 Let me go over the problems okay and briefly the solutions. So what is asked here that a particular geometry of a roller coaster track is given to us and we are told that a 1000 kg car starts from rest here and moves without friction on the track. So determine the force exerted by track at 0.2 and the minimum safe value of the radius of curvature at 0.3. Now the force exerted by the track at 0.2 is straight forward. What do we do? We apply a principle of conservation of energy. Quantity energy is 0 here. Take this as the datum. So the gravitational potential energy is what? Mgh or Mg into 12. So what do we know straight away? If you use the white board okay so Mg into 12 is V1 plus T1 is 0. T2 should be equal to V2 plus T2. Now V2 at the bottom using that bottom as the datum is 0 and T2 is half Mv square okay. This is potential energy. This is velocity square and from that we can find out what is the velocity of the cart at the bottom. And once we know that the radius of curvature of the track is given to us 6 meter. The normal acceleration okay so there are two components. One is a tangential acceleration okay and one is a normal acceleration. Now the tangential acceleration we cannot immediately figure out from whatever information is given to us okay. But if the track is frictionless then we realize that what we realize is this that look at the track here bottom. There are two possible forces. One is a friction force and other is a normal reaction okay that acts on the track. Now the friction force is 0 that is given to us that we are supposed to neglect friction and the normal force will be mass of this cart times the acceleration in the normal direction and what is that equal to mass into V square by rho. But V square we already found out from here and rho is given to us as 6 meter. So we can find out what is the normal reaction acting on it and how do we do that? Let us draw the free body diagram of the cart. So what does the forces acting on the cart is the normal reaction okay. What is the other force acting on it? Gravity and this should be equal to m times An. So m times An is equal to n-mg and from this we can find out what is the value of the normal reaction. Now second question we are asked is that when it comes to this point 3 what is the safe curvature that we can have. So the first thing is that the tangential velocity is 0 we do not bother about the tangential acceleration straight away because there is no force so no acceleration in the tangential direction okay at this particular point okay. No acceleration in the tangential direction but okay what is happening okay that by conservation of energy okay we can find out what is the velocity here but the difference between this part and this part is as follows. The curvature of this is opposite now here so this particle the acceleration is what in this direction okay m An is this but what are the forces acting on it? The forces acting on it are the gravitational force mg okay and the normal reaction from the surface n. So what do we know is that m An will be equal to mg-n or n will be equal to mg-n or n will be equal to mg-m An but what is this? This is equal to just g-v square by rho into m now note one thing that v square we cannot control it is controlled by the principle of conservation of energy but we can control rho and if the rho becomes very small then this quantity can become negative or it means that the surface has to provide a negative reaction if the rails are not guided okay then the surface can only provide a reaction in the vertical direction if it is just a track and as a result okay as a result that when the normal reaction n becomes 0 this vehicle will lose contact with the ground okay and will be unstable. So the maximum radius that we can the minimum radius we can have will become when g-v square by rho is equal to 0 or v is equal to square root of g times rho so that is the solution to this problem and why we have a critical curvature here that we cannot have a very sharp turn very sharp turn you will see that you will lose contact and you will see movies right for example when there is a car chase there is a bump in the road okay then suddenly the road there is a dip and there is a bump in the road you will see that when the car go over the bump and they are moving at very high velocities also okay then they lose contact with the ground and jump okay so whatever is seen in the movies okay that effect is coming from here that because the direction of curvature is this okay whenever the normal reaction has to become 0 the car will lose contact with the ground or the car will lose contact okay with this surface. I hope this is clear this problem is reasonably straight forward okay I will not go into significant details of that but what happens is that particles packages are thrown along the incline at a with a velocity of 1 meter per second the this slide here okay 2 point b okay and to a conveyor bed okay from a to b and from b to c where the coefficient of friction is 0.25 between these 2 surfaces and what we want is that that determine what is the distance d such that this packages when they slide down they are to arrive with a velocity of 2 meters per second in this direction is a very straight forward problem we have to simply use work energy theorem initial kinetic energy is given to us now what is the work done so the particle when it moves from a to