Answer:
(a) [tex]I_{system} = 1.014\ kg.m^{2}[/tex]
(b) [tex]\tau = 0.0179\ N-m[/tex]
Solution:
As per the question:
Mass of the object, m = 1.30 kg
Length of the rod, L = 0.780 m
Angular speed, [tex]\omega = 5010\ rev/min[/tex]
Now,
(a) To calculate the rotational inertia of the system about the axis of rotation:
Since, the rod is mass less, the moment of inertia of the rotating system and that of the object about the rotation axis will be equal:
[tex]I_{system} = ML^{2} = 1.30\times (0.780)^{2} = 0.791\ kg.m^{2}[/tex]
(b) To calculate the applied torque required for the system to rotate at constant speed:
Drag Force, F = [tex]2.30\times 10^{- 2}\ N[/tex]
[tex]\tau = FLsin\theta 90 = 2.30\times 10^{- 2}\times 0.780\times 1 = 0.0179\ N-m[/tex]
