Question

Consider a rod that is 0.220 cm in diameter and 1.50 m long, with a mass of 0.0400 kg.

1. Find the moment of inertia about an axis perpendicular to the rod and passing through one end.(kg*m^2 units please)

2. Find the moment of inertia about an axis along the length of the rod.

Express your answer in kilogram meters squared.

Answer 1

Solution:

Given:

Diameter (d) = 0.22 cm = 2.2 x 10^-3 m

Thus : r = 1.1 x 10^-3 m

Length (L) = 1.50 m

Mass (m) = 0.04 kg

the moment of inertia about an axis perpendicular to the rod and passing through its center = I_C = (1/2) ML²

^{Thus :} I_c = (0.5)(0.04)(1.50)² = 0.045 kg m²

Now : using parallel axis theorem :

the moment of inertia about an axis perpendicular to the rod and passing through one end (I_end) = I_C + m d²

I_end = (0.045) + M (L/2)²

I_end = (0.045) + (0.04) (1.50/2)²

I_end = 0.0675 kg m²

the moment of inertia about an axis along the length of the rod = (1/2) m r² = (0.5)(0.04)(1.1 x 10^-3 m)² = 2.42 x 10^-8 kg m²