Answer to Question #118120 in Classical Mechanics for Lizwi

Question

A slender uniform rod 100.00 cm long is used as a meter stick. Two parallel axes that are perpendicular to the rod are considered. The first axis passes through the 50-cm mark and the second axis passes through the 30-cm mark. What is the ratio of the moment of inertia through the second axis to the moment of inertia through the first axis?
1 : I2/I1 = 1.5
2 : I2/I1 = 1.7
3 : I2/I1 = 1.9
4 : I2/I1 = 2.1
5 : I2/I1 = 2.3

Answer 1

center of mass of the rod is at L/2 midpoint of the rod here 100/2 =50 cm

Moment of inertia of rod through center of mass (I₁)= $\frac{ml^2}{12}=\frac{m}{48}$

now let's calculate I₂

and now we are shifting the axis 20 cm from the center of mass

then moment of inertia is calculated by $\frac{ml^2}{12}+md^2=\frac{m}{48}+m\times(0.2)^2=$ $\frac{m}{48}+\frac{4m}{100}=\frac{73m}{1200}$

$\frac{I_2}{I_1}=$ 2.9