b the work done will be minus mu k into the normal reaction into the distance okay plus the component of gravity is same like the car problem that there are 2 forces which are acting on it one is a friction force which is opposing the motion so it will do negative work there is gravitational force the component in this direction which will be mg sin 30 which will do positive work and the sum of those 2 one positive gravitational one negative frictional force okay they will do work on this okay and how much will be the work the force into the distance and when it comes here this distance is given to us in this case the only work done will be the dissipative work coming from the friction which will be the 7 meters into minus mu into the normal reaction the normal reaction also will change here so the friction force also change here and ultimately what do we want that at point c the speed should be or velocity should be 2 meters per second and as a result of kinetic energy here is known okay so the initial kinetic energy plus work done from a to b plus work done from b to c should be equal to the final kinetic energy and we know all the components okay we just need to find out what is this distance d such that the final velocity is this so we get a simple linear equation we can solve for d now coming to the last problem it is a very interesting problem okay so what is told to us is this it is a combination that we need to use work energy theorem we need to use principle of conservation of momentum in normal direction tangential direction so when we solve this problem essentially all the concepts that we have discussed okay will become completely clear so what is given to us is this that 2 kg block is pushed up okay I will only qualitatively discuss this okay if there are any more questions okay the solution will be there in the slides but let me qualitatively discuss what is happening here 2 kg block b is pushed up against a spring compressing it at a distance of 0.1 meter so what do we know we know initially the potential energy of this entire system okay how do we measure okay so there is also there are 2 components of potential energy one component of potential energy comes from the spring and other component of potential energy comes from gravity so what we can do is as follows okay that initial so we can take this position as our reference position and measure the potential energy with respect to this reference position okay so at this particular position the potential energy is 0 what is the spring energy the spring energy is given by so in position 1 okay or if you want to just use simple work momentum the work energy theorem we know that this is position 1 okay then there is a work done to move this block from position 1 to 2 and that will give us T2 which is the kinetic energy at this position when this block hits the pendulum okay so this block is compressed it will slide down and then it will hit the pendulum so our first task is to find out that when the block hits this pendulum what is its velocity and how do we find that out what is this u1 to 2 that spring pushes against the block okay the displacement this is f spring and this is displacement so what is work done by the spring will be initially at x equal to 0 the spring force is 0 and this is x equal to 0.4 this is kx so the total work done by the spring will be equal to half kx square and this will be the work done 1 to 2 okay 1 to x from the spring will be positive now the second work is done by the frictional force okay which acts all the way distance between x and d so the total distance for which the friction is doing the work is x plus d and the work done will be negative and that will be equal to minus mu the normal reaction times x plus d how do we find a normal reaction we find out the normal reaction by equilibrium of this particle in the normal direction why because of the constraints the particle cannot have any acceleration in the normal direction so we get the normal reaction here so we then find out using this what is the final kinetic energy and from the final kinetic energy we know okay what is the momentum of this particle when it hits here now let us go to the next part of the question the next part of the question what does it involve that when this pendulum is about to be impacted this is ma and this is va what direction it is it is in the direction along the slope now we have to use impulse momentum theorem why because take this system this pendulum the impact has very short duration during the impact what are the forces that act on this mass the impact forces that act this okay this is the impact force given by f bar dt now on this pendulum what do we have an impact force will come from the tension okay and equal and opposite from this now the pendulum can have velocity in horizontal direction and the mass can have velocity only in this direction so what we can do is that we can immediately solve this problem okay by finding out what is the momentum balance for this system okay the impact momentum theorem what we do is we want to find out okay that when alpha is equal to 40 degrees this angle okay what we want is that that this starts from here compression in the spring gravity both will do the work and we can use the work energy theorem okay now we very well know how to use it to find out what is the velocity of this mass when it starts from here goes all the