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Parametric Differentiation

The velocitie...

Question

The velocities of the ends ' $A$ ' and ' $B$ " of the rod, namely $V_{A}$ and $V_{B}$ , respectively.

v_{A} = \frac{5}{32} v_{0}; v_{B} = \frac{1}{32} v_{0}

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v_{A} = \frac{7}{32} v_{0}; v_{B} = \frac{3}{32} v_{0}

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v_{A} = \frac{7}{32} v_{0}; v_{B} = \frac{1}{32} v_{0}

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v_{A} = \frac{5}{32} v_{0}; v_{B} = \frac{3}{32} v_{0}

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Solution

The correct option is C $v_{A} = \frac{7}{32} v_{0}; v_{B} = \frac{1}{32} v_{0}$
$V_{A} = \frac{v_{o}}{8} + w (O A)$
$V_{A} = \frac{v_{o}}{8} + \frac{3 v_{o}}{128 l} \times 4 l$

$V_{A} = \frac{7}{32} v_{o}$

$V_{B} = \frac{v_{o}}{8} - w (O B)$
$V_{B} = \frac{v_{o}}{8} - \frac{3 v_{o}}{128 l} \times 4 l$

$V_{B} = \frac{1}{32} v_{o}$

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The velocities of the ends 'A' and 'B" of the rod, namely VA and VB, respectively.

The correct option is C vA=732v0;vB=132v0VA=vo8+w(OA)VA=vo8+3vo128l×4lVA=732voVB=vo8−w(OB)VB=vo8−3vo128l×4lVB=132vo

The velocities of the ends ' $A$ ' and ' $B$ " of the rod, namely $V_{A}$ and $V_{B}$ , respectively.