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