183_notes:examples:rotational_angular_momentum_of_a_bicycle_wheel [Projects & Practices in Physics]

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===== 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: $I=MR^{2}$

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

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



=== Solution ===

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:

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