A door of width l = 1.00 m and mass M = 15.0 kg is attached to a door frame by t

Question

A door of width l = 1.00 m and mass M = 15.0 kg is attached to a door frame by two hinges. For this problem, you may ignore gravity, as we are interested in rotational motion around the vertical axis. There is no friction of any kind.

A person slams the door shut, by pushing at the middle of the door (l/2 from the hinges) with a force of F = 100 N, lasting a time t = 0.200 s.

The door is initially not rotating. The door can be taken to be a uniform rod, for the purpose of this exercise.

a) What is the angular acceleration of the door while it is being pushed?

b) What is the resulting angular velocity, angular momentum and rotational kinetic energy from this push?

c) Assuming that the person lets go of the door (leaving it to slam shut) at the moment in the motion when the door is perpendicular to the wall, how long does it take for the door to close?

d) What would be the result of a), b), c) and d) if the person had pushed not in the middle of the door but at the edge (l from the hinges)?

In addition to the force of the person, the hinges also provide force, both radial (centripetal) and tangential force while the person pushes (they also compensate for gravity to keep the door upright, but ignore that for now).

e) Combining angular acceleration and linear acceleration considerations, find the tangential force Fh supplied by the hinges as a function of d, the distance between the hinges and the point of application of the person (d was l/2 in part a) and b), and l in part c); now use a general d). Don’t put in explicit numbers, just find the equation.

f) For which d do the hinges not need to provide any tangential force?

Explanation / Answer

a) = l/2*F=50

=M*(l/2)2*

=> =/M*(l/2)2= 13.33 rad/s2

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b) =i+t = 2.7 rad/s

I=(1/12)*Ml2 = 1.25 kg*m2

L=I*=3.375 kg*m2/s

K=(1/2)*I*2=4.56 J

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c) =it+(1/2)t2

=> /2 = 2.7t + (1/2)13.33t2  

=> t=0.323 s

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d) = l*F=100

=M*l2*

=> =/M*l2 = 6.7 rad/s2

=i+t = 1.34 rad/s

I=(1/3)*Ml2 =5 kg*m2

L=I*=6.7 kg*m2/s

K=(1/2)*I*2=4.489 J

=it+(1/2)t2 /2 = 1.34t+(1/2)6.7t^2

=>t=0.512 s

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f) For d=l

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