 Let me just quickly go over to the solutions of the problems. In this case what we are told is that for this particular pulley there is an idler pulley and then what do we have? We have this belt that is AB going on over the pulley. The point of contact of the idler pulley with the belt is at B and what we are told is that that at the instant shown the velocity of point A is 375 mm per second to the left okay is in this direction and the velocity is in this direction 375 mm per second and acceleration is 225 mm per second square to the right. Now note that this belt is contiguous okay each point will move in the similar way so what we realize is that even at point B okay the velocity okay of the belt should be in the same direction okay as point A so it should have the same magnitude 150 375 mm per second and should be to the left and acceleration should be 225 mm per second square to the right okay. So 375 mm per second square Mexican is the velocity 225 mm per second square is the acceleration at this point because this point B is common to the belt and idler pulley and also in this mechanism we assume that there is no loss of contact between the belt and the pulley what we know is that that this pulley will undergo some angular acceleration about the center about this point of connection also so as a result omega times R okay will be the corresponding velocity the velocity has this direction so omega should be in the anticlockwise direction and what will that be V is equal to R omega from that we can find omega is equal to V by R similarly the angular acceleration of the idler pulley okay how do we find out because we know that 80 okay tangential acceleration will be nothing but R times alpha where alpha is the angular acceleration what should be the direction it should be in the clockwise direction because that will ensure that 80 is in this direction to the right so we know omega we know alpha now what we want to find out that for the point B on the pulley what are the components of acceleration and we can realize is that at this point B what are the motions it has okay point B has a tangential motion tangential velocity but the velocity also changes direction why because point B moves in along a circle only difference is that that for point B instantaneously the angular velocity is not constant there is also acceleration so the tangential compass so the normal component or the downward component of the acceleration at point B simply will be R omega square whatever we had discussed till now okay R omega square will be the normal component and the tangential component will be simply in this direction what is that component T is equal to 80 is equal to 225 that is given to us so at this point the normal component is equal to omega square R in the inward direction tangential component is already given to us and we are done okay we find out what is the resultant what is the angle and we have our final answer now the question is this what we are asked in this problem it is a somewhat complicated link mechanism but it is still a one degree of freedom problem okay think about it you will realize that it is still a one degree of freedom problem we have AB, BD, DE all the dimensions are given to us additionally we are also told that in the position that is shown to us bar AB has an angular velocity of 4 radians per second in the clockwise direction so we are asked to find out with this information determine the angular velocity of bars BD and bars DE so what we use is we use the simple trick that we have been using so far we know that velocity of A is 0 so velocity of B will be velocity of A plus what is this it will be omega what is omega in this case omega can be written as omega into k hat so let us look at point B what is the velocity of point B velocity of point B is velocity of point A which is 0 plus omega AB cross R of B with respect to A that the position vector of B with respect to A now what is omega AB note one thing that if this is our x axis this is y axis the rotation is clockwise so note the thumb goes in the plane whereas the z axis comes out of the plane okay so if you really want to use coordinate system this is one sure shot way of doing this problem is that this is x this is y this is the angular velocity in the clockwise direction so omega times minus k is the vectorial form for the angular velocity so we will see that angular velocity of AB will be minus 4 radians per second it is given to us minus k why minus k clockwise in the plane of the paper so R of B, A is what in this coordinate frame what is it is equal to minus 175 so let us fix the coordinate axis x y so if this is x then the coordinate of B is what minus 175 I is the unit vector in the x direction so straight away okay this is again this is simplest way of doing it there are multiple ways in which we can do it we can use the vectorial approach but it will be more complicated here additionally for example we can also use the virtual work kind of approach that we had discussed earlier but this is the simplest and most sure shot way of doing this problem so R of B with respect to A okay is nothing but minus 175 I so what is this VB is omega AB cross RB slash A position vector of B with respect to A just do the algebra and you will see that VB is 700 millimetres per second J upwards and it makes perfect sense because we know from our simple kinematics that if this is omega point B will only move upwards and the magnitude will be 175 times omega so we get VB now what we do is that we want to find out what is the rotation of this rod rotation of this rod so we work our way down now let us go to rod BD to rod BD what do we have let us say omega BD the vector omega BD is magnitude BD k hat okay so if it is anticlockwise it is positive clockwise omega BD okay this becomes negative what is position vector of D with respect to B note that our x y coordinate is how x is to the right y is upwards so the position vector of D with respect to B is simply just minus 200 J J is the unit vector along the y direction so what is VD this is a very mechanical procedure okay so if we make sure that all the coordinates are appropriately drawn all the signs are appropriately taken then in a very simple way we can get all the answers so what is velocity of point D will be nothing but VB vectorial plus omega BD what is omega BD is the magnitude k hat cross position vector of D with respect to B just substitute all the values here what will be see that velocity of point D will have two components clearly why will it have two components because B now is linked with rod DB is linked with point B because