Question

show that the curl of the velocity of any particle of a rigid body is equal to twice the angular velocity of the body.

Answer 1

Answer:

the vectors quantity

Explanation:

show that the curl of the velocity of any particle of a rigid body is equal to twice the angular velocity of the body.

Answer 2

SOLUTION

TO PROVE

The curl of the velocity of any particle of a rigid body is equal to twice the angular velocity of the body.

PROOF

We have

$\vec{v} = \vec{ \omega} \times \vec{r}$

Now

$curl \: \vec{v}$

$= \vec{ \nabla} \times \vec{v}$

$= \vec{ \nabla} \times ( \vec{ \omega} \times \vec{r})$

$= \vec{ \nabla} \times \displaystyle \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \\ \omega_1 & \omega_2 & \omega_3 \\ \\ x & y & z \end{vmatrix}$

$= \vec{ \nabla} \times \bigg[( \omega_2z - \omega_3y)\hat{i} + ( \omega_3x - \omega_1z) \hat{j} +( \omega_1y - \omega_2x) \hat{k} \bigg]$

$= \displaystyle \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \\ \frac{ \partial }{ \partial x} & \frac{ \partial }{ \partial y} & \frac{ \partial }{ \partial z} \\ \\ ( \omega_2z - \omega_3y) & ( \omega_3x - \omega_1z) &( \omega_1y - \omega_2x) \end{vmatrix}$

$= 2\bigg[ \omega_1\hat{i} + \omega_2 \hat{j} +\omega_3 \hat{k} \bigg]$

$= 2 \: \vec{\omega}$

Hence proved

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. write the expression for gradient and divergence

https://brainly.in/question/32412615

2. prove that the curl of the gradient of

(scalar function) is zero

https://brainly.in/question/19412715