is applied to the axis of rotation.
In this case, the force is applied at the outer edge of one section, and the radius of the door is given as 1.20 m. Since the force is directed perpendicular to the section, the perpendicular distance is equal to the radius.
The torque can be calculated as:
τ = rF
Substituting the given values:
τ = (1.20 m)(61 N)
τ = 73.2 N·m
Now, we need to calculate the moment of inertia for the door. The moment of inertia of a rectangular section about an axis perpendicular to its face and passing through its center is given by:
I = (1/12)ml^2
Where:
m is the mass of each section, and
l is the length of each section.
Substituting the given values:
I = (1/12)(86 kg)(l^2)
Since there are four sections making up the door, the total moment of inertia is four times the moment of inertia of each section:
I = 4[(1/12)(86 kg)(l^2)]
Now we can substitute the calculated torque and moment of inertia into the equation τ = Iα:
73.2 N·m = 4[(1/12)(86 kg)(l^2)]α
Solve this equation for α to determine the magnitude of the door's angular acceleration.