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Question

Physics Question on Motion in a plane

What is the value of linear velocity, if angular velocity is 3i^4j^+k^3 \hat i - 4 \hat j + \hat k and distance from the centre is 5i^6j^+6k^5 \hat i - 6 \hat j + 6 \hat k ?

A

6i^+6j^8k^6 \hat i + 6 \hat j - 8 \hat k

B

18i^13j^+2k^-18 \hat i - 13 \hat j + 2 \hat k

C

4i^13j^6k^4 \hat i - 13 \hat j - 6 \hat k

D

6i^2j^+8k^6 \hat i - 2 \hat j + 8 \hat k

Answer

18i^13j^+2k^-18 \hat i - 13 \hat j + 2 \hat k

Explanation

Solution

Linear vector v=ω×r v = \omega \times r
Given, ω=3i^4j^+k^  and  r=5i^6j^+6k^\omega = 3 \hat i - 4 \hat j + \hat k \ \ and \ \ r = 5 \hat i - 6 \hat j + 6 \hat k
v=i^j^k^ 341 566 \therefore v = \begin{vmatrix} \hat i & \hat j & \hat k \\\ 3 & -4 & 1 \\\ 5 & -6 & 6 \\\ \end{vmatrix}
=i^(24+6)j^(185)+k^(18+20)=\hat i (-24 + 6) - \hat j (18 - 5) + \hat k (-18 + 20)
=18i^13j^+2k^= -18 \hat i - 13 \hat j + 2 \hat k