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A rotating door is made from four rectangular sections, as indicated in the drawing. The mass of each section is 85 kg. A person pushes on the outer edge of one section with a force of F = 68 N that is directed perpendicular to the section. Determine the magnitude of the door's angular acceleration.

(r=1.2m of each door)

Added Thu, 23 Jul '15

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For one section the moment of inertia about the axis of rotation is

I0 = 1/3*m*h**2

where: m = mass of section, h = distance to outer edge

There are 4 sections so the combined inertia is

I= 4*I0 = 4/3*m*h**2

The Torque (M) applied to the door is

M = F*h

Then

M = I0*?

? = M/I0 = (F*h)/(4/3*m*h**2)

? = 3*F/(4*m*h)

We know F = 68 N and m = 85 kg, h=1.2m

?=3*68/(4*85*1.2)=0.5rad /s^2

Added Thu, 23 Jul '15

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