 Okay, welcome back, good afternoon and before I start with or continue with the lecture just one or two things which I want to clarify. So one of our colleagues pointed out that I use V is equal to r cross omega but it should be omega cross r, so that was a typo, so please note that V bar will be omega cross r in this scenario okay, so that will change the sign okay, so that was my typo please take a note of that and also another thing is that that because my zeal because I wanted to go into the details of virtual work and how does the mechanism happen okay, so I went into on to a tangent but I realized that that may have confused some of you regarding that how do we find out what is the rotation of this and what is the corresponding displacement, so let me quickly outline the process. So what we realize is that that this crank AB is rotating in the clockwise direction, so with respect to this point A which is stationary velocity of B is nothing but omega cross r which is given in this direction the magnitude will be omega into 75. So we know what is the magnitude of this velocity, now let us look at this okay, let us look at this mechanism what is happening is that from the point of view of rod BD, we need to find out what are first of all what is this angle BD, to do that we know realize that AB sin 4T should be the same as BD sin beta, so we just put this in the equation solving this equation getting a reverse of sin beta, you will see that beta is approximately 14 degrees, now entire motion of rod BD, it is not a pure translation neither is it rotation about a point, so what do we do is that we use a very simple concept that we had discussed before that this motion you break it down into 2 parts, one is translation okay along the direction of B with a velocity B plus 6 point B okay and rotate this with an unknown angular velocity we do not know what that unknown angular velocity is okay, such that okay that the velocity at point D with respect to B will be nothing but this omega cross omega cross L or the magnitude will be L into omega BD, now the idea is that what we want to figure out okay that what are these appropriate magnitudes, now let us we know that the velocity of point D has to be in the horizontal direction, we know that this velocity also has to be in this direction because it is known to us this angle this is VAB, so this angle has to be 50 degrees, this is VD which is unknown to us, this magnitude is also unknown to us but what we know is that that this angle is 76.05 degrees why because we saw in the previous slide that beta was 14 degrees and since beta was 14 degrees this has to be 76.05 90 minus that, so this angle we know what does the relative velocity V of D with respect to B makes with respect to the horizontal is 76.05 and now we know this magnitude we know all the angles, so we can simply use sign rule we know that VD divided by this angle sum of all the angles is 180 degrees, so we know this angle should be equal to VDB divided by sign 50 this is the angle sign 50 is equal to 15705 divided by sign 76.05, so this is the velocity or this is the speed that we had obtained from the previous step, so now we know all the velocity we know the velocity here we know all the angles, so we just use simple sign rule and from that we can figure out what is V of D which will come out to be 13.08 meter per second and V of DB this will again come out to be 12396 millimeter per second but we also know that V of DB is equal to length this arm BD multiplied by omega BD, so we divide V by L and from that we can find out what is the omega for this and omega is anticlockwise and magnitude comes from this simple relation, so this discussion I think it is much more easier to discuss and as simple as that to find out what is the rotation and what is the relative sliding of what is the sliding of point D, now let us move on to this very interesting concept which is the instantaneous center of rotation in plane motion, now recall okay recall that if we have just a sphere like this a circle like this which is rotating about point O which we say is the centroid for example or this is pinned about the bottom okay point P at the bottom and there is a rotation what is the rotation the rotation is about point P which is pure rotation about P okay, now note one thing that most of the motions that we had seen especially for the ladder and also for the last crank and arm problem the crank and piston problem we saw that the motion in general is not a simple translation or a simple fixed rotation about a point but it is a combination but there is this interesting concept that we can figure out what is called as the instantaneous center of rotation and what is this instantaneous rotation center of rotation is that that the velocities okay that when the body is moving each and every point has a velocity now we had seen that when each and every point has some velocity V bar we had seen that if the rotation is a pure if the if the motion is a pure rotation then V bar is equal to omega bar okay for point A cross R bar okay well it is a rotation about point O here and this R A is nothing but a position vector of point A and we know that the velocity is omega cross R A and what is the direction the direction is perpendicular to this you look at any point the direction will be perpendicular to this the direction will be perpendicular to this now think about it if I take the velocity direction here draw a perpendicular okay let us