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| 183_notes:curving_motion [2014/09/24 18:02] – [Relationship to the tangential and centripetal accelerations] caballero | 183_notes:curving_motion [2014/09/29 19:32] – pwirving |
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| [{{183_notes:phat_circle.png?200|While the magnitude of the momentum changes, the length of a unit vector is always 1. }}] | [{{183_notes:phat_circle.png?200|While the magnitude of the momentum changes, the length of a unit vector is always 1. }}] |
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| As [[183_notes:ucm|you have seen before]], you can track the angle $\theta$ through which the car moves on this circular track to determine how the direction is changing. The challenging part is determining the where the angle $\theta$ is measured from for the final momentum direction. In the figure below and to the left, you can see that the initial momentum unit vector is in the $+y$-direction ($\hat{p}_i = \langle 0, 1 \rangle$) and the final momentum unit vector has components in both the $-x$-direction and the $+y$-direction ($\hat{p}_f = \langle -\sin \theta, \cos \theta \rangle$). In this case, you are trying to determine the change in the direction of the unit vector over this interval. | As [[183_notes:ucm|you have seen before]], you can track the angle $\theta$ through which the car moves on this circular track to determine how the direction is changing. The challenging part is determining where the angle $\theta$ is measured from for the final momentum direction. In the figure below and to the left, you can see that the initial momentum unit vector is in the $+y$-direction ($\hat{p}_i = \langle 0, 1 \rangle$) and the final momentum unit vector has components in both the $-x$-direction and the $+y$-direction ($\hat{p}_f = \langle -\sin \theta, \cos \theta \rangle$). In this case, you are trying to determine the change in the direction of the unit vector over this interval. |
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| $$\dfrac{\Delta \hat{p}}{\Delta t} = \dfrac{\hat{p}_f-\hat{p}_i}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta \rangle-\langle 0, 1 \rangle}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta - 1\rangle}{\Delta t}$$ | $$\dfrac{\Delta \hat{p}}{\Delta t} = \dfrac{\hat{p}_f-\hat{p}_i}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta \rangle-\langle 0, 1 \rangle}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta - 1\rangle}{\Delta t}$$ |