Answer :
The magnitude of the rod's angular momentum is 165.92 [tex]Kg m^{2} s^{-1}[/tex].
The moment of inertia for the rod through an axis 'perpendicular' to one of its ends is given by :
[tex]I=\frac{Ml^{2} }{3}[/tex] (where M = mass of rod , l = length of rod)
Now, the angular momentum of the rod is given by;
L= Iω
= [tex]\frac{Ml^{2} }{3}[/tex]×ω
=[tex]\frac{12.7Kg * (3.15 m^{2})}{3}[/tex]* [tex]3.95 rad s^{-1}[/tex]
= 165.92 [tex]Kg m^{2} s^{-1}[/tex]
∴ The required angular momentum is: 165.92 [tex]Kg m^{2} s^{-1}[/tex]
As the rotating equivalent of linear momentum in physics, angular momentum (rarely a moment of momentum or rotational momentum) is used. It is a conserved quantity, meaning that in a closed system, the total angular momentum stays constant, making it a significant physical quantity. The magnitude and direction of angular momentum are both conserved. The conservation of angular momentum is responsible for the useful characteristics of bicycles, motorcycles, frisbee-lead bullets, and gyroscopes. Hurricanes develop spiral structures and neutron stars rotate rapidly due to the conservation of angular momentum. Conservation generally sets a limit on a system's potential motion but does not determine it in any way.
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