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 revolving door having four separate parts with a radius of 1.20

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 rotational analog of Newton's second law, which states that the net torque acting on an object is equal to the moment of inertia multiplied by the angular acceleration.

The moment of inertia for a rectangular section rotating around its center can be calculated using the formula: I = (1/12) * m * (h^2 + w^2), where m is the mass of the section, h is the height of the section, and w is the width of the section.

In this case, each section has the same mass of 86 kg. We are given the radius of the door, which is 1.20 meters. Assuming that the height and width of the section are equal, we can use h = w = 1.20 meters.

Calculating the moment of inertia for one section:
I = (1/12) * 86 kg * (1.20^2 + 1.20^2)
= (1/12) * 86 kg * (2.88 + 2.88)
= (1/12) * 86 kg * 5.76
= 36.48 kg.m^2 (moment of inertia for one section)

Since the door is made up of four sections, we can calculate the total moment of inertia using the formula: I_total = 4 * I
= 4 * 36.48 kg.m^2
= 145.92 kg.m^2 (total moment of inertia)

Next, we need to calculate the net torque acting on the door. The torque exerted by the person pushing on the outer edge of one section can be calculated using the formula: τ = r * F, where r is the radius and F is the force applied.

τ = 1.20 m * 61 N
= 73.2 N.m (torque exerted by one person)

Since there are four sections, the total net torque is given by: τ_total = 4 * τ
= 4 * 73.2 N.m
= 292.8 N.m (total net torque)

Now we can use the formula for rotational analog of Newton's second law to determine the angular acceleration:

τ_total = I_total * α, where τ_total is the total net torque, I_total is the total moment of inertia, and α is the angular acceleration.

Rearranging the equation to solve for α:
α = τ_total / I_total
= 292.8 N.m / 145.92 kg.m^2
≈ 2.00 rad/s^2

Therefore, the magnitude of the door's angular acceleration is approximately 2.00 rad/s^2.