The dimensions for the mechanisms shown in the gures are given in the tables. Th
ID: 1849435 • Letter: T
Question
The dimensions for the mechanisms shown in the gures are given in the tables. The driver link AB rotates with a constant angular speed of n rpm. For the instant when theta = theta 1 = pi/4 find:
the velocity and acceleration of point B on link 2 (vB1 and aB1 );
the angular velocity of link 2 (w2);
the relative velocity and at B (relative linear velocity at B);
the Coriolis acceleration at B;
the relative angular velocity and acceleration of link 2 with respect to link 1;
the angular velocity of link 4 (w4).
AB = 0.20 ; % m
AD = 0.30 ; % m
CD = 0.60 ; % m
CE = 0.30 ; % m
yE = 0.40; % m
phi = pi/4; % rad
n = 60/pi; % (rpm)
i have the answers but i just need to understand the steps. any help will be greatly appreciated
Explanation / Answer
4.2 VELOCITY AND ACCELERATION ANALYSIS OF MECHANISMS-1 Velocity and acceleration analysis of mechanisms can be performed vectorially using the relative velocity and acceleration concept. Usually we start with the given values and work through the mechanism by way of series of points A, B, C, etc. Solving vector equations in the form: VB = VA+ VB/A VC= VB+ VC/B etc. for velocity and aB= aA+atB/A+anB/A aC= aB+atC/B+anC/B etc for acceleration. The points that one has to use are usually the revolute joint axes between the links since these are the points where the relative velocity or acceleration between the two coincident points on two different links are zero and they have equal velocity and accelerations. If we are to determine the velocity of a point on a link we must first determine the velocity of the points located at the joint axes. For the graphical solution, we can draw the vector equations for the velocity and acceleration of point C directly on the velocity and acceleration polygons drawn for the loop equations. Note that the velocity and acceleration of point C cannot be determined before solving the velocity and acceleration loop equations. From the tip of the vector VB, if we draw the velocity vector VC/B, the vector joining the starting point of VB to the tip of the vector VC/B will give us the velocity of point C. Similarly, The acceleration of point C can be determined by drawing the acceleration vectors anC/B and atC/Bstarting from the tip of the vector aB .The acceleration vector from the starting point of aB to the tip of the vector atC/B , will give us the acceleration of point C. Alternatively, note that the position of point C could have been written as: (15) by differentiation: (16) (17) Which are velocity and acceleration vector equations: VC=VA+VC/A (18) and aC = atA + anA + atC/A + anC/A (19) (15) Let us label the tips of the vectors of the velocity and acceleration polygon by a lower case letter corresponding to the point whose velocity is represented (in case of acceleration the acceleration of point A and C are the sum of two or more acceleration vectors). The tips of the velocity vector VA,VB, VC and aA,aB, aC form triangles abc. We have an important theorem known as Mehmke's Theorem or "The principle of the velocity and acceleration