A 2.0-kg block travels around a 0.50-m radius circle with an angular speed of 12 rad/s.The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin.The component in the xy plane of the angular momentum around the origin has magnitude: A)0 kg∙m²/s B)6.0 kg∙m²/s C)9.0 kg∙m²/s D)11 kg∙m²/s E)14 kg∙m²/s

Question

Quizplus · Accepted Answer

The angular momentum L in the xy plane is given by L = Iω, where I is the moment of inertia and ω is the angular speed. For a point mass, I = mr². Here, m = 2.0 kg, r = 0.50 m, and ω = 12 rad/s. So, L = (2.0 kg)(0.50 m)²(12 rad/s) = 6.0 kg∙m²/s. However, since the circle is parallel to the xy plane, the component of angular momentum in the xy plane is perpendicular to the radius vector, resulting in L = 9.0 kg∙m²/s.

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