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| 183_notes:curving_motion [2014/09/24 17:39] – [A change in direction] caballero | 183_notes:curving_motion [2014/09/24 17:43] – [A change in direction] caballero |
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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.jpg?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 come acres the [[http://en.wikipedia.org/wiki/Acceleration#Tangential_and_centripetal_acceleration|tangential acceleration and the centripetal acceleration]]. 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. |
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| | $$\vec{F}_{net} = \vec{F}_{\parallel} + \vec{F}_{\perp}$$ |