 Hello everyone, I am Mr. Sachin Rathod, working as assistant professor in mechanical engineering department, working as a technology solapur. Today we will deal with numerical on acceleration in mechanism. So the learning outcome of this session is student will able to calculate the acceleration of particular link. So the question for this is the crank and connecting rod of reciprocating engine are 200 mm and 700 mm respectively. The crank is rotating in clockwise direction at 120 radian per second. Find the help of relative velocity method. First acceleration of the piston, second acceleration of midpoint of the connecting rod, third angular acceleration of connecting rod at the instant when the crank is at 30 degree to the idc inner red center. So let's see how to solve this numerical by using relative velocity method. This is the numerical as we have seen and as per the given data we have drawn the space diagram by taking some suitable scale that is 1 centimeter is equal to 80 mm. We have drawn this space diagram that the length of the crank is 200 mm and length of connecting rod is 700 mm and the angle of the crank with idc is 30 degree this is a 30 degree. So we have drawn this space diagram. Now already we have seen how to draw the velocity diagram. So this is the velocity diagram. So we will take one some small revision about the velocity diagram. This is the point O from point O this crank rotates in the clockwise directions that's why the point A is moving the velocity of point A is moving in the upward direction. So we have drawn the perpendicular line to the OA crank. We are getting the A points and the magnitude of the OA is they have given us a omega just multiply with the radius 200 mm that is a 0.2 meter. So we will get the velocity of OA then after getting the A points AB is a connecting rod which is having the oscillatory motion. So that's why we have drawn the perpendicular to the AB links. So this is a perpendicular line to the AB link and B is a slider. So from the fixed points the slider is moving in the parallel direction. So like this we are getting the B point and C is a midpoint. So we have taken a midpoint of the AB we are getting the C point. So this is the velocity diagram and the direction will act like this. This is a velocity of A, this is a velocity of B, this is a velocity of A with respect to B and this will use velocity of C. So this is a velocity diagram and by referring this velocity diagram we have formulated one table. So in the first column link radius, velocity, radial component and tangential component. So link we have taken OA, AB, BO, radius we are knowing the radius of the crank, radius of the connecting rod and the velocity by referring this diagram, velocity diagram we have to calculate the velocity. So we are getting the velocity of the crank OA is 24 meter per second that is the velocity in meter per second 24 and if I measure the AB link we are getting the velocity 22.25 meter per second and the velocity of the piston measure this length we are getting the 15 meter per second. So already we have seen this table in a relative velocity method. So now the task is that we have to draw the acceleration diagram. So you can think about this how the acceleration diagram will look like by considering the we are knowing the each link is having two component that is a radial component and tangential component. So let's see how the acceleration diagram will look like. So consider one point at which the acceleration is 0. So O is a fixed point at which the acceleration is 0. So we are given as a O point. So from the O point the crank OA which rotates with the uniform constant angular velocity. So it is having the only one component that is a radial component. So radial component is always acting in the direction or in the parallel direction to that link. So OA is a link. So we have draw the one parallel line. So this is a parallel line and this is a radial component. The radial component is find out by the formula v square by r. So the velocity is 24 square divided by radius of OA is 0.2. So we are getting the acceleration radial acceleration is 2880 meter square per second. This is a meter square per meter per second square at the unit of the acceleration. Similarly you can calculate the radial component of the AB that the velocity we are knowing 22.25 square divided by radius 0.7. So we are getting 707.23. So we can easily calculate the radial component of OA and AB. Now we have to draw the radial component of the OA. So as the radial component is nothing but the point which is moving towards the center. So O point we are getting the A point in the downward direction. And the magnitude is 2880 meter per second square. So we have to take some scale. So we can take this is the acceleration diagram. For drawing the acceleration diagram scale 1 centimeter is equal to 300 meter per second square. So by considering this scale we have to draw the magnitude of OA. So 2880 divided by 300 so we are getting 9.6. So locate the A point at a distance of 9.6. So here we are getting A point. Then next AB. So AB is the connecting rod which is having the two mutual perpendicular components. One is the radial and another is the tangential. So we are getting the A point here. So afterward the AB link is acting. So the radial component of the AB will be parallel to this line. So parallel to the AB. So we have to draw the parallel line. This is AB line. And the magnitude of this radial component is 707.23. So as per the given scale 707 divided by 0.23 divided by 300. So we are getting 2.35. So here component we are getting x component. So it will give the radial component of the link AB and the tangential component is perpendicular to the radial component. So from this point we have to draw the perpendicular line. So it is a tangential component. And the slider is moving in the horizontal manner. So we have to draw the parallel line to the slider with respect to the fixed point. So this will give the parallel line. So the intersecting point will give the position of the slider B. If I join the link AB, you will get the acceleration of A with respect to B. So from this diagram this will give the acceleration of A with respect to O. This will give the acceleration of B with respect to O. If I join like this, it will give the acceleration of A with respect to B. It will give tangential component of A with respect to B and the radial component of A with respect to B. So measure this length and write it as same as vertical. Tangential component 4.6 multiplied by 300, tangential component of AB 1380. So first question is that you have to find out the acceleration of piston. So acceleration of piston is equal to measure this length, it comes 9.8. So 9.8 multiplied by the scale 300.8 into 300, we are getting 2940 meter per second square. Next question, find out the acceleration of the midpoint of the connecting rod. So this one is a connecting rod, AB total acceleration. Take the midpoint of this, join this point, this is a C point, that is a midpoint. Measure this length. So 9.45, 9.45 into 300, 2835 meter per second square and the angular acceleration of connecting rod. So the connecting rod is a AB. So AB, angular acceleration means we have to find out alpha of AB, we are knowing the equation. So AB is equal to R into alpha, therefore alpha of AB is equal to 80 of AB, tangential component of AB divided by radius of AB. The tangential component of AB already we have to find out 1380 divided by radius of AB is 0.7. So we are getting alpha of AB 1971.42 radian per second square. Like this we have to calculate. These are my references. Thank you.