Question 90: Centre of Mass and Rotational Motion

Question

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius $R$ and length $L(\mathrm{R}<\mathrm{L})$ about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as $M$ ) :

Accepted Answer

Total mass of the square loop is M which is made of 4 identical solid cylinders.
So, mass of each cylinder $=m=\frac{M}{4}$
The length of each cylinder is L and radius is R .
Axis of rotation passes through the midpoints of two opposite sides.
Let's assume the square lies in the xy -plane and the axis bisects the square vertically.
For the cylinder B and D
The axis passes through the center of these cylinders and is perpendicular to their length.
The Moment of Inertia ( $\mathrm{I}_{\mathrm{B}}, \mathrm{I}_{\mathrm{D}}$ ) of a solid cylinder about a transverse axis passing through its center is:
$$ \mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{D}}=\frac{\mathrm{mL}^2}{12}+\frac{\mathrm{mR}^2}{4} $$
$$\Rightarrow $$ $$ \mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{D}}=2\left(\frac{\mathrm{~mL}^2}{12}+\frac{\mathrm{mR}^2}{4}ight)=\left(\frac{\mathrm{mL}^2}{6}+\frac{\mathrm{mR}^2}{2}ight) $$
For the cylinders A and C :
The axis is parallel to the length of these cylinders.
The axis is at a distance $\mathrm{d}=\frac{\mathrm{L}}{2}$ from the center of each cylinder.
The Moment of Inertia ( $\mathrm{I}_{\mathrm{CM}}$ ) of a solid cylinder about its own longitudinal axis is $\frac{1}{2} \mathrm{mR}^2$.
Using the Parallel Axis Theorem ( $\mathrm{I}=\mathrm{I}_{\mathrm{CM}}+\mathrm{md}^2$ ), the inertia of one such cylinder about the central axis is:
$$ \mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{C}}=\frac{1}{2} \mathrm{mR}^2+\mathrm{m}\left(\frac{\mathrm{~L}}{2}ight)^2=\frac{\mathrm{mR}^2}{2}+\frac{\mathrm{mL}^2}{4} $$
$$\Rightarrow $$ $$ \mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{C}}=2 	imes\left(\frac{\mathrm{mR}^2}{2}+\frac{\mathrm{mL}^2}{4}ight)=\left(\mathrm{mR}^2+\frac{\mathrm{mL}^2}{2}ight) $$
Hence, the total moment of inertia of system about the given axis of rotation is,
$$ \mathrm{I}_{	ext {total }}=\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{D}} $$
$$\Rightarrow $$ $\mathrm{I}_{	ext {total }}=\left(\frac{\mathrm{mL}^2}{6}+\frac{\mathrm{mR}^2}{2}ight)+\left(\mathrm{mR}^2+\frac{\mathrm{mL}^2}{2}ight)$
$$\Rightarrow $$ $\mathrm{I}_{	ext {total }}=\left(\frac{1}{6}+\frac{1}{2}ight) \mathrm{mL}^2+\left(\frac{1}{2}+1ight) \mathrm{mR}^2$
$$\Rightarrow $$ $\mathrm{I}_{	ext {total }}=\frac{4}{6} \mathrm{~mL}^2+\frac{3}{2} \mathrm{mR}^2$
$$\Rightarrow $$ $\mathrm{I}_{	ext {total }}=\frac{2}{3} \mathrm{~mL}^2+\frac{3}{2} \mathrm{mR}^2$
$$\Rightarrow $$ $\mathrm{I}_{	ext {total }}=\frac{2}{3}\left(\frac{\mathrm{M}}{4}ight) \mathrm{L}^2+\frac{3}{2}\left(\frac{\mathrm{M}}{4}ight) \mathrm{R}^2$
$$\Rightarrow $$ $\mathrm{I}_{	ext {total }}=\frac{2}{12} \mathrm{ML}^2+\frac{3}{8} \mathrm{MR}^2$
$$\Rightarrow $$ $$ \mathrm{I}_{	ext {total }}=\frac{1}{6} \mathrm{ML}^2+\frac{3}{8} \mathrm{MR}^2 $$
Hence, the correct option is (D).

Answer

$\frac{3}{4} M R^2+\frac{1}{6} M L^2$

Answer

$\frac{3}{8} M R^2+\frac{7}{12} M L^2$

Answer

$\frac{3}{4} M R^2+\frac{7}{12} M L^2$

Answer

$\frac{3}{8} M R^2+\frac{1}{6} M L^2$

Centre of Mass and Rotational Motion

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length L(\mathrm{R}<\mathrm{L}) about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as M ) :

Choose the correct answer: