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In a downtown office building, you notice each of the four sections of a rotating door has a mass of 75 kg. What is the width, in meters, of each section of the door if a force of 56 N applied to the outer edge of a section produces an angular acceleration of 0.420 rad/s2?

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Answer:

each door has a width of 2.666 meters

Step-by-step explanation:

from Newton's second law applied to rotational motion:

ζ = I α

where ζ= torque , I = moment of inertia , α = angular acceleration

the moment of inertia for a flat plate around its central axis is

I = 1/12 m a² , where m= mass, a= total width = 2L

therefore the moment of inercia for a flat plate with length 2L ( 2 doors, one in each side of the central axis) is

I1 = 1/12 m (2L)² = 1/3 m L²

if we have 4 doors , that is 2 flat plates with length of 2L perpendicular to each other:

I = Ix + Iy = 2*I1 = 2/3 m L²

thus

ζ = I α

4* F * L = 2/3 * (4*m) L² * α

L = 3/2* F/ ( m*α) = 3/2* 56 N / ( 75 Kg * 0.420 rad/seg²) = 2.666 m