A rigid sculpture, consisting of thin hoop (of mass $m$ and radius $R = 0.15 m$ ) and two thin rods (each of mass $m$ and length $L = 2.0 R$ ), is arranged as shown in figure. The sculpture can be pivot around a horizontal axis in the plane of the hoop, passing through its centre. Starting from rest, the sculpture rotates around the rotation axis from the initial upright orientation of figure. What is its angular speed about the axis when it is inverted?

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$12 rad / s$

$14 rad / s$

$8 rad / s$

$9 . 5 rad / s$

answer is A.

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Detailed Solution

In the process mechanical energy remains constant. Therefore

$K_{i} + U_{i} = K_{f} + U_{f} 0 + [m g R + m g (R + R)] = \frac{1}{2} I ω^{2} + [- m g R - m g (R + R)] H e r e I = \frac{1}{2} m R^{2} + m {(2 R)}^{2} + [\frac{1}{12} m {(2 R)}^{2} + m {(2 R)}^{2}] = \frac{53}{6} m R^{2} After substituting the values and solving, we get ω = 9.5 rad / s$