A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and rotates at 226 rev/min. Calculate (a) its rotational inertia about the axis of rotation and (b) the magnitude of its angular momentum about that axis.

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