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
