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At a given instant of time the position vector of a particle moving in a circle with a velocity $3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ is $\hat{\mathbf{i}}+9 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$. Its angular velocity at that time is
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The correct answer is:
$\frac{(13 \hat{\mathbf{i}}-29 \hat{\mathbf{j}}-31 \hat{\mathbf{k}})}{146}$
Angular momentum,
$\begin{aligned}
& \mathbf{L}=m \mathbf{r} \times \mathbf{v} \\
& \text { but } \quad \mathbf{L}=I \omega \\
& \therefore \quad m r^2 w=m \mathbf{r} \times \mathbf{v} \\
& \omega=\frac{\mathbf{r} \times \mathbf{v}}{r^2}=\frac{\mathbf{r} \times \mathbf{v}}{|\mathbf{r}|^2} \\
& \mathbf{r}=\hat{\mathbf{i}}+9 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}, \mathbf{v}=3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} \\
& \mathbf{r} \times \mathbf{v}=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & 9 & -8 \\
3 & -4 & 5
\end{array}\right| \\
& =13 \hat{\mathbf{i}}-29 \hat{\mathbf{j}}-31 \hat{\mathbf{k}} \\
& \therefore \quad \omega=\frac{13 \hat{\mathbf{i}}-29 \hat{\mathbf{j}}-31 \hat{\mathbf{k}}}{\left[\sqrt{1^2+9^2+(-8)^2}\right]^2} \\
& =\frac{13 \hat{\mathbf{i}}-29 \hat{\mathbf{j}}-31 \hat{\mathbf{k}}}{146} \\
&
\end{aligned}$
$\begin{aligned}
& \mathbf{L}=m \mathbf{r} \times \mathbf{v} \\
& \text { but } \quad \mathbf{L}=I \omega \\
& \therefore \quad m r^2 w=m \mathbf{r} \times \mathbf{v} \\
& \omega=\frac{\mathbf{r} \times \mathbf{v}}{r^2}=\frac{\mathbf{r} \times \mathbf{v}}{|\mathbf{r}|^2} \\
& \mathbf{r}=\hat{\mathbf{i}}+9 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}, \mathbf{v}=3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} \\
& \mathbf{r} \times \mathbf{v}=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & 9 & -8 \\
3 & -4 & 5
\end{array}\right| \\
& =13 \hat{\mathbf{i}}-29 \hat{\mathbf{j}}-31 \hat{\mathbf{k}} \\
& \therefore \quad \omega=\frac{13 \hat{\mathbf{i}}-29 \hat{\mathbf{j}}-31 \hat{\mathbf{k}}}{\left[\sqrt{1^2+9^2+(-8)^2}\right]^2} \\
& =\frac{13 \hat{\mathbf{i}}-29 \hat{\mathbf{j}}-31 \hat{\mathbf{k}}}{146} \\
&
\end{aligned}$
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