way here by traveling a distance of d plus x equal to 0.1 meter okay d is also given to as 1.5 meter all the distance are given we can find out what is the velocity of this when it is about to impact this this mass at b now we see we assume for the time being okay as a simple model that this is a central impact what do you mean by central impact central impact means as we had seen that there is a well defined normal okay at the point of contact and the and the and the and the forces or the the the deforming and the restitution forces they act along that line of contact is what we mean by a central contact problem now in this central contact problem what we realize is that that for the pendulum there can be an impact which is provided by the tension which acts in the string at the point of contact there can be okay and there will be an impact force okay or or or an impulse f dt which acts on the top mass and equal and opposite impulse acts on the bottom cylinder what is the momentum for this the momentum for this is m va which we found out from the work energy theorem for this initial momentum is 0 and after the contact what will happen is that that this will have some velocity momentum which is m va prime after the contact okay this is after and this pendulum can have only a momentum in the horizontal direction why because the string will preventing it from having momentum in the vertical direction so it can have a momentum mb vb prime okay now what we realize is that for this combined system okay for this combined system of this mass a and pendulum b this internal impulse okay it goes off okay it will cancel out from each other and the only effective external impulse like in the in the cart and the package problem is acting in the perpendicular direction to the horizontal so what we can say is that we can use conservation of momentum for this and this a and b in the horizontal direction or in the x direction okay and from that we will get one equation okay because this velocity is va prime this is vb prime so we will get one equation which is relating va prime and vb prime with va but we need the second equation we need a second equation and that second equation now will be provided by the coefficient of restitution but coefficient of restitution has to be applied along the line of impact so what we can do is we can find out what is the component of velocity for this and this in the along the line of contact and then and we can get a second equation and this one and two equations can be solved together to find out what is the effective speed of this okay what is the effective velocity of this after the impact it can only be along this and what is the effective velocity of b just after the impact okay it can be only along the horizontal direction and we will get it from these two equations now the third question is when this angle alpha is equal to 40 degrees what is the tension in the string now we again come back to principle of conservation of energy what we realize is that this pendulum now has an initial velocity here so we can use conservation of energy how do we use that we say that let us take a reference here and we can find out what is the potential energy of this mass b kinetic energy is given so T1 v1 is known to us in this configuration again the potential energy is very well known to us because the vertical distance is known to us and then we can use T1 v1 is equal to T2 v2 T1 v1 v2 are known to us and from that we can find out what is T1 and T1 will give us okay what is the velocity square and the tension in the string okay can be found out by the logic we had used previously which is what is the tension in the string you look here this is T this is the path of the particle so the normal acceleration for this particle is in this direction what is that equal to v square which we find out from energy conservation principle divided by rho which is nothing but the length of the string L which is given to us and then what then this is gravity G this angle alpha is given to us okay mg T alpha and what do we know that man is the acceleration in the normal direction that has to be equal to T-mg cos alpha and then you can find out what is this tension because we know an this is known and then only thing that is unknown is tension so this problem if you can solve it fully can understand it properly then essentially this entire topic on work energy theorem impulse momentum theorem okay including what is meant by impact okay what is meant by impulse what is meant by coefficient of restitution what is work energy theorem what is principle of conservation of energy all these things are put together in this problem and if you can understand this then you are done essentially anything and everything to know about work energy theorem and impulse momentum theorem for a particle okay okay why particle because we are not still worried about the angular momentum here which we are going to do in the lecture on kinetics of rigid bodies which we will start discussing okay very soon in a few moments but if you do this then anything and everything to deal with particle is done so we discuss Newton's laws of motion okay for particles done we discussed impulse momentum theorem done we discussed work energy theorem okay so anything and everything okay that can be done on a particle okay on kinetics and kinematics of particle we have done we have finished it