Question

What is the magnitude of the angular momentum (in kgm2/s) of a 40 g golf ball flying through the air and spinning at 4300 rpm after having been hit? The golf ball can be modelled as a uniformly dense sphere with a radius of 2.5 cm.

Answer 1

Answer:

$L=0.0045\ kg-m^2/s$

Explanation:

Given that,

The mass of a golf ball, m = 40 g = 0.04 kg

Its angular velocity, $\omega=4300\ rpm=450.29\ rad/s$

The radius of the sphere is 2.5 cm or 0.025 m

We need to find the magnitude of the angular momentum of the ball. It is given by the formula as follows:

$L=I\omega$

Where I is moment of inertia

For sphere, $I=\dfrac{2}{5}mr^2$

$L=\dfrac{2}{5}mr^2\omega\\\\L=\dfrac{2}{5}\times 0.04\times (0.025)^2\times 450.29\\\\L=0.0045\ kg-m^2/s$

So, the magnitude of the angular momentum of the sphere is $0.0045\ kg-m^2/s$.