Question

A thin, hollow sphere of radius

r = 0.550 m
and mass
m = 16.0 kg
turns counterclockwise about a vertical axis through its center (when viewed from above), at an angular speed of 2.90 rad/s. What is its vector angular momentum about this axis? (Enter the magnitude in kg · m2/s.)
magnitude?

Answer 1

The magnitude of the angular momentum of the hollow sphere about the given axis is approximately 15.672 kg · m^2/s.

The angular momentum (L) of an object rotating about a fixed axis is given by the formula:

L = I * ω

Where:

L is the angular momentum vector,

I is the moment of inertia of the object,

ω is the angular velocity vector.

For a hollow sphere, the moment of inertia (I) is given by the formula:

I = (2/3) * m * r^2

Let's calculate the magnitude of the angular momentum using the given values:

r = 0.550 m (radius)

m = 16.0 kg (mass)

ω = 2.90 rad/s (angular velocity)

First, calculate the moment of inertia (I):

I = (2/3) * m * r^2

I = (2/3) * 16.0 kg * (0.550 m)^2

I = 2.7088 kg * m^2

Now, calculate the magnitude of the angular momentum (L):

L = I * ω

L = 2.7088 kg * m^2 * 2.90 rad/s

L ≈ 15.672 kg · m^2/s

Therefore, the magnitude of the angular momentum of the hollow sphere about the given axis is approximately 15.672 kg · m^2/s.