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--- 183_notes:curving_motion [2014/09/24 17:12] – [A change in direction] caballero
+++ 183_notes:curving_motion [2014/09/24 17:51] – [Relationship to the tangential and centripetal accelerations] caballero
@@ Line 39: / Line 39: @@
 [{{183_notes:phat_circle.png?200|While the magnitude of the momentum changes, the length of a unit vector is always 1. }}]
-As [[183_notes:ucm|you have seen before]], we 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 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.
 $$\dfrac{\Delta \hat{p}}{\Delta t} = \dfrac{\hat{p}_f-\hat{p}_i}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta \rangle-\langle 0, 1 \rangle}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta - 1\rangle}{\Delta t}$$
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 And thus you find that the change in the direction of the unit vector is given by:
-$$\dfrac{\Delta \hat{p}}{\Delta t} = \dfrac{R\theta}{v_{avg}} \langle -\sin \theta, \cos \theta - 1\rangle$$
+$$\dfrac{\Delta \hat{p}}{\Delta t} = \dfrac{v_{avg}}{R\theta} \langle -\sin \theta, \cos \theta - 1\rangle$$
+As [[183_notes:ucm|you read before]], you can take the limit of smaller and smaller arc lengths, and thus smaller $\theta$, to arrive at the differential form:
+$$\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{v}|}{R\theta} \langle -\theta, 0\rangle = \dfrac{|\vec{v}|}{R} \langle -1, 0\rangle$$
+where you are using the approximation for small $\theta$ in both the cosine ($\cos \theta \approx 1$) and sine ($\sin \theta \approx \theta$). Here, the average velocity is replaced by the instantaneous because you are looking at infinitesimally short time interval. This vector points in the $-x$-direction, which is toward the turn and perpendicular to the direction of the momentum vector. This result generalizes to:
+$$\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{v}|}{R} \hat{n}$$
+[{{183_notes:mi3e_05-026.jpg?250|The perpendicular component of the net force points inward towards the turn.}}]
+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.
+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.
+$$\vec{F}_{net,\perp} = |\vec{p}|\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{p}||\vec{v}|}{R} \hat{n} =\dfrac{mv^2}{R} \hat{n} $$
+==== Relationship to the tangential and centripetal accelerations ====
+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.
+$$\vec{F}_{net} = \vec{F}_{\parallel} + \vec{F}_{\perp}$$
+$$\vec{F}_{\parallel}  = m\vec{a}_{t} = m{a}_{t}\hat{p} \qquad \vec{F}_{\perp} = m\vec{a}_{c} = m{a}_{c}\hat{n}$$