Question 477: Centre of Mass and Rotational Motion

Question

The moment of inertia of a thin uniform rod of mass ' $M$ ' and length ' $L$ ', about an axis perpendicular to length of the rod and at a distance ' $L / 4$ ' from one end is

Accepted Answer

The moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is given by,
$$ \mathrm{I}_0=\frac{\mathrm{ML}^2}{12} $$
A point at a distance $\frac{\mathrm{L}}{4}$ from its end will also be at a distance $\frac{\mathrm{L}}{4}$ from the centre.
Hence by parallel axis theorem,
$$ \mathrm{I}=\mathrm{I}_0+\mathrm{M}\left(\frac{\mathrm{~L}}{4}\right)^2=\frac{\mathrm{ML}^2}{12}+\frac{\mathrm{ML}^2}{16}=\frac{7 \mathrm{ML}^2}{48} $$

Answer

$\frac{\mathrm{ML}^2}{6}$

Answer

$\frac{\mathrm{ML}^2}{12}$

Answer

$\frac{7 \mathrm{ML}^2}{24}$

Answer

$\frac{7 \mathrm{ML}^2}{48}$

Centre of Mass and Rotational Motion

The moment of inertia of a thin uniform rod of mass ' M ' and length ' L ', about an axis perpendicular to length of the rod and at a distance ' L / 4 ' from one end is

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