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183_notes:examples:rotational_angular_momentum_of_a_bicycle_wheel [2014/11/20 07:06] – pwirving | 183_notes:examples:rotational_angular_momentum_of_a_bicycle_wheel [2014/11/20 16:32] (current) – pwirving | ||
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=== Approximations & Assumptions === | === Approximations & Assumptions === | ||
+ | No friction in the bearings therefore angular speed is constant | ||
+ | Ignore the spokes of the bicycle | ||
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Equation for moments of inertia for a hoop: $I=MR^{2}$ | Equation for moments of inertia for a hoop: $I=MR^{2}$ | ||
- | $\omega = \frac{2\pi}/{T}$ | + | $\omega = \frac{2\pi}{T}$ |
+ | $\vec{L}_{rot} = I \vec{\omega}$ | ||
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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: | ||
$\vec{L}_{rot} = \langle 0, -0.69, 0 \rangle kg \cdot m^2/s$ | $\vec{L}_{rot} = \langle 0, -0.69, 0 \rangle kg \cdot m^2/s$ |