Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a mass of 5.00*10-2 {kg}.1. About an axis perpendicular to the rod and passing through its center. (in I=kg*m^2) 2. About an axis perpendicular to the rod and passing through one end. 3. About an axis along the length of the rod.

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 * 10 ^ (-2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg. m²

B) About an axis perpendicular to the rod and passing through one end in kg. m²

C) About an axis along the length of the rod in kg. m²

Answer:

A) I = 0.012 kg. m²

B) I = 0.048 kg. m²

C) I = 8.1 * 10^ (-8) kg. m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 * 10^ (-2) m

Length; L = 1.7m

Mass; m = 5 * 10^ (-2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 * 10^ (-2) * 1.7²) / 12

I = 0.012 kg. m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 * 10^ (-2) * 1.7²) / 3

I = 0.048 kg. m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 * 10^ (-2) m, thus radius; r = (0.36 * 10^ (-2)) / 2 = 0.18 * 10^ (-2) m

I = (5 * 10^ (-2) * (0.18 * 10^ (-2)) ^2) / 2

I = 8.1 * 10^ (-8) kg. m²