 Alright, so let us take up one final question before we start the next concept. You have a hemisphere, okay draw a hemisphere, this is a hemisphere, the radius of the hemisphere is r, fine radius is r, okay the hemisphere is smooth, alright you are keeping a very small mass over here, okay very small mass is keep you are keeping there and then you are slightly nudging it you just slightly touched it and it starts going down getting it so what will happen it will move down along the sphere alright and at some angle it will leave the sphere fine so it may not move all the time along the sphere only so there is a point where it leaves the hemisphere fine so let's say that that angle is theta with the vertical fine you need to find what is the angle theta at which this mass m leaves the hemisphere, okay try to do this, okay you need to find the angle theta at which the mass m will leave the hemisphere, Bharat got one of the answers others those who agree with Bharat can say that okay Bharat's answer is correct yeah I will solve it everybody tried this okay what is the condition when the mass just tell me the condition when it will leave the hemisphere no Bharat yes all the contact forces goes to 0 because it is losing the contact right so since it is losing the contact so contact forces will become equal to 0 fine so what are the contact forces here there is only one contact force that is normal reaction out here so normal reaction when it becomes 0 that's the point when it leaves the hemisphere okay so all the contact forces even if there is a friction even friction force will become 0 at that moment fine okay now if suppose it has a velocity v at that moment let's say it has a velocity v at that moment okay then since it is moving in a circle before leaving the hemisphere it will have an acceleration of v square by r towards the center so this one will be the acceleration so if you draw the free body diagram of this mass you will have mg like this and then you will have a centipede acceleration v square by r getting it and this angle will also be theta fine so if you draw this mg force this angle will be equal to that angle alright so the force along the direction of radius is what mg cos theta fine this force should be equal to mass time acceleration so mv square by r alright so I am just ignoring the normal reaction normal reaction goes to 0 when it leaves the surface fine so this equation should be true at the point when it leaves the surface okay if it is not leaving the surface then what would be the equation mg cos theta minus normal reaction this is the net force this net force will become mv square by r but at this moment normal force is becoming 0 so mg cos theta is mv square by r alright so this is the condition for which it will leave the hemisphere now what I will do I will write work in the theorem between 0.1 and 0.2 okay so work done by all the forces will be equal to change in the kinetic energy what is mg cos theta mg cos theta is the component of mg along the direction of acceleration component of mg along the direction of acceleration is mg cos theta fine as it was let me answer if anybody asks me so mg is acting downwards its component along this direction is mg cos theta anyways so work done by all the forces will be equal to change in kinetic energy which is half mv square half mv square is the final kinetic energy right and initial kinetic energy is 0 fine so this is equation 1 and that is equation number 2 alright now I need to find the work done who all are doing work is normal reaction doing any work when it is sliding down normal reaction is perpendicular to the displacement all the way okay displacement is tangential normal reaction is perpendicular to that so work done by normal reaction is 0 the only force that is doing work here is work done by gravity fine this is equal to mg multiplied by this displacement which is along the direction of gravity okay let's call it as y so y will become how much this total length is r fine and this much is r cos theta fine you can see that there is a right angle triangle over here this length from here to here this length is r cos theta so y is equal to r minus r cos theta so 1 minus cos theta times r fine so work done by gravity is mg r 1 minus cos theta fine so all you have to do is just substitute there mg r 1 minus cos theta is the work done by gravity this is equal to half mv square all right and how much should be half mv square so if you look at here you know half mv square should be equal to mg r cos theta by 2 okay I am just utilizing this equation so I would like to write mv square in terms of mg r because mg r will get cancelled out from left hand side okay this is equal to mg r by 2 into cos theta fine so now you just have to solve so cos theta will come out to be 2 by 3 fine so theta is cos inverse 2 by 3 fine any doubt with respect to this question please type in okay how I get y first of all you need to understand why is what y is a displacement of this mass when it goes there along the direction of mg because in order to find the work done by gravity I need to find how much it has displaced along the direction of gravity fine so you can see that there is a right angle triangle okay from here to here this length this length is r okay this angle is theta and since this is 90 degree this length is r cos theta getting it so I know from here to there top most point the distance is r and from there to here the distance is r cos theta fine so the y is equal to r minus r cos theta right so that's how you get the value of r so that's how you get the value of y okay okay you just have to solve this guys you'll get cos inverse 2 by 3 okay so cos theta will come out to be 2 minus 2 cos theta so 2 cos theta you get it that side you'll get 3 cos theta is equal to 2 and cos theta is equal to 2 by 3 how you got the value of work done by mg see work done by mg is force mg multiplied by displacement along the direction of force which is r into 1 minus cos theta okay so we have got the value of y just like that fine so I'll just make this figure a little bit more clearer so I'll draw it here so mass goes along an arc like this fine so this is r and that is also r and this is theta fine so if I drop a perpendicular from here this is 90 degree this length will be what r cos theta this length okay and I need to find that length which is y right the total length is so y is equal to r minus r cos theta that y is the displacement along the direction of gravity so like this you get this particular work done okay okay shall we move to next let's move to next concept now we are talking about conservative and non-conservative forces right now as a by the way that work energy theorem that we have derived which is work