Question 573: 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 passing through a point at a distance $$\frac{L}{4}$$ from one of its ends and perpendicular to the length of the rod 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

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

Answer

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

Answer

$$\frac{5 \mathrm{ML}^2}{48}$$

Answer

$$\frac{9 \mathrm{ML}^2}{48}$$

Centre of Mass and Rotational Motion

The moment of inertia of a thin uniform rod of mass 'M' and length 'L' about an axis passing through a point at a distance $\frac{L}{4}$ from one of its ends and perpendicular to the length of the rod is

Choose the correct answer: