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.

First, we need to find the moment of inertia of the block. Since the circle is parallel to the xy plane, we can use the formula for the moment of inertia of a thin hoop: I = 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 kg)(0.50 m)^2 = 0.50 kg·m^2

Next, we can use the given angular speed to find ω:

ω = 12 rad/s

Then, we can use the formula for angular momentum to find L:

L = Iω = (0.50 kg·m^2)(12 rad/s) = 6 kg·m^2/s

Finally, to find the magnitude of L, we just take the absolute value:

|L| = 6 kg·m^2/s

Thus, the best choice is C: 11 kg·m^2/s.

As a 20-Kg Block Travels Around a 0