Rotating Rigid Body

Geometry Level 5

Mayank rotates a rigid body about an axis passing through the point ( 3 , 1 , 2 ) (3,-1,-2) .

Akul finds that the particle at the point ( 4 , 1 , 0 ) (4,1,0) has the velocity 4 i 4 j + 2 k 4i-4j+2k and that at the point ( 3 , 2 , 1 ) (3,2,1) has the velocity 6 i 4 j + 4 k 6i-4j+4k .

Find the magnitude of the angular velocity of the body.

Wanna have more fun with Mayank and Akul. This question is a part of the set Mayank and Akul

3 4 1 5 2

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1 solution

Mark Hennings
May 13, 2019

We need ω × ( 1 2 2 ) = ( 4 4 2 ) ω × ( 0 3 3 ) = ( 6 4 4 ) \mathbf{\omega} \times \left(\begin{array}{c}1 \\ 2 \\ 2 \end{array}\right) \; = \; \left(\begin{array}{c} 4 \\ -4 \\ 2 \end{array}\right) \hspace{2cm} \mathbf{\omega} \times \left(\begin{array}{c} 0 \\ 3 \\ 3 \end{array}\right) \; = \; \left(\begin{array}{c} 6 \\ -4 \\ 4 \end{array}\right) and hence ω × ( 1 0 0 ) = ( 0 4 3 2 3 ) ω × ( 0 1 1 ) = ( 2 4 3 4 3 ) \omega \times \left(\begin{array}{c}1 \\ 0 \\ 0 \end{array}\right) \; = \; \left(\begin{array}{c} 0 \\ -\frac43 \\ -\frac23 \end{array}\right) \hspace{2cm} \omega \times \left(\begin{array}{c} 0 \\ 1 \\ 1 \end{array}\right) \; = \; \left(\begin{array}{c} 2 \\ -\frac43 \\ \frac43 \end{array}\right) Thus ( 0 ω 2 ω 3 ) = ω ( ω i ) i = i × ( ω × i ) = ( 1 0 0 ) × ( 0 4 3 2 3 ) = ( 0 2 3 4 3 ) \left(\begin{array}{c} 0 \\ \omega_2 \\ \omega_3\end{array}\right) \; = \; \omega - \left(\omega \cdot \mathbf{i}\right) \mathbf{i} \; = \; \mathbf{i} \times (\mathbf{\omega} \times \mathbf{i}) \; = \; \left(\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right) \times \left(\begin{array}{c} 0 \\ -\frac43 \\ -\frac23 \end{array}\right) \; = \; \left( \begin{array}{c} 0 \\ \frac23 \\ -\frac43 \end{array}\right) and hence ( 2 4 3 4 3 ) = ω × ( 0 1 1 ) = ( ω 1 2 3 4 3 ) × ( 0 1 1 ) = ( 2 ω 1 ω 1 ) \left(\begin{array}{c} 2 \\ -\frac43 \\ \frac43 \end{array}\right) \; = \; \mathbf{\omega} \times \left(\begin{array}{c} 0 \\ 1 \\ 1 \end{array}\right) \; = \; \left(\begin{array}{c} \omega_1 \\ \frac23 \\ -\frac43 \end{array}\right) \times \left(\begin{array}{c} 0 \\ 1 \\ 1 \end{array}\right) \; = \; \left(\begin{array}{c} 2 \\ -\omega_1 \\ \omega_1 \end{array}\right) so that ω 1 = 4 3 \omega_1 = \frac43 and hence ω = ( 4 3 2 3 4 3 ) \mathbf{\omega} \; = \; \left(\begin{array}{c} \frac43 \\ \frac23 \\ -\frac43 \end{array} \right) which has magnitude 2 \boxed{2} .

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