Q122 / 160 · TS EAMCET 2005 · Physics

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

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

Solution

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

Hence, the correct answer is option: 'B'.

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