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--- 183_notes:curving_motion [2014/09/24 17:18] – [A change in direction] caballero
+++ 183_notes:curving_motion [2014/09/24 17:40] – [A change in direction] caballero
@@ Line 53: / Line 53: @@
 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:
-$$\dfrac{d \hat{p}}{d t} = \dfrac{v_{avg}}{R\theta} \langle -\theta, 0\rangle = \dfrac{v_{avg}}{R} \langle -1, 0\rangle$$
+$$\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{v}|}{R\theta} \langle -\theta, 0\rangle = \dfrac{|\vec{v}|}{R} \langle -1, 0\rangle$$
-where you are using the approximation for small $\theta$ in both the cosine ($\cos \theta \approx 1$) and sine (#\sin \theta \approx \theta$). This vector points in the $-x$-dreiction, which is toward the turn and perpendicular to the direction of the momentum vector. This result generalizes to:
+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:
-$$\dfrac{d \hat{p}}{d t} = \dfrac{v_{avg}}{R} \hat{n}$$
+$$\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{v}|}{R} \hat{n}$$
-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.
+[{{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.
+$$\vec{F}_{net,\perp} = |\vec{p}|\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{p}||\vec{v}|}{R} \hat{n} =\dfrac{mv^2}{R} \hat{n} $$