Physics, asked by indiangamers98675, 1 month ago

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​

Answers

Answered by suryakantighosh2002
1

Answer:6√2 m/s

Explanation:

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

Answered by nirman95
5

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

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