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Moment of inertia of a two-particle system
Concepts tested here
- reduced-mass-moment-of-inertia
All Questions
2016 Phase II 1 question
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A light rod of length l 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:
$I = \mu l^2$ with reduced mass μ.
$r_1 = \frac{m_2l}{m_1 + m_2}$, $r_2 = \frac{m_1l}{m_1 + m_2}$
$I_{cm} = m_1r_1^2 + m_2r_2^2 = \frac{m_1m_2}{m_1 + m_2}l^2$
