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183_notes:curving_motion [2014/09/24 17:39] – [A change in direction] caballero | 183_notes:curving_motion [2015/09/27 15:50] – [Relationship to the tangential and centripetal accelerations] caballero |
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The motion of objects is not limited to [[183_notes:displacement_and_velocity|straight line motion]]. As you read earlier, [[183_notes:momentum_principle|forces can change the momentum of objects]] (including the direction of that momentum). These interactions can produce [[183_notes:localg|projectile motion]], [[183_notes:ucm|circular motion]], [[183_notes:springmotion|oscillations]], or more generalized trajectories. In these notes, you will read about how to model more generalized motion using the [[183_notes:momentum_principle|momentum principle]]. | The motion of objects is not limited to [[183_notes:displacement_and_velocity|straight line motion]]. As you read earlier, [[183_notes:momentum_principle|forces can change the momentum of objects]] (including the direction of that momentum). These interactions can produce [[183_notes:localg|projectile motion]], [[183_notes:ucm|circular motion]], [[183_notes:springmotion|oscillations]], or more generalized trajectories. In these notes, you will read about how to model more generalized motion using the [[183_notes:momentum_principle|momentum principle]]. |
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| ==== Lecture Video ==== |
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| {{youtube>wm2NbUDoAV0?large}} |
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==== The Derivative form of the Momentum Principle ==== | ==== The Derivative form of the Momentum Principle ==== |
[{{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.png?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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[{{183_notes:mi3e_05-026.jpg|The perpendicular component of the net force points inward towards the turn.}}] | |
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The change in the direction of the momentum is the result of the component of the net force that is perpendicular to the direction of motion (momentum). This component is referred to as "F net perpendicular" or $\vec{F}_{net,\perp}$. Some also call this force the "centripetal component of the net force"; it is always perpendicular to the tangent to the trajectory of the object. So, we have a relationship between this force component and the change in the magnitude of the momentum. | The change in the direction of the momentum is the result of the component of the net force that is perpendicular to the direction of motion (momentum). This component is referred to as "F net perpendicular" or $\vec{F}_{net,\perp}$. Some also call this force the "centripetal component of the net force"; it is always perpendicular to the tangent to the trajectory of the object. So, we have a relationship between this force component and the change in the magnitude of the momentum. |
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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 have come across the [[http://en.wikipedia.org/wiki/Acceleration#Tangential_and_centripetal_acceleration|tangential acceleration]] ($\vec{a}_{t}$) and the [[http://en.wikipedia.org/wiki/Acceleration#Tangential_and_centripetal_acceleration| centripetal acceleration]] ($\vec{a}_{c}$). These 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 (or opposite it in the case of negative acceleration). 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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| $$F_{\parallel} = m{a}_{t} = \dfrac{d|\vec{p}|}{dt} = \dfrac{d|m\vec{v}|}{dt} = m\dfrac{d|\vec{v}|}{dt} \qquad\longrightarrow\qquad {a}_{t} = \dfrac{d|\vec{v}|}{dt}$$ |
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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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| $$F_{\perp} = m{a}_{c} = \dfrac{|\vec{p}||\vec{v}|}{R} = \dfrac{mv^2}{R} = m\dfrac{v^2}{R}\qquad\longrightarrow\qquad {a}_{c} = \dfrac{v^2}{R}$$ |
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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. |