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Torque
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=== Derivatives of torque === In [[physics]], '''rotatum''' is the derivative of [[torque]] with respect to [[time]]<ref>{{Cite journal |title=Survey of Human–Robot Collaboration in Industrial Settings: Awareness, Intelligence, and Compliance |date=2021 |doi=10.1109/TSMC.2020.3041231 |url=https://ieeexplore.ieee.org/abstract/document/9302892/keywords#keywords |last1=Kumar |first1=Shitij |last2=Savur |first2=Celal |last3=Sahin |first3=Ferat |journal=IEEE Transactions on Systems, Man, and Cybernetics: Systems |volume=51 |pages=280–297 |url-access=subscription }}</ref><blockquote><math>\mathbf P = \frac{\mathrm d \boldsymbol \tau}{\mathrm d t},</math></blockquote>where '''τ''' is torque. This word is derived from the [[Latin]] word {{lang|la|rotātus}} meaning 'to rotate'. The term ''rotatum'' is not universally recognized but is commonly used. There is not a universally accepted lexicon to indicate the successive derivatives of rotatum, even if sometimes various proposals have been made. Using the cross product definition of torque, an alternative expression for rotatum is: <blockquote><math>\mathbf{P} = \mathbf{r} \times \frac{\mathrm d \mathbf{F}}{\mathrm d t} + \frac{\mathrm d \mathbf{r}}{\mathrm d t} \times \mathbf{F}.</math></blockquote> Because the rate of change of force is yank <math display="inline">\mathbf{Y}</math> and the rate of change of position is velocity <math display="inline">\mathbf{v}</math>, the expression can be further simplified to: <blockquote><math>\mathbf{P} = \mathbf{r} \times \mathbf{Y} + \mathbf{v} \times \mathbf{F}.</math></blockquote>
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