done is equal to k2 minus k1 okay you can apply this theorem at a frame of reference where the reference frame is at rest okay or the reference frame can move with the constant velocity also fine so if reference frame moves with a constant velocity then you observe relative velocity so the measurement of kinetic energy will change if you are observing the object with respect to a frame that is moving with constant velocity fine not only kinetic energy will change but the work then will also change because then you'll see a different displacement getting it so that is slightly more involved when you solve questions you will encounter certain scenarios where if you change the frame then the question gets solved easily fine so we'll take up questions like that changing frame of reference to find the kinetic energy later part of when we revise or when we revisit this chapter we'll take up those scenarios okay all right but then your frame of reference cannot accelerate if it accelerate there'll be pseudophores and pseudophores also does the work fine so pseudophores is another concept that I will have to take separately that I have not yet discussed fine so I'm just telling you so that you know if you are like if you want to move ahead you know don't hesitate you can ask me individually also in case you want to okay Bharat is asking something yeah I just discussed that all right so we are talking about conservative and non-conservative forces now the forces can be you know broadly classified into conservative and non-conservative forces depending on how you calculate or how the work done by the forces will behave okay so if write down if work done by a force independent of frame independent of of the path taken and only depends on initial and final points okay then the force is conservative fine so you'll soon understand why we are defining conservative and non-conservative force right now just take it as a definition okay now suppose this is point one okay and that is point two fine now you are going from point one and two suppose you are going like this and then you are reaching point two like this getting it so it doesn't matter what path you take if initial and final points they are same then the work done by the force does the same work I mean it the work done by the force will be same in both the path let's say this is path one and this is path two okay although the displacement is independent of path okay displacement will be what from point one and two displacement will be this only this is the displacement okay and if it is a constant force fine so if the force is constant anyways you know it will be independent of the path because displacement you have to take the dot product with the force fine so displacement is independent of path and force is a constant force so f dot displacement will give you the work done fine but I am talking about in a generic sense even if the force can change its magnitude and direction both in that case also there are few forces okay for which the work done by the force is independent of the path it only depends on initial and final points okay so in that case that forces the conservative force fine but in a case where the work done by a force depends on the path taken by the particle or an mass or a mass then that force is not a conservative force fine now how you test whether a force is conservative or not you test whether a force is conservative or not by just finding the work done just find out work done by a force if an object moves in a loop what do you think the work done by conservative force will be if an object moves in a loop let's say object moves in a circle okay like this so in one full circle how much will be the work done by the conservative force it will be zero because initial and final points are same okay so work done by the conservative force in a closed loop will always be zero getting it fine now I'll just take one example of non-conservative force so that you can understand the difference take an example of this this is an enclosure it's a spherical enclosure it is a hollow sphere you're taking a mass from here okay you're taking a mass all right you are pushing it all right from both sides from top and the bottom and this object moves like this goes to the top when it goes to the top you're pushing it like this okay and you're making sure it completes one full circle it starts from one and comes to the same point so initial and final point both are the same fine so is the work done by friction zero work done by friction is it zero it is not zero right friction will do some work friction will do some work all right so friction is a force which is you can say an example of non-conservative force fine so it depends on what path you have taken when you calculate the work done by the friction so you'll see that all the while when it is sliding up that is sliding up friction is always in opposite direction of displacement so you should add up the small small work done by the friction all the small work done by the friction will be a negative quantity so all the negative quantity cannot add up and equal to zero fine so there will be a net work done by the friction which is negative because all the small small work done by the friction they are negative so you just add them up it will never become zero fine so work done by the friction is not zero when it moves in a closed loop so friction is an example of non-conservative force okay now take an example of gravity what is the gravity force mg mg is the gravity force and it is always acting vertically down okay so what is the work done by the gravity force work done by mg force what it is right now how much is a displacement along the direction of gravity it will be zero in case your connection is failed please come live and watch it okay you can revisit what you have left later on okay niranjan is saying minus mg into 2r niranjan if you are talking about point one and let's say this is point three so if the object goes from one the bottom most point to the top most point then the work done by the gravity is minus mg into 2r because a displacement along the dirt it is now against the direction of gravity actually this is a displacement which is 2r and gravity is acting downwards fine so mg is downward and displacement is 2r against it so minus mg into 2r when the object goes from one to three fine but if i'm saying that object comes back the displacement becomes zero it goes up by 2r and comes back by 2r so displacement is zero okay so work done by the gravity is zero fine so gravity could be an example of and conservative force fine so we'll take up a couple of more examples and we'll ascertain that the gravity is the conservative force and then we'll analyze it further