A uniform rod with a mass of 0.50 kg and a length of 1 m has a...

The moment of inertia of a uniform rod about an axis passing through its center of mass and perpendicular to its length is given by:

$I_{cm} = \dfrac{mL^2}{12}$

Substituting the given values $m = 0.50 \text{ kg}$ and $L = 1 \text{ m}$:

$I_{cm} = \dfrac{0.50 \times (1)^2}{12} = \dfrac{1}{24} \text{ kg-m}^2$

According to the parallel axis theorem, the moment of inertia $I$ about any parallel axis is:

$I = I_{cm} + md^2$

where $d$ is the distance from the axis to the center of mass (middle point) of the rod.

Given $I = 0.10 \text{ kg-m}^2$, substituting the values gives:

$0.10 = \dfrac{1}{24} + 0.50 \times d^2$

$\dfrac{1}{10} - \dfrac{1}{24} = \dfrac{1}{2} d^2$

$\dfrac{12 - 5}{120} = \dfrac{1}{2} d^2$

$\dfrac{7}{120} = \dfrac{1}{2} d^2$

$d^2 = \dfrac{7}{60} \approx 0.1167 \text{ m}^2$

$d = \sqrt{0.1167} \approx 0.34 \text{ m}$