How long would it take Exena to close the door IF SHE DOUBLED THE MAGNITUDE OF T

Question

How long would it take Exena to close the door IF SHE DOUBLED THE MAGNITUDE OF THE FORCE APPLIED TO THE DOOR?

Explanation / Answer

The door performs a rotational motion, accelerated by the torque exerts on the edge. J·a = F·w => a = F·w/J where J moment of inertia, a angular acceleration, w width of the door (= length of lever arm of the force F) Moment of inertia of plane rectangular object, rotating around axis through on edge (see link) J = (1/3)·m·w since mass equals weight W divided by acceleration due to earth gravity g J = (1/3)·(W/g)·w² So the angular acceleration of the door is: a = F·3·g / (W·w) = 220N · 3 · 9.81m/s² / (750N · 1.25m) = 6.90624 s^-2 Integrating twice with initial conditions ?(t=0) = 0 and a(t=0) = 0 you find angular displacement of the door: a = d?/dt => ? = ? a dt = a·t ? = df/dt => f = ? ? dt = ? a·t dt = (1/2)·a·t² So time elapsed until door closed at an angle of f' = 90° = (1/2)·p is t' = v( 2·f'/a ) = v( p / 6.90624 s^-2 ) = 0.67s

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