Answer:
The angular momentum is
[tex]\vec{L} = I\vec{\omega}\\[/tex]
So, we need to find the moment of inertia of the barbell.
The barbell consist of three parts, two masses of 25 kg and a rod of 15 kg.
The moment of inertia of the barbell can be found by summing the moment of inertia of the rod and two masses.
[tex]I = I_{rod} + 2I_{mass} = \frac{1}{12}ML^2 + 2*mr^2 \\I = \frac{1}{12}15(1.6)^2 + 2(25)(0.8)^2 = 35.2[/tex]
We need to convert 10 rpm to rad/s.
[tex]10 ~rpm = \frac{10}{60} ~rev/sec = \frac{10*2\pi}{60}~rad/sec = 1.047 ~ rad/s[/tex]
The angular momentum can now be calculated.
[tex]L = I\omega = 35.2\times 1.047 = 36.87~kg* m^2/s[/tex]
