Answer
given,
length of rod (L)= 6 m
weight of the rod (W)= 10 N
rotates at = 240 rev/min
= [tex]240\times \dfrac{2\pi}{60}[/tex]
= 25.13 rad/s
a) Rotational inertia of the rod about axis of rotation
[tex]I = I_{CM} + m d^2[/tex]
d is the distance from the center of gravity.
d = L/2
[tex]I = \dfrac{1}{2}mL^2+ m (\dfrac{L}{2})^2[/tex]
[tex]I = \dfrac{3}{4}mL^2[/tex]
[tex]I = \dfrac{3}{4}\times \dfrac{10}{9.8}\times 6^2[/tex]
I = 27.55 kg.m²
rotational moment of inertia of the rod is equal to I = 27.55 kg.m²
b)angular momentum of rod
L = I ω
L =27.55 x 25.13
L = 692.36 Kg.m²/s
angular momentum of rod is equal to L = 692.36 Kg.m²/s
