Question

A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, $A$ is the point of contact, $B$ is the centre of the sphere and $C$ is its topmost point. Then,

• A. ${\overrightarrow{V}}_C-{\overrightarrow{V}}_B={\overrightarrow{V}}_B-{\overrightarrow{V}}_A$
• B. $|{\overrightarrow{V}}_C-{\overrightarrow{V}}_A|=4|{\overrightarrow{V}}_B|$
• C. ${\overrightarrow{V}}_C-{\overrightarrow{V}}_A=2({\overrightarrow{V}}_B-{\overrightarrow{V}}_C)$
• D. $|{\overrightarrow{V}}_C-{\overrightarrow{V}}_A|=2|{\overrightarrow{V}}_B-{\overrightarrow{V}}_C|$

Answer 1

Answer: (A), (D)

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

Assume $R$ is the radius of the sphere and $\omega$ is angular velocity (clockwise).


Then,
${\overrightarrow{V}}_C={\overrightarrow{V}}_B+{\overrightarrow{V}}_{C,B}$
$\Rightarrow {\overrightarrow{V}}_C-{\overrightarrow{V}}_B=\omega R\hat{i}...\left(i\right)$
[${\overrightarrow{V}}_{C,B}=\omega (-\hat{k})\times R\left(\hat{j}\right)=\omega R\hat{i}$]

${\overrightarrow{V}}_A={\overrightarrow{V}}_B+{\overrightarrow{V}}_{A,B}$
$\Rightarrow {\overrightarrow{V}}_A-{\overrightarrow{V}}_B=\omega R(-\hat{i})$
[${\overrightarrow{V}}_{A,B}=\omega (-\hat{k})\times R(-\hat{j})=\omega R(-\hat{i})$]
$\Rightarrow {\overrightarrow{V}}_B-{\overrightarrow{V}}_A=\omega R\hat{i}...\left(ii\right)$

From Eq. (i) and (ii) $\Rightarrow$ (i) + (ii)
${\overrightarrow{V}}_C-{\overrightarrow{V}}_A=2\omega R\hat{i}...\left(iii\right))$
From Eq. (1) ${\overrightarrow{V}}_B-{\overrightarrow{V}}_C=\omega R(-\hat{i})...\left(iv\right)$

From equation (i), (ii), (iii) & (iv) we can say
${\overrightarrow{V}}_C-{\overrightarrow{V}}_A\ne 2({\overrightarrow{V}}_B-{\overrightarrow{V}}_C)$
Option A is incorrect.
${\overrightarrow{V}}_C-{\overrightarrow{V}}_B={\overrightarrow{V}}_B-{\overrightarrow{V}}_A$ (from Eq (1) & (ii))
Option B is correct.
$|{\overrightarrow{V}}_C-{\overrightarrow{V}}_A|=2\omega R$ and $|{\overrightarrow{V}}_B-{\overrightarrow{V}}_C|=\omega R$
So, $|{\overrightarrow{V}}_C-{\overrightarrow{V}}_A|=2|{\overrightarrow{V}}_B-{\overrightarrow{V}}_C|$
Option C is correct.

Given, rolling without slipping
So, $V_A=0$
$V_B-\omega R=0$
$V_B=\omega R$
$|{\overrightarrow{V}}_B|=\omega R...\left(v\right)$
Thus $|{\overrightarrow{V}}_C-{\overrightarrow{V}}_A|\ne 4|{\overrightarrow{V}}_B|$
Option D is incorrect.