Example: Rotational Angular Momentum of a Bicycle Wheel

A bicycle wheel has a mass of 0.8kg and a radius of 32cm. If the wheel rotates in the xz plane, spinning clockwise when viewed from the +y axis, and making one full revolution in 0.75 seconds, what is the rotational angular momentum of the wheel?

Facts

Mass of bicycle wheel = 0.8kg.

Bicycle wheel has a radius of 32cm.

Bicycle wheel is spinning clockwise when viewed from the +y axis.

Bicycle wheel rotates in the xz plane.

Bicycle wheel completes one full revolution in 0.75 seconds.

Lacking

The rotational angular momentum of the wheel

Approximations & Assumptions

No friction in the bearings therefore angular speed is constant

Ignore the spokes of the bicycle

Representations

Equation for moments of inertia for a hoop: $I = MR^{2}$

$\omega = \frac{2\pi}{T}$

${\overset{\rightarrow}{L}}_{rot} = I\overset{\rightarrow}{\omega}$

Solution

We know from the right hand rule that because the wheel is moving clockwise in xz plane that the direction of $\overset{\rightarrow}{\omega}$ is -y.

We are trying to find the rotational angular momentum and to do so we must find $I$ and $\overset{\rightarrow}{\omega}$ to fill into the following equation: ${\overset{\rightarrow}{L}}_{rot} = I\overset{\rightarrow}{\omega}$

We can find $I$ by knowing the mass of the wheel and radius of the wheel.

$I = MR^{2} = (0.8kg)(0.32m)^{2} = 0.082kg \cdot m^{2}$

We can find $\omega$ because we know that one revolution is equal to $2\pi$ and that this revolution is completed in 0.75seconds.

$\omega = \frac{2\pi}{0.75s} = 8.38s^{- 1}$

We now have values for $I$ and $\omega$ and can find the rotational velocity by filling into ${\overset{\rightarrow}{L}}_{rot} = I\overset{\rightarrow}{\omega}$

$\mid {\overset{\rightarrow}{L}}_{rot} \mid$ = $(0.082kg \cdot m^{2})(8.38s^{- 1}) = 0.69kg \cdot m^{2}/s$

Therefore the rotational angular momentum is equal to:

${\overset{\rightarrow}{L}}_{rot} = \langle 0, - 0.69,0\rangle kg \cdot m^{2}/s$