If position vector of a particle is mathop rlimits^ to = (3hat{i} + 4hat{j}) meter and i

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3-2.Motion in Plane

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If the position vector of a particle is $\mathop r\limits^ \to = (3\hat i + 4\hat j)$ $meter$ and its angular velocity is $\mathop \omega \limits^ \to = (\hat j + 2\hat k)$ $rad/sec$ then its linear velocity is (in $m/s$)

$(8\hat i - 6\hat j + 3\hat k)$

$(3\hat i + 6\hat j + 8\hat k)$

$ - (3\hat i + 6\hat j + 6\hat k)$

$(6\hat i + 8\hat j + 3\hat k)$

Solution

$\overrightarrow {v\,} = \overrightarrow \omega \times \overrightarrow {r\,} $

$(3\hat i + 4\hat j + 0\hat k) \times (0\hat i + \hat j + 2\hat k)$ $\overrightarrow v = \left| {\,\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\3&4&0\\0&1&2\end{array}\,} \right|\, = \,8\hat i – 6\hat j + 3\hat k$

Standard 11

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(AIEEE-2010)

If the position vector of a particle is $\mathop r\limits^ \to = (3\hat i + 4\hat j)$ $meter$ and its angular velocity is $\mathop \omega \limits^ \to = (\hat j + 2\hat k)$ $rad/sec$ then its linear velocity is (in $m/s$)

Solution

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