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How do I prove whether a force perpendicular to the motion is conservative and $\mathbf{F}=\mathbf{F_{0}}\sin(at)$ conservative, where $\mathbf{F_{0}}$ is a constant vector.

I knew that for a force to be conservative, it's $\nabla \times \mathbf{F}=0$ everywhere or the work done around a closed path without including origin should be zero.

Self-learning from Kleppner and Kolenkow's book.

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  • $\begingroup$ Why would you exclude the origin? $\endgroup$
    – mikuszefski
    Commented Mar 8, 2017 at 7:57
  • $\begingroup$ @mikuszefski : Let me define a force $F=\frac{A}{r}\hat{\theta}$, if I calculate the work done around a closed path including origin is $W=A(\theta_{2}-\theta_{1})=2\pi A$, which is not equal to zero. $\endgroup$
    – 147875
    Commented Mar 8, 2017 at 8:21
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    $\begingroup$ The Force $F=F_{0}\sin(at)$ is not even a vector as it's written, can you please be more specific? With $\nabla XF=0$ do you mean $\nabla \times \mathbf{F}=0$? $\endgroup$
    – DrManhattan
    Commented Mar 8, 2017 at 8:33
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    $\begingroup$ @Rumplestillskin Conservative doesn't mean conserved, so what you have written has nothing to do with the question. To Gopal: a vector is usually denoted with a bold character $\mathbf{F}$ or with an arrow $\vec{F}$, otherwise it's considered a scalar. Anyway, what is $t$? Just a parameter? Because in this case is trivial to prove that $\nabla \times \mathbf{F}=0$ $\endgroup$
    – DrManhattan
    Commented Mar 8, 2017 at 8:47
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    $\begingroup$ Your example above is non-zero, because it is non-conservative. Still no reason to exclude the origin. $\endgroup$
    – mikuszefski
    Commented Mar 8, 2017 at 10:11

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If a force is perpendicular to the motion then $\mathbf{F}\cdot \dot{\mathbf{x}}=0 \quad\forall t$, then the work

$$ W=\oint_\ell \mathbf{F}\cdot d\mathbf{x}=0$$

So it's conservative.

If $t$ is time, then the force does not depend on the position, then all the derivatives are zero and its trivial to say that $\nabla \times \mathbf{F}=0$.

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  • $\begingroup$ Does this mean viscous force is not conservative? $\endgroup$
    – 147875
    Commented Mar 8, 2017 at 8:48
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    $\begingroup$ Tipically all the friction/dumping forces are not conservative. $\endgroup$
    – DrManhattan
    Commented Mar 8, 2017 at 8:50

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