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--- 183_notes:curving_motion [2014/09/24 15:24] – caballero
+++ 183_notes:curving_motion [2014/09/29 19:32] – pwirving
@@ Line 31: / Line 31: @@
 ==== A change in direction ====
-The second half of the momentum change describes how the direction of the momentum is changing (${d\hat{p}}/dt$). This vector will point in the direction that the momentum is changing. This direction is always perpendicular to the tangent of the path, and pointing in the direction of the turn. Let's see how that's the case.
+The second half of the momentum change describes how the direction of the momentum is changing (${d\hat{p}}/dt$) and will always point in the direction that the momentum is changing. This direction is always perpendicular to the tangent of the path, and pointing in the direction of the turn. Let's see how that's the case.
-[{{183_notes:speeding_up.png?200|A car driving on a circle track speeding up: v' is larger than v.}}]
+[{{ 183_notes:speeding_up.png?200|A car driving on a circle track speeding up: v' is larger than v.}}]
+Consider a car that is speeding up as it drives on a circular track (Figure to the right). In this case, the velocity vector is always tangent to the path (or the track), but it gets longer; the car speeds up. However, the unit vector for the velocity is always the same length and just changes direction. Because [[183_notes:momentum|the velocity and the momentum always point in the same direction]], the momentum unit vector also just changes direction (Figure below and to the left).
+[{{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]], 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.
+$$\dfrac{\Delta \hat{p}}{\Delta t} = \dfrac{\hat{p}_f-\hat{p}_i}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta \rangle-\langle 0, 1 \rangle}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta - 1\rangle}{\Delta t}$$
+[[183_notes:ucm|As before]], we can find the time ($\Delta t$) that it takes for the object to move through the arc length ($R\theta$) -- but, in this case, the velocity at which it does this is the average between the initial and final location (because the object is speeding up).
+$$v_{avg} = \dfrac{R\theta}{\Delta t} \qquad \rightarrow \qquad \Delta t = \dfrac{R\theta}{v_{avg}}$$
+And thus you find that the change in the direction of the unit vector is given by:
+$$\dfrac{\Delta \hat{p}}{\Delta t} = \dfrac{v_{avg}}{R\theta} \langle -\sin \theta, \cos \theta - 1\rangle$$
+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.
+$$\vec{F}_{net} = \vec{F}_{\parallel} + \vec{F}_{\perp}$$
+$$\vec{F}_{\parallel}  = m\vec{a}_{t} = m{a}_{t}\hat{p} \qquad \vec{F}_{\perp} = m\vec{a}_{c} = m{a}_{c}\hat{n}$$
+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.
+$$F_{\parallel}  = m{a}_{t} = \dfrac{d|\vec{p}|}{dt} = \dfrac{d|m\vec{v}|}{dt} = m\dfrac{d|\vec{v}|}{dt} \longrightarrow {a}_{t} = \dfrac{d|\vec{v}|}{dt}$$
+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.
+$$F_{\perp}  = m{a}_{c} = \dfrac{|\vec{p}||\vec{v}|}{R} = \dfrac{mv^2}{R} = m\dfrac{v^2}{R} \longrightarrow {a}_{c} = \dfrac{v^2}{R}$$
+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.
-[{{183_notes:phat_circle.png?200|While the magnitude of the momentum changes, the length of a unit vector is always 1.}}]
-Consider a car driving on a circular track that is speeding up.