Question

please use graphical instant center method

Problem 7: Given the velocity of A is 10 m/sec in the direction shown, find the angular velocities of link 4 and link 5, and the angular velocity of link 4 w.r.t. link 2. Solve using the traditional graphical method and the instant centers method. (Do not use intermediate steps: i.e. do not use or w2 in order to find 2w4 when solving the problem using the instant centers method.) Assume that the mechanism is drawn in the correct proportions and that the link 2 is 1 m long. Show all intermediate necessary steps. Keep the solutions separate and make sure to label the drawing with the terms that you use in your solutions. Clearly list all values from your geometric measurements. A2 4

Answer 1

GIVEN :- THE CONFIGURATION DIAGRAM

VELOCITY OF POINT A = V= 10m/s

LENGTH OF LINK 2 = 1m

TO FIND:- ANGULAR VELOCITY OF LINK 4 AND LINK 5 AND ALSO ANGULAR VELOCITY OF LINK 4 WITH RESPECT TO LINK 2.

SOLUTION:- To solve this problem , first draw the configuration diagram to the scale and consider fixed link as link 1 .

Let the fixed link points be O1 and O2 ( link2 and fixed link joint is O1 , link 4 and fixed link joint is O2)

The velocity of point A is given , using which we can find out w2 (refer image):

V=w2*length of link 2 ,

putting values of length of link 2 and velocity , we get w2=10rad/s in counter clockwise direction ,

Now , to solve by using I center method , first we have to calculate number of I centers in this problem :-

6 links are given , so the number of I centers will be 6*(6-1)/2=15 I centers , but we need not locate them all , only the ones we require to solve the problem.

Now locate the I centers using the following steps :-

STEP 1- Since the turning pair between two links is itself the I center of those two links , so we can easily locate I12 at O1 , I23 at A , I34 at B , I14 at O2 , I45 and I56 . Now for I16 , since link 6 is a slider w.r.t the fixed link 1 , so it can be assumed to be rotating in a circle of infinite radius , hence I16 lies at infinity passing through the slider and perpendicular to the line of motion of slider .

Now to locate the I centers , draw a regular polygon of number of sides equal to links i.e. 6 , so we draw a regular hexagon and name its vertices 1,2,3,4,5 and 6.

By using kennedy's theorem which says that if three links are in relative motion then their relative I centers lie in a straight line , we can find out the remaining I centers i.e to locate I13 , it will lie on the line containing I12 and I23 , and also I34 and I14 , finding the intersection of the two lines , will give I13 . Similarly we can locate I24 , I15 and I25 .

After locating the I centers , we can easily find out angular velocity of link 4 and link 5 by using theorem of angular velocities as shown below :

1. To find the angular velocity of link 4

Velocity of I24 can be written in two ways , i.e w.r.t link 2 and also w.r.t. link 4 and by equating them , we get:

w2*(length of line segment I 24 I 12) = w4*(length of line segment I 24 I 14)

putting the values of lengths from the diagram , we get w4=4.167 rad/s

and since I 12 and I 14 lie on the same side of I 14 in the diagram (refer image) , we can say that w2 and w4 have the same direction i.e. both are in counter clockwise direction .

2. To find the angular velocity of link 5  

Velocity of I 25 can be written in two ways i.e w.r.t. link 2 and also w.r.t link 5 and by equating them , we get :

w2*(length of line segment I 25 I 12) = w5 *( length of line segment I 25 I 15)

putting the values of lengths from the diagram , we get w5= 1.103 rad/s

and since I 12 and I 25 lie on the same side of I 25 in the diagram (refer image) , we can say that w2 and w5 have same sense of rotation i.e. in counterclockwise(ccw) direction.

Now to find out angular velocity of link 4 w.r.t link 2 , taking the relative angular velocity is w4 -w2 , we get -5.833 rad/s in counter clockwise i.e +5.833 in clockwise direction , i.e. link 4 is rotating clockwise w.r.t link 2 with angular speed 5.833 rad/s.