- #1
RIgel
- 4
- 0
currently working on format.. sor i was not preparedHi
I think this question would be much related to calculus more than physics cause it seems I'd lost my way cause of calculus... but anyway! it says,
[itex]Q=- \frac{\partial{U}}{\partial{q}}+\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )[/itex]
but i don't get why ∑[itex]\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq[/itex] is a work...
reason why I'm asking this is...
For it says no matter which dimension [itex]Q[/itex] have , ∑[itex]Q⋅δq[/itex] must have a dimension of work,then because [itex]Q=- \frac{\partial{U}}{\partial{q}}+\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )[/itex]
so [itex]∑Q⋅δq=∑- \frac{\partial{U}}{\partial{q}}⋅δq+\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq[/itex]
should have dimension of work too.
I know [itex]- \frac{\partial{U}}{\partial{q}}⋅δq [/itex] part is work indeed, but i don't get why the latter part [itex]\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq[/itex] is having a dimension of work.I kinda have a feeling that i had lost my way because of calculus because i neither can't figure out why [itex]\frac{d}{dt}(\frac{∂r}{∂q})=(\frac{∂\dot{r}}{∂q})[/itex]is right or how it goes like that.
I think this question would be much related to calculus more than physics cause it seems I'd lost my way cause of calculus... but anyway! it says,
[itex]Q=- \frac{\partial{U}}{\partial{q}}+\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )[/itex]
but i don't get why ∑[itex]\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq[/itex] is a work...
reason why I'm asking this is...
For it says no matter which dimension [itex]Q[/itex] have , ∑[itex]Q⋅δq[/itex] must have a dimension of work,then because [itex]Q=- \frac{\partial{U}}{\partial{q}}+\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )[/itex]
so [itex]∑Q⋅δq=∑- \frac{\partial{U}}{\partial{q}}⋅δq+\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq[/itex]
should have dimension of work too.
I know [itex]- \frac{\partial{U}}{\partial{q}}⋅δq [/itex] part is work indeed, but i don't get why the latter part [itex]\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq[/itex] is having a dimension of work.I kinda have a feeling that i had lost my way because of calculus because i neither can't figure out why [itex]\frac{d}{dt}(\frac{∂r}{∂q})=(\frac{∂\dot{r}}{∂q})[/itex]is right or how it goes like that.
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