Sunday, November 27, 2016

Having fun with revolving doors almost as much as Buddy the Elf

In our hotel there is a revolving door. And I wanted to consider the different torques each family member could give to the door.


To use this model for problems, lets assume that the door weighs a total of 100 kg
And lets say each arm of the door is 3m in length.

So if I, standing 1.5 m from the center push on the door with a force of 50N how much torque do I create?
T=F×r×sinΘ
T= 50N × 1.5m × sin (90°)
T=75Nm

Then, lets say my dad, who can push at a force of 60 N, wanted to match my torque. How far away would he have to be from the center to equal my torque?
T=F×r×sinΘ
75Nm= 60N × x m × sin (90°)
x= 1.25 m

And finally, lets say my mom, who pushes at the same force that I do (50 N), was pushing in the same direction but at 2 m. what would be the doors angular acceleration?




ΣT=Iα
Sum of Torques= T(me) + T(mom)
ΣT=75Nm + (50N)(2m)
ΣT= 175Nm

Moment of Inertia for a revolving door= MR2/3
I= (100kg)(3m)2/3
I= 300 kgm2

Then plugging these into the first equation we can find the angular acceleration (assuming no air resistance)
ΣT=Iα
175Nm= 300 kg2m × α

α=0.58 rad/sec2

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