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| 183_notes:curving_motion [2014/09/24 17:23] – [A change in direction] 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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| $$\dfrac{\Delta \hat{p}}{\Delta t} = \dfrac{\hat{p}_f-\hat{p}_i}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta \rangle-\langle 0, 1 \rangle}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta - 1\rangle}{\Delta t}$$ | $$\dfrac{\Delta \hat{p}}{\Delta t} = \dfrac{\hat{p}_f-\hat{p}_i}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta \rangle-\langle 0, 1 \rangle}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta - 1\rangle}{\Delta t}$$ |
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| $$\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{v}|}{R} \hat{n}$$ | $$\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{v}|}{R} \hat{n}$$ |
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| | [{{183_notes:mi3e_05-026.jpg?250|The perpendicular component of the net force points inward towards the turn.}}] |
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| where the unit vector, $\hat{n}$, always points inward towards the turn. For more general trajectories, the value of $R$ is the radius of curvature of the arc, that is, it is the radius of the circle that has exactly the same curvature at the location of interest. | where the unit vector, $\hat{n}$, always points inward towards the turn. For more general trajectories, the value of $R$ is the radius of curvature of the arc, that is, it is the radius of the circle that has exactly the same curvature at the location of interest. |
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| $$\vec{F}_{net,\perp} = |\vec{p}|\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{p}||\vec{v}|}{R} \hat{n} =\dfrac{mv^2}{R} \hat{n} $$ | $$\vec{F}_{net,\perp} = |\vec{p}|\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{p}||\vec{v}|}{R} \hat{n} =\dfrac{mv^2}{R} \hat{n} $$ |
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| | ==== Relationship to the tangential and centripetal accelerations ==== |
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| | In your previous studies, you might come acres the [[http://en.wikipedia.org/wiki/Acceleration#Tangential_and_centripetal_acceleration|tangential acceleration ($\vec{a}_{t}$) and the centripetal acceleration ($\vec{a}_{c}$)]]. This are directly connected to the definitions of the parallel and perpendicular components of the net force. You can write the net force as the sum of these parallel and perpendicular components, which arise from the tangential and centripetal accelerations. |
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| | $$\vec{F}_{net} = \vec{F}_{\parallel} + \vec{F}_{\perp}$$ |
| | $$\vec{F}_{\parallel} = m\vec{a}_{t} = m{a}_{t}\hat{p} \qquad \vec{F}_{\perp} = m\vec{a}_{c} = m{a}_{c}\hat{n}$$ |
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| | The direction of each of these accelerations is the same as their corresponding forces. The tangential acceleration is tangent to the path, and this points in the $\hat{p}$ direction. The centripetal acceleration is perpendicular to the path and points in the $\hat{n}$ direction. You can use the magnitudes of each force component to determine formulae for the accelerations. |
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| | $$F_{\parallel} = m{a}_{t} = \dfrac{d|\vec{p}|}{dt} = \dfrac{d|m\vec{v}|}{dt} = m\dfrac{d|\vec{v}|}{dt} \longrightarrow {a}_{t} = \dfrac{d|\vec{v}|}{dt}$$ |
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| | The tangential acceleration tells you how the speed of the object changes, just as the parallel component of the net force is responsible for this speeding up and slowing down. |
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| | $$F_{\perp} = m{a}_{c} = \dfrac{|\vec{p}||\vec{v}|}{R} = \dfrac{mv^2}{R} = m\dfrac{v^2}{R} \longrightarrow {a}_{c} = \dfrac{v^2}{R}$$ |
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| | The centripetal acceleration tells you how the direction of the object's motion changes, just as the perpendicular component of the net force is responsible for this directional change. |
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