Question
Asked Nov 8, 2019
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The door shown in the figure is equipped with a torsional spring of stiffness 25 Nm/rad and a torsional viscous damper c so that it automatically returns to its closed position after being opened. The door has a mass of 60 kg and a centroid moment of inertia about axis AB parallel to the door's axis (Oz) of 7.2 kgm^2.
 
a. What is the damping coefficient such that the system is critically damped?
b. What is the initial angular velocity of the door so that the door is opened 70 degrees? How long will the door return to 5 degrees of completely closing?
c. Repeat a and b if the door is designed with a damping ratio of 1.3.
0.9 m
В
Ө
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0.9 m В Ө

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Expert Answer

Step 1

a.

Consider I as the moment of inertia and Zeta as the damping ratio. The damping coefficient is,

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C =25RI For critically damped system 1 Thus, calculate the damping coefficient. C 2x1x25 Nm/radx 7.2 kg m2 26.83 N.m/rad

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Step 2

b.

Calculate the angular velocity.

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25 Nm/rad 7.2 kg m2 =1.863 rad/s

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Step 3

Calculate the time taken for the...

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0= ax rad |(1.863 rad's ) 700 180° 4 =0.6558 s Similarly, for 5 0= cot2 rad 180° = (1.863 rad/s)t 50 t, = 0.0468 s

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