Hello students welcome back to the chapter circular motion in the previous module we discussed about centripetal force which is necessary for a particle to move in a circular motion we need a uniform circular motion or non-uniform circular motion the value of the centripetal force is equal to mV square by r or M Omega square R we also discussed that the particle would be acted upon by a tangential  force which is M alpha r non-uniform circular motion in this particular module will discuss about the examples of the same the statement of the first example says that the particle is projected to the speed u at an angle theta  with the horizontal and we are required to find out the radius of curvature at the highest point of the trajectory for the project right now we know that in this particular case the particle is going in a projectile motion it is not a circular motion but every point in the path can be treated as a point of circle so we can find out the radius of curvature in this particular case as well so what we know is that the particle is initially thrown at an angle theta  that will be horizontal with a speed u if we try to resolve the speed will get the value of horizontal speed to be ucos theta and the vertical speed here is  usin theta the trajectory  that the particle would follow would look something like this and at the highest point we know that the particle would have a horizontal speed the vertical speed vanishes and sends the horizontal speed is ucos theta it does not change if air resistance is neglected at the highest Point the particle would have a horizontal speed of ucos theta think that now we know that for particle moving in circle the velocity is always tangential  so therefore at the highest point what we can say is the horizontal direction is a Tangential  direction that is e theta  a gap and the vertical direction is a radial direction that is er cap let’s say that the center of the circle  here is C and the radius is R .we are required to find out this value of R now i will recall these steps that we have to follow to solve this kind of question is first we have to identify the forces and we do that by drawing  the FBD in this particular case the particle is thrown as a projectile and therefore it is going under the influence of gravity that’s the only force that acts on the particle is mg and that is also what Vertically  down that means it is towards the center and the net force which is required for a particle in circular motion towards the center is equal to mV square by r thus this mg is the only force and we can conclude that mg here is providing the necessary centripetal force so equating this we get mg is equal to mV square by r which on solving gives us the value of R to be equal to v square by G and we’re putting down the value of instantaneous speed at the highest point which is ucos theta so we get finally that the radius of curvature at the highest point as required in the question is u square cos square to theta divided by G this is the answer to this particular example let us move forward with the another example which says that there’s a bead which is placed the over a rod which is hinge at one end now this bead is at a distance of R from the hinge and the Rod is rotating with a constant angular acceleration of alpha we are also given the this coefficient of friction between the Rod and the bead and the values nil we have to find out the value of time and we angle after which the bead starts slipping and we have to neglect the gravity in this particular question let us try to visualize what is happening in this particular case the rod actually rotating horizontally in this fashion and we can conclude that The bead would be moving in a circle of radius R before it starts slipping so if we try to view this particular motion from the top end we’ll get the motion of the Rod looks like something like so we can conclude that the bead is moving in a circle of radius R and for a particle to move in a circle what is required is a radial force and this radial force should be equal to M Omega square R that means if it has a angular acceleration then we can conclude that the bead will also experienced and then tangential force which is equal to m into alpha are now what provides the bead this radial force there is no gravity and the normal reaction would provide the tangential force  so the friction in this particular case provides the necessary centripetal force the radial force therefore we can write down and m omega square is equal to F which is equal to meu n even let us mark this as equation 1 also we discuss the tangential forces provided by the normal reaction and thus we can say that m alpha r is equal to n,  this is a second equation now if we divided the equation one and two what we get is the value of meu the coefficient of friction is equal to omega square r / alpha r.  now what we can conclude from here is that the value of omega is equal to under root mu into alpha so this is the value of Omega where the bead would actually start slipping why because after this Omega the friction would not be able to provide the necessary centripetal force  the value of friction would be less than the required value so what we’ll do is now we have to find out the value of time and angle will use the fact that we can use the three equation of motion for the constant angular acceleration where the first equation is omega is equal to Omega naught plus alpha t which gives us the value of T to be equal to under root of meu by alpha now use the third equation we get omega square is equal to omega  knot square plus 2 alpha theta which gives us the value of angle to be equal to under root of meu by 2 these two values are the final answer to this particular example i hope u understood the module will continue R discussion on circular motion thank u