Question

At a given instant of time the position vector of a particle moving in a circle with a velocity $3 \hat{ i }-4 \hat{ j }+5 \hat{ k }$ is $\hat{ i }+9 \hat{ j }-3 \hat{ k }$. Its angular velocity at that time is

• A. $\frac{(13 \hat{ i }+29 \hat{ j }-31 \hat{ k })}{\sqrt{146}}$
• B. $\frac{(13 \hat{ i }-29 \hat{ j }-31 \hat{ k })}{146}$
• C. $\frac{(13 \hat{ i }+29 \hat{ j }-31 \hat{ k })}{146}$
• D. $\frac{(13 \hat{ i }+29 \hat{ j }+31 \hat{ k })}{146}$

Answer 1

Answer: (B)

Angular momentum,
$L =m r \times v$
but $L =I \omega $
$\therefore m r^{2} \omega =m r \times v$
$\omega=\frac{ r \times v }{r^{2}}=\frac{ r \times v }{| r |^{2}} $
$r =\hat{ i }+9 \hat{ j }-8 \hat{ k }, v =3 \hat{ i }-4 \hat{ j }+5 \hat{ k }$
$r \times v =\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 9 & -8 \\ 3 & -4 & 5\end{vmatrix}$
$=13 \hat{ i }-29 \hat{ j }-31 \hat{ k }$
$\therefore \omega =\frac{13 \hat{ i }-29 \hat{ j }-31 \hat{ k }}{\left[\sqrt{1^{2}+9^{2}+(-8)^{2}}\right]^{2}} $
$=\frac{13 \hat{ i }-29 \hat{ j }-31 \hat{ k }}{146} $