Moment of Inertia of Two Masses on a Rod – Rankers Physics

Moment of Inertia of Two Masses on a Rod

A light rod of length $\ell$ has two masses $m_1$ and $m_2$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is: (2016 - II)

$(m_1 + m_2)\ell^2$
$\sqrt{m_1 m_2}\ell^2$
$\frac{m_1 m_2}{m_1 + m_2}\ell^2$
$\frac{m_1 + m_2}{m_1 m_2}\ell^2$

Solution:

The moment of inertia about the center of mass uses reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$, resulting in $I = \mu \ell^2 = \frac{m_1 m_2}{m_1 + m_2}\ell^2$.

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