This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision Next revisionBoth sides next revision |
183_notes:curving_motion [2014/09/24 17:39] – [A change in direction] caballero | 183_notes:curving_motion [2014/09/24 17:40] – [A change in direction] caballero |
---|
$$\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{v}|}{R} \hat{n}$$ | $$\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{v}|}{R} \hat{n}$$ |
| |
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.}}] |
| |
[{{183_notes:mi3e_05-026.jpg|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. | 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. |
| |
$$\vec{F}_{net,\perp} = |\vec{p}|\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{p}||\vec{v}|}{R} \hat{n} =\dfrac{mv^2}{R} \hat{n} $$ | $$\vec{F}_{net,\perp} = |\vec{p}|\dfrac{d \hat{p}}{d t} = \dfrac{|\vec{p}||\vec{v}|}{R} \hat{n} =\dfrac{mv^2}{R} \hat{n} $$ |