ece105: add rolling motion
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@ -153,3 +153,44 @@ This is the same as linear momentum.
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$$\vec L = \vec r\times\vec p$$
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$$\vec L = I\vec\omega$$
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$$\vec L =\vec\tau t$$
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## Rolling motion
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!!! definition
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- **Slipping** is spinning without sliding.
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- **Skidding** is sliding without spinning.
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Pure rolling motion is **only true if** the tangential velocity of the centre of mass is equal to its rotational velocity.
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$$v_{cm}=R\omega$$
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In pure rolling motion, the point at the top is moving at two times the velocity while the point at the bottom has no tangential velocity.
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<img src="https://upload.wikimedia.org/wikipedia/commons/8/8d/Velocitats_Roda.svg" width=500>(Source: Wikimedia Commons)</img>
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For any point in the mass:
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$$
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v_{net} = v_{trans} + v_{rot} \\
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v_{net} = v_{cm} + \vec R \times\vec\omega \\
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E_{k roll} = E_{k trans} + E_{k rot}
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$$
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Alternatively, the rolling can be considered as a rotation about the pivot point between the disk and the ground, allowing rolling motion to be represented as rotational motion around the pivot point. The **parallel axis theorem** can be used to return it back to the original point.
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$$\sum\tau_b=I_b\alpha$$
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At least one external torque and one external force is required to initiate pure rolling motion because the two components are separate.
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If an object is purely rolling and then it moves to:
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- a flat, frictionless surface, it continues purely rolling
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- an inclined, frictionless surface, external torque is needed to maintain pure rolling motion
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- an inclined surface with friction, it continues purely rolling
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Where $c$ is the coefficient to the moment of inertia ($I=cMR^2$), while rolling down an incline:
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$$
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v_{cm} = \sqrt{\frac{2}{1+c}gh} \\
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a_{cm} = \frac{g\sin\theta}{1+c}
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$$
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