Both sides previous revision Previous revision Next revision | Previous revision Next revisionBoth sides next revision |
183_notes:curving_motion [2014/09/24 17:51] – [Relationship to the tangential and centripetal accelerations] caballero | 183_notes:curving_motion [2014/09/29 19:32] – pwirving |
---|
[{{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. }}] |
| |
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. |
| |
$$\dfrac{\Delta \hat{p}}{\Delta t} = \dfrac{\hat{p}_f-\hat{p}_i}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta \rangle-\langle 0, 1 \rangle}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta - 1\rangle}{\Delta t}$$ | $$\dfrac{\Delta \hat{p}}{\Delta t} = \dfrac{\hat{p}_f-\hat{p}_i}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta \rangle-\langle 0, 1 \rangle}{\Delta t} = \dfrac{\langle -\sin \theta, \cos \theta - 1\rangle}{\Delta t}$$ |
$$\vec{F}_{net} = \vec{F}_{\parallel} + \vec{F}_{\perp}$$ | $$\vec{F}_{net} = \vec{F}_{\parallel} + \vec{F}_{\perp}$$ |
$$\vec{F}_{\parallel} = m\vec{a}_{t} = m{a}_{t}\hat{p} \qquad \vec{F}_{\perp} = m\vec{a}_{c} = m{a}_{c}\hat{n}$$ | $$\vec{F}_{\parallel} = m\vec{a}_{t} = m{a}_{t}\hat{p} \qquad \vec{F}_{\perp} = m\vec{a}_{c} = m{a}_{c}\hat{n}$$ |
| |
| 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. |
| |
| |