Question

A uniform solid sphere of radius r = 0.490 m and mass m = 13.5 kg turns counterclockwise about a vertical axis through its center (when viewed from above). Find its vector angular momentum about this axis when its angular speed is 2.9 rad/s.

Answer 1

Answer:

Vector angular momentum about this axis of the sphere is:

L= 3.76$\hat k$ kg-m²/sec

Explanation:

The formula for the moment of inertia of a sphere is:

$I=\frac{2}{5}\times MR^2$

Given:

Mass of the sphere = 13.5 kg

Radius of the sphere = 0.490 m

Thus, moment of inertia :

$I=\frac{2}{5}\times 13.5\times (0.490)^2 kg\ m^2$

$I=1.29654 kg\ m^2$

The expression for the angular momentum is:

L=I×ω

Given:

Angular speed(ω) = 2.9 rad/s

I, above calculated = 1.29654 kgm⁻²

Thus, angular momentum is:

L= 1.29654×2.9 kg-m²/sec

L= 3.76 kg-m²/sec

Given, the sphere is turning counterclockwise about the vertical axis. Thus, the direction of the angular momentum will be on the upper side of the plane. ( $+\hat k$ ).

Thus, angular momentum with direction is:

L= 3.76$\hat k$ kg-m²/sec