Answer:
a. Rotational inertia: 5.21kgm²
b. Magnitude of it's angular momentum: 123.32kgm²/s
Explanation:
Length of the rod = 3.46m
Weight of the rod = 12.8 N
Angular velocity of the rod= 226 rev/min
a. Rotational Inertia (I) about its axis
The formula for rotational inertia =
I = (1/12×m×L²) + m × ( L ÷ 2)²
Where L = length of the rod
m = mass of the rod
Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.
Acceleration due to gravity = 9.81m/s²
Mass of the rod = 12.8N/ 9.81m/s²
Mass of the rod = 1.305kg
Rotational Inertia =
(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²
Rotational Inertia = 1.3019115 + 3.9057345
Rotational Inertia = 5.207646kgm²
Approximately = 5.21kgm²
b. The magnitude of the rod's angular momentum about the rotational axis is calculated as
Rotational Inertia about its axis × angular speed of the rod.
Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60
= (226×2π) /60
= 23.67 rad/s
Rotational Inertia = 5.21kgm²
The magnitude of the rod's angular momentum about the rotational axis
= 5.21kgm²× 23.67 rad/s
= 123.3207kgm²/s
Approximately = 123.32kgm²/s
