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?

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.

The rotational angular momentum of the wheel

No friction in the bearings therefore angular speed is constant

Ignore the spokes of the bicycle

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

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

$\vec{L}_{rot} = I \vec{\omega}$

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

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

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

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

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

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

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

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

Therefore the rotational angular momentum is equal to:

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