1. Understanding the Kinematics of Rolling Motion

Let the sphere have a radius $R$. The center of the sphere, point B, moves with a linear velocity $\overrightarrow{V}_B$. As it's moving on a horizontal plane, let's assume it moves to the right. We can write its velocity as $\overrightarrow{V}_B = v \hat{i}$, where $v$ is the speed of the center of mass and $\hat{i}$ is the unit vector in the horizontal direction.

For the sphere to roll without slipping, there is a condition relating its linear velocity $v$ and its angular velocity $\omega$: $v = \omega R$.

As the sphere moves to the right, it rotates clockwise. In a coordinate system where $\hat{i}$ is to the right and $\hat{j}$ is upwards, the angular velocity vector is $\vec{\omega} = -\omega \hat{k}$, where $\hat{k}$ is the unit vector pointing out of the plane.

2. Calculating Velocities of Points A, B, and C

The velocity of any point P on the rolling sphere can be found using the relation: $\overrightarrow{V}_P = \overrightarrow{V}_B + \vec{\omega} \times \overrightarrow{r}_{BP}$ where $\overrightarrow{r}_{BP}$ is the position vector from the center B to the point P.

• Velocity of Point A (Point of Contact): The position vector from B to A is $\overrightarrow{r}_{BA} = -R \hat{j}$. $\overrightarrow{V}_A = \overrightarrow{V}_B + \vec{\omega} \times \overrightarrow{r}_{BA} = v \hat{i} + (-\omega \hat{k}) \times (-R \hat{j})$ $\overrightarrow{V}_A = v \hat{i} + \omega R (\hat{k} \times \hat{j})$ Since $\hat{k} \times \hat{j} = -\hat{i}$, we have: $\overrightarrow{V}_A = v \hat{i} - \omega R \hat{i}$ Using the no-slip condition $v = \omega R$: $\overrightarrow{V}_A = v \hat{i} - v \hat{i} = 0$

• Velocity of Point B (Center): This is the translational velocity of the sphere: $\overrightarrow{V}_B = v \hat{i}$

• Velocity of Point C (Topmost Point): The position vector from B to C is $\overrightarrow{r}_{BC} = R \hat{j}$. $\overrightarrow{V}_C = \overrightarrow{V}_B + \vec{\omega} \times \overrightarrow{r}_{BC} = v \hat{i} + (-\omega \hat{k}) \times (R \hat{j})$ $\overrightarrow{V}_C = v \hat{i} - \omega R (\hat{k} \times \hat{j}) = v \hat{i} - \omega R (-\hat{i})$ $\overrightarrow{V}_C = v \hat{i} + \omega R \hat{i}$ Using the no-slip condition $v = \omega R$: $\overrightarrow{V}_C = v \hat{i} + v \hat{i} = 2v \hat{i}$

Summary of velocities:

• $\overrightarrow{V}_A = 0$
• $\overrightarrow{V}_B = v \hat{i}$
• $\overrightarrow{V}_C = 2v \hat{i}$

3. Evaluating the Options

Now, we test each option using the derived velocities.

A: ${\overrightarrow V _C} - {\overrightarrow V _A} = 2({\overrightarrow V _B} - {\overrightarrow V _C})$

• Left Hand Side (LHS): ${\overrightarrow V _C} - {\overrightarrow V _A} = 2v \hat{i} - 0 = 2v \hat{i}$.
• Right Hand Side (RHS): $2({\overrightarrow V _B} - {\overrightarrow V _C}) = 2(v \hat{i} - 2v \hat{i}) = 2(-v \hat{i}) = -2v \hat{i}$.
• LHS $\neq$ RHS. Thus, option A is incorrect.

B: ${\overrightarrow V _C} - {\overrightarrow V _B} = {\overrightarrow V _B} - {\overrightarrow V _A}$

• LHS: ${\overrightarrow V _C} - {\overrightarrow V _B} = 2v \hat{i} - v \hat{i} = v \hat{i}$.
• RHS: ${\overrightarrow V _B} - {\overrightarrow V _A} = v \hat{i} - 0 = v \hat{i}$.
• LHS = RHS. Thus, option B is correct.

C: $|{\overrightarrow V _C} - {\overrightarrow V _A}| = 2|{\overrightarrow V _B} - {\overrightarrow V _C}|$

• LHS: $|{\overrightarrow V _C} - {\overrightarrow V _A}| = |2v \hat{i} - 0| = |2v \hat{i}| = 2|v|$.
• RHS: $2|{\overrightarrow V _B} - {\overrightarrow V _C}| = 2|v \hat{i} - 2v \hat{i}| = 2|-v \hat{i}| = 2|v|$.
• LHS = RHS. Thus, option C is correct.

D: $|{\overrightarrow V _C} - {\overrightarrow V _A}| = 4|{\overrightarrow V _B}|$

• LHS: $|{\overrightarrow V _C} - {\overrightarrow V _A}| = |2v \hat{i} - 0| = 2|v|$.
• RHS: $4|{\overrightarrow V _B}| = 4|v \hat{i}| = 4|v|$.
• LHS $\neq$ RHS. Thus, option D is incorrect.