take the case of pure rotation let us take two points here so this is the there is a pure rotation of this body about point O this is R bar A this is R bar B for point B note that because it is a pure rotation this is the velocity direction which is perpendicular to the the radius vector this is the velocity vector which is perpendicular okay to the radius vector now note one thing if I know the directions of the velocity vectors okay this is the direction of the velocity vector what happens if I draw a normal to it if I draw a normal to the velocity I will get this one line if I draw a normal to this second velocity vector of point B I will get another line and this two will intersect at point O about we the rotation was happening now we extend a similar concept that even though the body is not undergoing pure translation or pure rotation okay it is a combination of rotation and translation what we say is that that there will exist one point okay there will exist one point such that you draw a perpendicular okay that at point A suppose this is velocity you draw a perpendicular to that then this distance times omega okay is nothing but the instantaneous velocity at point A and this center C then is called as the instantaneous center of rotation similarly if I want to find out that if there is another point B here we join this C with B and ask ourselves okay that what is the instantaneous velocity or the velocity at that time for point B what do we do this is the length multiplied by omega will be the magnitude and what is the direction it will be omega cross R so this will be the corresponding direction of velocity at point B now this particular point okay which may keep shifting for example if you have a rolling cylinder okay the cylinder rolls without slip this is point A after some time the cylinder will roll point A will go here and this will be point B in this case you say that point A will be instantaneous center of rotation why because this point A is stationary okay this point A is stationary it has no velocity in any rotational motion the point about which the rotation happens is stationary so that this criteria is satisfied and you will see that if you want to find out velocity at any point you just join here multiplied by omega okay this will be the direction and this will be the magnitude so this is the instantaneous point and this point can keep shifting for example this A will go here after some time and will be replaced by B okay so this is the concept of instantaneous center of rotation now we ask ourselves that if we know some velocity patterns can we find out what is the instantaneous center of rotation now let us say for this rigid body at 2 points A and B we know that this is the velocity at A both the direction and the magnitude at point B we know the velocity both the magnitude and the direction now we say okay let us drop a perpendicular to this okay it will be a line let us drop another perpendicular from this line to other line they have to intersect if they are not parallel at point C now when they intersect at point C what we realize is that that the overall velocity at point B okay will be nothing but this distance CB multiplied by omega velocity of point A is nothing but this distance multiplied by omega in this direction or if you want to think about it it is omega cross this position vector or omega cross this position vector and this point C will be instantaneously stationary okay which means that it is as if this entire object is rotating about point C what does that mean that velocity at any point on this object is nothing but you join this call that as r cross omega is the velocity of any point and we just did a back calculation to find out what is that point so if in your rigid body if you can find out 2 points who have different directions of velocity then the instantaneous center of rotation is immediately determined without even bothering about the magnitude just draw a perpendicular just draw a perpendicular this point is the instantaneous center of rotation now a case where if those 2 points have velocity in the same direction then the way to find the instantaneous center of rotation is that draw this arrow of the magnitude draw this arrow of the magnitude draw a perpendicular line to both A and B and outside the outside tip of point A or point B just join here that will be the center of rotation what is the justification the justification is that that look at this that the magnitude okay the magnitude of the velocity is proportional to the momentum and so these 2 are similar triangles so this by this will be nothing but this length divided by this length and that is our justification in saying that this C is the instantaneous center of rotation take any other point okay take any other point and you will see that the velocity at that point will be nothing but that position vector cross the omega cross that position vector or it will be this magnitude multiplied by omega direction will be perpendicular to the line line joining it okay so that in simple words is the instantaneous center of rotation now let us look about this thing how do we figure out that if the velocity at this point A is given then what is the corresponding velocity at point B we know one thing we know that the direction of velocity at point A has to be this the direction of velocity at this point has to be vertical so we just draw a line perpendicular to this velocity draw a line perpendicular to this