Question 72: Centre of Mass and Rotational Motion

Question

A solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The smaller part having mass $M / 8$ is converted into a sphere of radius $r$ and the larger part is converted into a circular disc of thickness $t$ and radius $2 R$. If $I_1$ is moment of inertia of a sphere having radius $r$ about an axis through its centre and $I_2$ is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia $I_2 / I_1=$ $\_\_\_\_$

Accepted Answer

Vectors are categorized into two types based on their symmetry :
1. Polar Vectors : These represent actual physical displacements. They flip their sign (direction) when the coordinate system is inverted ( $\vec{A} ightarrow-\vec{A}$ ).
2. Axial Vectors (Pseudo-vectors) : These usually represent rotations or are the result of a cross product of two polar vectors. They do not flip their sign under coordinate inversion $(\overrightarrow{\mathrm{B}} ightarrow \overrightarrow{\mathrm{B}})$.
Velocity is the rate of change of position: $\vec{v}=\frac{d \vec{r}}{d t}$
Since $\vec{r}$ changes to $-\vec{r}$, then $\frac{d(-\vec{r})}{d t}=-\vec{v}$.
So, when position vector changes sign then velocity vector also get flipped.
Momentum is defined as $\overrightarrow{\mathrm{p}}=\mathrm{m} \overrightarrow{\mathrm{v}}$.
Since $\overrightarrow{\mathrm{v}}$ flips to $-\overrightarrow{\mathrm{v}}$, then $\overrightarrow{\mathrm{p}}$ becomes $\mathrm{m}(-\overrightarrow{\mathrm{v}})=-\overrightarrow{\mathrm{p}}$.
So the linear momentum also flips if position vector changes sign.
Acceleration is the rate of change of velocity: $\vec{a}=\frac{d \vec{v}}{d t}$.
Since $\overrightarrow{\mathrm{v}}$ flips to $-\overrightarrow{\mathrm{v}}$, then $\frac{\mathrm{d}(-\overrightarrow{\mathrm{v}})}{\mathrm{dt}}=-\overrightarrow{\mathrm{a}}$.
So, the acceleration also gets flipped when position vector changes sign.
Angular momentum is defined by the cross product :
$\overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} 	imes \overrightarrow{\mathrm{p}}$.
If we apply the sign change :
$$ \overrightarrow{\mathrm{L}^{\prime}}=(-\overrightarrow{\mathrm{r}}) 	imes(-\overrightarrow{\mathrm{p}}) $$
$$\Rightarrow $$ $$ \overrightarrow{\mathrm{L}^{\prime}}=(\overrightarrow{\mathrm{r}}) 	imes(\overrightarrow{\mathrm{p}}) $$
So, it does not flip. It is an axial vector. Hence, the correct option is (D).

Answer

35

Answer

70

Answer

140

Answer

210

Centre of Mass and Rotational Motion

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