It's all about practice. Resources like these videos may help clarify key concepts, but in a complex course like calculus-based physics, there's just no way you can "get it" without diving into problems, labs, etc. Some problems I'd do with my students in my class take us 30-40 minutes -- much longer than is reasonable to do in a video, but it's all about the practice. The more you do it, the more you learn to organize as you go, and you'll save yourself time and be neat. Good luck!
Indicates direction... Pull a spring apart (displacement x), the force wants to go back the other way (-F). See videos on dot product / work for info on dot dr.
sir i m a great fan of physics and your videos because they help me understand it better.but sir in the recent physics test i found out that i was horrifyingly slow in completing the test because i cared more about a neat test than a complete test. isnt ther away through which i can increase my speed , in physics .plz do reply.
Here the term Gm1m2/r2 is taken with a negative sign,then why the term dr hasn't been taken with a negative sign as both of them are directed towards the negative direction, i.e from infinity towards r.
It has been taken into account when we look at our limits of integration. Because we're integrating from infinity to r, we've already taken that sign into account in our work integral (F * dr)
Thanks for the explanation. But while solving the above equation vectorially like vectorF.vectordr=|F||dr|cos0=Fdr and then integrating it from intifity to r with the negative sign before the integration,it produces a result with a positive gravitational potential energy.Why is it so.
dinesh ram Keep in mind that gravitational potential energy always has to be referenced to something. We're setting an arbitrary reference at r=infinity as zero. You could, however, pick any reference you want as long as you maintain consistency with that reference point.
Sorry sir , but you haven’t got my problem. If we follow the definition that change in potential energy in a conservative field is negative of the work done the conservative force and then use calculus to evaluate it with the equation ∆U=-∆Wgravity , now F=-(Gm1m2/r2)r̂ and dr=-(dr)r̂ , so work done by gravity ∆Wgravity = F.dr=|F||dr|cos0=Fdr and then using ∆U=-∆Wgravity and integrating ,we get ∆U=-∫∞rFdr=-∫∞rGm1m2/r2)dr= Gm1m2/r ,therefore Ur = Gm1m2/r, which is positive, but as we have taken the gravitational potential energy at infinity =0,so gravitational potential energy should have come negative according to the law of conservation of energy. This seems to contradict the fact that gravitational potential energy at a given point in a conservative field is negative if we consider it 0 at infinity. That’s the problem, have I done something wrong in calculation.
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It's all about practice. Resources like these videos may help clarify key concepts, but in a complex course like calculus-based physics, there's just no way you can "get it" without diving into problems, labs, etc. Some problems I'd do with my students in my class take us 30-40 minutes -- much longer than is reasonable to do in a video, but it's all about the practice. The more you do it, the more you learn to organize as you go, and you'll save yourself time and be neat. Good luck!
You are at awesome level 9000. Thank you for your videos.
Thanks!
That was fast lol. It's awesome to see teachers so tech savvy haha.
Why is everything negative like F=-kx? Also, what is dr mean in the F dot dr?
Indicates direction... Pull a spring apart (displacement x), the force wants to go back the other way (-F). See videos on dot product / work for info on dot dr.
non-singular potential function U(x) what is mean to be the force conservative
Farhoud Farhoud Conservative forces are defined in the video at 3:00.
sir i m a great fan of physics and your videos because they help me understand it better.but sir in the recent physics test i found out that i was horrifyingly slow in completing the test because i cared more about a neat test than a complete test. isnt ther away through which i can increase my speed , in physics .plz do reply.
Practice
Here the term Gm1m2/r2 is taken with a negative sign,then why the term dr hasn't been taken with a negative sign as both of them are directed towards the negative direction, i.e from infinity towards r.
It has been taken into account when we look at our limits of integration. Because we're integrating from infinity to r, we've already taken that sign into account in our work integral (F * dr)
Thanks for the explanation.
But while solving the above equation vectorially like vectorF.vectordr=|F||dr|cos0=Fdr and then integrating it from intifity to r with the negative sign before the integration,it produces a result with a positive gravitational potential energy.Why is it so.
dinesh ram Keep in mind that gravitational potential energy always has to be referenced to something. We're setting an arbitrary reference at r=infinity as zero. You could, however, pick any reference you want as long as you maintain consistency with that reference point.
Sorry sir , but you haven’t got my problem. If we follow the definition that change in potential energy in a conservative field is negative of the work done the conservative force and then use calculus to evaluate it with the equation ∆U=-∆Wgravity , now F=-(Gm1m2/r2)r̂ and dr=-(dr)r̂ , so work done by gravity ∆Wgravity = F.dr=|F||dr|cos0=Fdr and then using ∆U=-∆Wgravity and integrating ,we get ∆U=-∫∞rFdr=-∫∞rGm1m2/r2)dr= Gm1m2/r ,therefore Ur = Gm1m2/r, which is positive, but as we have taken the gravitational potential energy at infinity =0,so gravitational potential energy should have come negative according to the law of conservation of energy. This seems to contradict the fact that gravitational potential energy at a given point in a conservative field is negative if we consider it 0 at infinity. That’s the problem, have I done something wrong in calculation.
dinesh ram Did you take into account cos 180 = -1?
Great help!!! THX
+Ze Rui You're welcome.