For a particle,the angular velocity is vec{om | vedclass

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Question

(A) The linear velocity $\vec{v}$ is given by the cross product of angular velocity $\vec{\omega}$ and position vector $\vec{r}$:$\vec{v} = \vec{\omeg

Vedclass · Accepted Answer

$-5\hat{i} + 2\hat{j} + 3\hat{k}$

Vedclass · Answer

(A) The linear velocity $\vec{v}$ is given by the cross product of angular velocity $\vec{\omega}$ and position vector $\vec{r}$:$\vec{v} = \vec{\omega} 	imes \vec{r}$$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 1 & 1 & 1 \end{vmatrix}$Expanding the determinant:$\vec{v} = \hat{i}((-2)(1) - (3)(1)) - \hat{j}((1)(1) - (3)(1)) + \hat{k}((1)(1) - (-2)(1))$$\vec{v} = \hat{i}(-2 - 3) - \hat{j}(1 - 3) + \hat{k}(1 + 2)$$\vec{v} = -5\hat{i} - \hat{j}(-2) + 3\hat{k}$$\vec{v} = -5\hat{i} + 2\hat{j} + 3\hat{k}$

For a particle,the angular velocity is $\vec{\omega} = \hat{i} - 2\hat{j} + 3\hat{k}$ and the position vector is $\vec{r} = \hat{i} + \hat{j} + \hat{k}$. What is the linear velocity $\vec{v}$?

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