Answer:
526 kg-m2/sec
Explanation:
The radius = length of the rod= 3.05 m
angular velocity = 3.60 rad/s
mass = 15.7 kg
L = I ω
Where;
L = angular momentum
I = moment of inertia = mr^2
ω = angular velocity
substituting;
L= 15.7 * (3.05)^2 * 3.60
L = 526 kg-m2/sec
The magnitude of the rod's angular momentum will be 526 Kgm²/sec.
The rotational analog of linear momentum is angular momentum also known as moment of momentum or rotational momentum.
It is significant in physics because it is a conserved quantity. the total angular momentum of a closed system remains constant. Both the direction and magnitude of angular momentum are conserved.
The given data in the problem is;
r is the radius of the rod =3.05 m
l is the length of rod =3.05 m
[tex]\rm \omega[/tex] is the angular velocity = 3.60 rad/s
m is the mass = 15.7 kg
The angular momentum is given by;
[tex]\rm L=I \omega \\\\ \rm L=mr^ 2 \times \omega \\\\ \rm L=15.07 \times (3.05)^2 \times 3.60 \\\\ \rm L=526 \ Kg m^2 /sec[/tex]
Hence the magnitude of the rod's angular momentum will be 526 Kgm²/sec.
To learn more about the angular momentum refer to the link;
https://brainly.com/question/15104254
