Question

If angular velocity of a particle having position vector
r = (2i – 2j - 2k) m about the origin is
omega = (2i – 2j - k) rad/s then magnitude of linear
velocity of the particle will be
6√2 m/s
4√2 m/s
2√2 m/s
4√3 m/s​

Answer 1

Answer:6√2 m/s

Explanation:

Just find out the cross product of (2î -2j - k ) × (2î-2j + 2k )

Answer 2

Given:

Angular velocity of a particle having position vector r = (2i – 2j - 2k) m about the origin is omega = (2i – 2j - k) rad/s.

To find:

Linear velocity of particle?

Calculation:

In vector notation, we can say:

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

$\rm \implies\vec{v} = ( 2 \hat{i} - 2 \hat{j} - 2 \hat{k})\times(2 \hat{i} - 2 \hat{j} - \hat{k})$

$\rm\implies\vec{v}=\left[\begin{array}{ccc}i&j&k\\2&-2&-2\\2&-2&-1\end{array}\right]$

$\rm \implies \vec{v} = - 2 \hat{i} - 2 \hat{j} + 0 \hat{k}$

$\rm \implies | \vec{v}| = \sqrt{ {( - 2)}^{2} + {( - 2)}^{2} + {(0)}^{2} }$

$\rm \implies | \vec{v}| = \sqrt{ {4 + 4} }$

$\rm \implies | \vec{v}| = \sqrt{8}$

$\rm \implies | \vec{v}| = 2 \sqrt{2} \: m/s$

So, linear velocity is 2√2 m/s