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| 183_notes:examples:rotational_angular_momentum_of_a_bicycle_wheel [2014/11/16 21:41] – pwirving | 183_notes:examples:rotational_angular_momentum_of_a_bicycle_wheel [2014/11/20 07:17] – pwirving | ||
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| === Facts === | === 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 === | === Lacking === | ||
| + | The rotational angular momentum of the wheel | ||
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| === Representations === | === Representations === | ||
| + | Equation for moments of inertia for a hoop: $I=MR^{2}$ | ||
| + | $\omega = \frac{2\pi}{T}$ | ||
| + | $\vec{L}_{rot} = I \vec{\omega}$ | ||
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| === Solution === | === Solution === | ||
| - | The direction of $\vec{\omega}$ is -y. | + | 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. | ||
| $I = MR^{2} = (0.8kg)(0.32m)^2 = 0.082 kg \cdot m^2$ | $I = MR^{2} = (0.8kg)(0.32m)^2 = 0.082 kg \cdot m^2$ | ||
| + | |||
| + | We can find $\omega$ because we know that one revolution is equal to $2\pi$ and that this revolution is completed in 0.75seconds. | ||
| $\omega = \frac{2\pi}{0.75s} = 8.38 s^{-1}$ | $\omega = \frac{2\pi}{0.75s} = 8.38 s^{-1}$ | ||
| + | |||
| + | We now have values for $I$ and $\omega$ and can find the rotational velocity by filling into $\vec{L}_{rot} = I \vec{\omega}$ | ||
| $\mid\vec{L}_{rot}\mid$ = $(0.082 kg \cdot m^2)(8.38 s^{-1}) = 0.69 kg \cdot m^2/s$ | $\mid\vec{L}_{rot}\mid$ = $(0.082 kg \cdot m^2)(8.38 s^{-1}) = 0.69 kg \cdot m^2/s$ | ||
| + | |||
| + | Therefore the rotational angular momentum is equal to: | ||
| $\vec{L}_{rot} = \langle 0, -0.69, 0 \rangle kg \cdot m^2/s$ | $\vec{L}_{rot} = \langle 0, -0.69, 0 \rangle kg \cdot m^2/s$ | ||