B moved upwards okay clearly it can have a vertical component and because of the rotation it can have a sideways component now what is the sideways component okay 200 omega BD I now omega BD we said anticlockwise is positive so if this is anticlockwise 200 into omega clearly in the positive direction should be the x direction velocity what is the upward velocity clearly is the same as point B so our logic okay so all this mechanistic calculation we can also reinterpret and make sure that our calculations are fine so it is 700 J which is nothing but the velocity at point B and this 200 omega BD is nothing but because of the rotation in the anticlockwise direction for this this is the velocity in the plus x direction or to the right now this is an unknown we do not know what this quantity is let us work our way down further let us go to rod DE so what is rod DE okay UZCA omega is omega DE is equal to magnitude okay again if it is anticlockwise plus clockwise minus so k hat position vector of D with respect to E how do we write look here so D with respect to E okay so it is going in this direction what is the x plus 75 I okay plus 100 plus 175 which is minus 275 I plus 75 J this is to the left so the minus sign upwards so plus sign so we know VD is equal to omega D cross RDE just substitute everything together what do you see that VD will be equal to minus 2.75 DE with respect to J okay omega DE J minus 75 omega DE I will be the velocity of D with respect to E why because absolute value at E is equal to 0 so we are done now okay so everything is now given to us okay does it make sense it makes perfect sense why because if this is omega DE we know that rod DE will have pure rotation okay about point E okay in the anticlockwise direction and as a result of velocity will have a direction which is perpendicular to the direction of the rod we can take a dot product and immediately verify that this velocity vector is perpendicular to the direction of this rod now what do we do we just look here that VD okay chi expression comes out to be omega DE cross RD with respect to E but VD is common both to the top rod and to the bottom rod and both should have the same velocities because otherwise this rod will split open and the only way to have the split velocity the same velocities is we equate these two and immediately see that this will be equal to 700 so from that we can find out that omega DE will be minus 700 divided by 275 so omega DE assume to be in the anticlockwise direction minus means it is actually in the clockwise direction and omega BD okay straight forward okay we got omega DE equate these two omega DE is known from that we can find out what is omega or the angular rotation rotational speed of BD so from this top down approach you start from here go all the way till here we know that absolute values of A and E are 0 and from that we can figure out knowing this is the angular velocity what is the angular velocity of each and every point okay so I hope this is clear okay so now let us move on the third problem is reasonably straight forward okay you can have a look at it later on so fourth problem is the problem on instantaneous center of rotation what I asked to find out in the position shown bar AB has an angular velocity of 4 radians per second clockwise okay determine the angular velocity of bars BD and DE and let us use the concept of instantaneous center of rotation now note one thing is that look at point DE what is the only way this D can move okay what can be the instantaneous velocity of point D in what direction the only direction it can be can be only along the line which is perpendicular to D because that is what it is this rod DE is pinned along point E and it can it can only have a pure rotation about point E and as a result okay the angular velocity if it is clockwise motion if the rotation is clockwise then the angular velocity the velocity of point D is in this direction if the rotation is in the anticlockwise direction the velocity at point D is perpendicular to this but in the inward direction so we know what is the line along which the velocity should lie what about at point B point AB okay point B is common is common to rod AB and BD so when you view this from rod AB point of view AB undergoes a pure rotation about A okay what is the angular velocity 4 radians per second clockwise but point B can only move okay in the downward direction clockwise we can only move vertically and downward so we know that there are two points okay there are there is one point here so for rod BD let us now think about rod BD beca velocity should be the same should act downwards beca velocity should act in this perpendicular direction so when we now look at rod BD we know two points and we know the direction of velocity at two points and those two directions are not parallel to each other so what do we do to get the instantaneous center of rotation this is the velocity for rod BD we are talking about rod BD velocity has only this direction here the velocity can have plus or minus direction okay but let us say it has this direction join this perpendicular line join this perpendicular line and what you will see is that this C okay this is velocity at D join this velocity at B join this okay this C will be the instantaneous center of rotation this angle is beta and as a result this angle can also be seen to be equal to beta now what do we know this is 100 millimeters so what do we know that CB into omega is equal to VB omega is given to us from that we can find out okay so we do not know what is omega but what we know from the previous portion is that that the velocity at this point B will be 4 radians per second into 250 so we know the magnitude of the velocity but that magnitude of velocity is also equal to omega of this rod multiplied by distance CB how do you find distance CB we know this distance okay so 0.1 divided by tan beta will get BC so this distance BC multiplied by omega BD will be equal to velocity what was that velocity it was 4 radians per second into 0.250 meters which is 1 meter per second substitute that we get what is omega for this rod to find out that the magnitude is this the direction of velocity here is this but what is the magnitude what is the magnitude here simply is equal to omega times this distance CD how do you get CD this distance is known so 0.1 divided by tan 21.8 sorry so 0.25 divided by cos beta we can immediately find out okay what is from simple geometry here what is this distance DC and once we know DC omega already found out omega times DC will give us what is the velocity at point D so this concludes our tutorial