Let's consider a point like object with mass $m$ upon which acts a force $\vec{F} = \vec{c} \times \vec{v}$ ($\vec{c}$ is supposed to be a constant vector). Given that $\vec{F}$ is perpendicular to $\vec{v}$ and therefore to $\vec{r}$, the work $W$ done by $\vec{F}$ should be zero. So if $W$ is always zero, then $\oint \vec{F}d\vec{r} = 0$ and $\vec{F}$ should be a conservative force, right? And if $\vec{F}$ is conservative, then we have conservation of energy.
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Related: https://physics.stackexchange.com/q/528336/2451 and links therein. – Qmechanic Jan 14 '23 at 21:52
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$\vec F$ perpendicular to $\vec v$ does not necessarily mean $\vec F$ is perpendicular to $\vec r$. Look at uniform circular motion. – John Darby Jan 14 '23 at 21:58
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There's a measure of freedom in the way you define a conservative force, but the usual definition is that the vector can be written as a gradient, in other words you can define a scalar field that depends on position only so that: $$\vec{F}=-\vec{\nabla}V$$ Within this definition, a force perpendicular to velocity can be non-conservative.
A force with an expression like the one you mention is the magnetic part of the Lorentz force: $$\vec{F}=q\vec{V}\times\vec{B}$$ But think about forces like the tension of the string for a pendulum, or the reaction of the ground on an object. Those forces don't work, but I really wouldn't consider them conservative.
Miyase
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