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Q.
Moment of inertia of a uniform rod of length $L$ and mass $M$, about an axis passing through $L/4$ from one end and perpendicular to its length is
JIPMERJIPMER 2013System of Particles and Rotational Motion
Solution:
Moment of inertia of a uniform rod of length $L$ and mass $M$ about an axis passing through the centre and perpendicular to its length is given by
$I_0 = \frac{ML^{2}}{12}$ ..(i)
According to the theorem of parallel axes, moment of inertia of a uniform rod of length $L$ and mass $M$ about an axis passing through $L/4$ from one end and perpendicular to its length is given by
$I = I_{0} +M \left(\frac{L}{4}\right)^{2} = \frac{ML^{2}}{12} +\frac{ML^{2}}{16} = \frac{7ML^{2}}{48}$ (Using (i))