velocity both will intersect at point C and that will be our instantaneous center of rotation now if I ask you that at that instant when the system is in this configuration this is velocity VA okay what will be velocity VB and what will be omega the answer is really straightforward what you realize is that that because this is the instantaneous center of rotation if VA is like this the only way you can have VA in this direction at point A is that if this rod has instantaneously an anticlockwise rotation about point B how much anticlockwise rotation omega such that omega times this distance C A should be equal to V of A so then we immediately can find out if geometrically we can locate this point C then we immediately know what is the corresponding omega now similarly we go to point B we need to find out what is the velocity of point B we found out omega from this first step then what we do we recognize what is this distance BC BC will be nothing but L sin theta and what is the velocity here the velocity here is that the magnitude will be omega times BC what will be the direction look at this this is acting in the anticlockwise direction so the velocity need to have a downward direction so by finding out what is the instantaneous center of rotation okay we can immediately figure out what is the motion of any part of the body and we can immediately even figure out what is the corresponding rotation and note that rotation will remain constant okay if you take rotation about BA or C the magnitude omega will never change but what will change okay is with respect to point B if the rotation was happening with respect to point B what is the relative velocity of A with respect to B or if we model the problem like we did previously some of velocity at point A plus relative velocity of B at point B B with respect to A is the total velocity at point B okay so then in that case the rotation happens at this point and a relative velocity changes but omega always remains the same and with this instant instantaneous center of rotation concept you can easily find out okay for example even what is the velocity at the center so what do we do we just join this line draw perpendicular to that that is a direction magnitude is how much omega times that distance straight away so using this is a very useful concept and we will see how this concept can be very gainfully used for the problem we had discussed previously what we realize is what is the velocity direction of point B because link AB has pure rotation about A B has a direction which is in this direction the rotation is clockwise so B has a direction perpendicular to this here magnitude is 75 times omega what is the velocity at point D we do not know the magnitude but we know the direction the direction has to be along this then what do we do we try to locate what is the instantaneous center of rotation how do we do that to this line we draw a perpendicular let it go to this horizontal line we draw another perpendicular these two lines will intersect at point C and this C will be the instantaneous center of rotation now the question is how do we find out okay that we found out the instantaneous rotation geometrically but what are the corresponding magnitudes what is length CD what is length CB and what we recognize is that that the angle this angle beta from our previous calculation was 13.95 close to 14 and now we note one thing is this that this angle was 40 to begin with this was 13.95 so in an absolute sense this angle DBC is equal to 40 plus 13.5 53.95 look at this angle how much is this angle BDC will be nothing but 90 minus 13.95 which will be 76.05 this angle is given so now what do we know for this triangle we know this angle this angle and this angle and one side so simply use sign rule and then what we get we get what is BC we get what is CD okay but we also know that this length is given to us which is 200 this angle is given to us we immediately find out what is BC we immediately find out what is CD and then what we know that BC okay what should be the omega okay because omega times BC is the velocity of point B but we already know from previous calculation that the omega of rod AB multiplied by distance AB was the corresponding velocity so we just equate them and from that find out what is omega for this rod and once we get to omega for this rod which find simply just from this calculation what do we do CD multiplied by omega note here that if the omega here is clockwise for it to get the same velocity for instantaneous rotation of this rod BD about the instantaneous center of rotation C the velocity need to have the angular velocity needs to have a direction which is in the counter clockwise direction because if this is clockwise you will get velocity like this about this center if the velocity has to be like this okay it will be BC into a clockwise direction omega okay so we immediately see from instantaneous center of rotation that if this is clockwise this has to be anti clockwise similarly if this omega is known clockwise sorry anti clockwise then what do we see that CD times omega is the velocity in this direction at point D it will be outwards because the velocities in the angular velocities in the anti clockwise direction times CD and we are done because omega we found out from the previous calculation CD we got from the geometry we immediately could find out what is the velocity of point D so this is the concept of center of rotation and this is how we can use it.