QJEEAdv09Phf99c: JEE Advanced 2009 Physics - Class 11 JEEAdv09Phf99c Solution: A sphere is rolling without slipping on a fixed horizontal…

We have

$\overrightarrow {{V_C}} = \overrightarrow {{V_{CM}}} + \overrightarrow {r\omega } = 2\overrightarrow {{V_{CM}}}$

$\overrightarrow {{V_B}} = \overrightarrow {{V_{CM}}}$

$\overrightarrow {{V_A}} = 0$

Therefore,

$\overrightarrow {{V_C}} - \overrightarrow {{V_B}} = \overrightarrow {r\omega } = \overrightarrow {{V_{CM}}}$

$\overrightarrow {{V_B}} - \overrightarrow {{V_A}} = \overrightarrow {{V_{CM}}}$

$\overrightarrow {{V_C}} - \overrightarrow {{V_A}} = \overrightarrow {{V_{CM}}} + \overrightarrow {r\omega } = 2\overrightarrow {{V_{CM}}}$

$\overrightarrow {{V_B}} - \overrightarrow {{V_C}} = - \overrightarrow {r\omega }$

For pure rolling, we have

$|\overrightarrow {{V_{CM}}} | = |\overrightarrow {r\omega } |$

Therefore,

$|\overrightarrow {{V_C}} - \overrightarrow {{V_A}} | = 2{V_{CM}}$

$\Rightarrow 2|\overrightarrow {{V_B}} - \overrightarrow {{V_C}} | = 2r\omega = 2{V_{CM}}$