As a 2.0-kg block travels around a 0.50-m radius circle it has 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 magnitude of its angular momentum around the origin is: A)6.0 kg∙m²/s B)9.0 kg∙m²/s C)11 kg∙m²/s D)14 kg∙m²/s E)20 kg∙m²/s

Question

Quizplus · Accepted Answer

The formula for angular momentum is L = Iω, where I is the moment of inertia and ω is the angular speed.

We can find the moment of inertia of the block using the formula for the moment of inertia of a uniform solid sphere rotating about an axis passing through its center:

I = (2/5)MR^2

where M is the mass of the block and R is the radius of the circle. Plugging in the given values, we get:

I = (2/5)(2 kg)(0.5 m)^2 = 0.5 kg·m^2

Then, we can calculate the angular momentum using the formula:

L = Iω = (0.5 kg·m^2)(12 rad/s) = 6 kg·m^2/s

However, we need to find the magnitude of the angular momentum around the origin. Since the circle is centered on the z-axis and parallel to the xy-plane, the angular momentum vector is perpendicular to the xy-plane and has a magnitude equal to its z-component:

Lz = Iωz = Iω = (0.5 kg·m^2)(12 rad/s) = 6 kg·m^2/s

Therefore, the magnitude of the angular momentum around the origin is 6 kg·m^2/s, which is closest to choice C).

As a 20-Kg Block Travels Around a 0