Q. A particle moves in a circle with a uniform speed. When it goes from a point $A$ to a diametrically opposite point $B$, the momentum of the particle changes by ${\overrightarrow{\mathbf{p}}}_{A} - {\overrightarrow{\mathbf{p}}}_{B} = 2\ kg\ ms^{- 1}(\widehat{\mathbf{j}})$ and the centripetal fore acting on it changes by ${\overrightarrow{\mathbf{F}}}_{A} - {\overrightarrow{\mathbf{F}}}_{B} = 8$ N$\left( \widehat{\mathbf{i}} \right)$ where $\widehat{\mathbf{i}}$ and $\widehat{\mathbf{j}}$ are unit vectors along $X$ and $Y$ axes respectively. The angular velocity of the particle is
A) Dependent of its mass
B) 4 $rad\ s^{- 1}$
C) $\frac{2}{\pi}$ $rad\ s^{- 1}$
D) $16\ \mu$ rad $s^{- 1}$
Explanation
(b) ${\overrightarrow{\mathbf{p}}}_{B} - {\overrightarrow{\mathbf{p}}}_{A} = m\left( {\overrightarrow{\mathbf{v}}}_{B} - {\overrightarrow{\mathbf{v}}}_{A} \right) = mv(\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{i}})$
$= 2mv\widehat{\mathbf{j}} = 2kg\ ms^{- 1}$ …(i)
${\overrightarrow{\mathbf{F}}}_{B} - {\overrightarrow{\mathbf{F}}}_{A} = \frac{mv^{2}}{R}\left( - \widehat{\mathbf{i}} \right) - \frac{mv^{2}}{R}( + \widehat{\mathbf{i}})$
$= \frac{2mv^{2}}{R}\left( - \widehat{\mathbf{i}} \right) = 8N$ …(ii)
Dividing Eq. (ii) by Eq. (i), $\frac{v}{R} = \omega = 4$ rad $s^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.