Question

15.
If the position vector of a particle is r = (3i +4j) meter and its angular velocity is w = (j+ 2K)
rad/sec then its linear velocity is (in m/s)
(a) (8i – 6j +3k) (b) (3i + 6j +8k) (C) –(3i +6j + 6k) (d) (6i + 8j + 3k)
Parallel​

Answer 1

Hi there,

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(Fair Deal - pls mark me brainliest)

letters v, r and ω in bold are vectors:

Vector formula:-

v = ω x r  [cross product is involved]

where v = linear velocity

          ω = angular velocity

           r = radius

now.... here I have represented the vectors in the matrice form where the first row indiactes i direction, second row j and so on.......

so... given

r =$\left[\begin{array}{ccc}3\\4\\0\end{array}\right]$ m

and

ω =$\left[\begin{array}{ccc}0\\1\\2\end{array}\right]$ rad/sec

and now using formula we get

v =$\left[\begin{array}{ccc}0\\1\\2\end{array}\right]$x$\left[\begin{array}{ccc}3\\4\\0\end{array}\right]$ m/s

and solving the cross product

    | i        j          k |

ω  | 0      1          2 |        =        -8i + 6j - 3k

r   | 3      4         0 |

option a) -ve

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cheers :)