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Question

(B) The relationship between linear velocity $\vec{v}$,angular velocity $\vec{\omega}$,and position vector $\vec{r}$ for a particle in circular motion

Vedclass · Accepted Answer

$\frac{13 \hat{i} - 29 \hat{j} - 31 \hat{k}}{146}$

Vedclass · Answer

(B) The relationship between linear velocity $\vec{v}$,angular velocity $\vec{\omega}$,and position vector $\vec{r}$ for a particle in circular motion is given by $\vec{v} = \vec{\omega} 	imes \vec{r}$.Since $\vec{\omega}$ is perpendicular to both $\vec{r}$ and $\vec{v}$,we can use the relation $\vec{\omega} = \frac{\vec{r} 	imes \vec{v}}{|\vec{r}|^2}$.Given $\vec{r} = \hat{i} + 9 \hat{j} - 8 \hat{k}$ and $\vec{v} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$.First,calculate the cross product $\vec{r} 	imes \vec{v}$:$\vec{r} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 9 & -8 \ 3 & -4 & 5 \end{vmatrix} = \hat{i}(45 - 32) - \hat{j}(5 - (-24)) + \hat{k}(-4 - 27) = 13 \hat{i} - 29 \hat{j} - 31 \hat{k}$.Next,calculate $|\vec{r}|^2 = 1^2 + 9^2 + (-8)^2 = 1 + 81 + 64 = 146$.Therefore,$\vec{\omega} = \frac{13 \hat{i} - 29 \hat{j} - 31 \hat{k}}{146}$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

At a given instant of time,the position vector of a particle moving in a circle with a velocity $\vec{v} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$ is $\vec{r} = \hat{i} + 9 \hat{j} - 8 \hat{k}$. Its angular velocity $\vec{\omega}$ at that time is:

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