Problem #5 The moment of inertia I of a cheap door of mass M=4.00 kg (about an axis going through the hinges at the door frame) is I= (1/3) M · R², where R= 0.960 m is the width of the door. The door is initially open and at rest. The door suddenly is struck by a huge and heavy dart of mass m= 0.300 kg traveling perpendicular to the plane of the door at a speed v = 20.0 m/s. The dart perforates the wooden door getting permanently stuck at the point of impact, which happened to be right next to the free vertical edge of the door (close to the handle). Because the dimensions of the dart are so small (even though they are exaggerated in the picture for clarity) compared to its distance R to the rotational axis (the distance from the hinge to the free vertical edge next to which the dart strikes the door, which is the width of the door R) we can treat the dart as a point masS I remind %3D %3D

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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Problem #5
The moment of inertia I of a cheap door of mass M = 4.00 kg (about an axis going through the hinges at the
door frame) is I = (1/3) M · R?, where R = 0.960 m is the width of the door. The door is initially open and at
rest. The door suddenly is struck by a huge and heavy dart of mass m = 0.300 kg traveling perpendicular to
the plane of the door at a speed vi = 20.0 m/s. The dart perforates the wooden door getting permanently stuck at
the point of impact, which happened to be right next to the free vertical edge of the door (close to the handle).
Because the dimensions of the dart are so small (even though they are exaggerated in the picture for clarity)
compared to its distance R to the rotational axis (the distance from the hinge to the free vertical edge next to
which the dart strikes the door, which is the width of the door R) we can treat the dart as a point mass. I remind
you that we learned in class that the moment of inertia of a point mass is Im= m·R², where R is the
perpendicular distance from the point mass to the rotational axis.
a) What is the final angular speed wf of the door (and dart stuck in it) after the dart finishes perforating it?
b) At what final linear speed vf are the free edge of the door (close to the handle) and the dart moving together
(relative to the ground) after the collision?
Vi
m
R
M
Page 7 of 7
Transcribed Image Text:Problem #5 The moment of inertia I of a cheap door of mass M = 4.00 kg (about an axis going through the hinges at the door frame) is I = (1/3) M · R?, where R = 0.960 m is the width of the door. The door is initially open and at rest. The door suddenly is struck by a huge and heavy dart of mass m = 0.300 kg traveling perpendicular to the plane of the door at a speed vi = 20.0 m/s. The dart perforates the wooden door getting permanently stuck at the point of impact, which happened to be right next to the free vertical edge of the door (close to the handle). Because the dimensions of the dart are so small (even though they are exaggerated in the picture for clarity) compared to its distance R to the rotational axis (the distance from the hinge to the free vertical edge next to which the dart strikes the door, which is the width of the door R) we can treat the dart as a point mass. I remind you that we learned in class that the moment of inertia of a point mass is Im= m·R², where R is the perpendicular distance from the point mass to the rotational axis. a) What is the final angular speed wf of the door (and dart stuck in it) after the dart finishes perforating it? b) At what final linear speed vf are the free edge of the door (close to the handle) and the dart moving together (relative to the ground) after the collision? Vi m R M Page 7 of 7
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