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One end of a thin rod is attached to a pivot about which it can rotate without friction. Air resistance is absent. The rod has a length of 0.48 m and is uniform. It is hanging vertically straight downward. The end of the rod nearest the floor is given a linear speed V so that the rod begins to rotate upward about the pivot. What must be the value of V such that the rod comes to a momentary halt in a straight up orientation, exactly opposite to its initial orientation?

Sagot :

ANSWER:

5.32 m/s

STEP-BY-STEP EXPLANATION:

We can determine the velocity from the following equation (initial kinetic energy:

[tex]\frac{1}{6}\cdot v^2_0=g\cdot L[/tex]

Where L is the length of the rod and g would be gravity, we replace and calculate the value of the initial velocity:

[tex]\begin{gathered} v^2_0=6\cdot9.81\cdot0.48 \\ v^{}_0=\sqrt[]{28.2528} \\ v_0=5.32\text{ m/s} \end{gathered}[/tex]

The velocity is 5.32 m/s

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