Question

1. Use the velocity triangle method to determine the angular velocity of the follower or rocker (Link 4 or 04C) at the time instant marked with the current position and configuration as illustrated in the following figure. Note that all lengths depicted for link 1 to 4 are depicted as O2B=1.875 in, BC-2 in, CO:=1.25 in, 020.=3.5 in with the driver (Link 2) angular velocity 02=1 rad/sec. Identify the degrees of freedom (DOF) and instant center between Link 3 and Link 1. (10 points)

Answer 1

In the images provided below, a velocity triangle is constructed based on the configuration presented above. Based on the velocities of the links obtained, the angular velocity of link 4 is calculated using the relation with the tangential velocity that is calculated using the velocity triangle. Further, the degree of freedom of the mechanism is calculated using Kutzbach Equation.

Given: O2 B-1.875in, BC= 2in; CO = 1.25n; 904= 3.5in, W2=/rade B 2 1.875inAW, in 1.25in -3.5int 04 Vo = Q x B O₂ = 1x1.875 inch rad/sec = 1.875 inch rad/sec Using Velocity diagram, = 2.84in o z veb-2.61 2.84 in Ware - Veo - 2.8t in = 2.27rad/sec Therefore angular velocity of follower/rocker is 2.27 rad/sec with respect to worrent position.

Degree of Freedom (F)= 3(1-1)-2;-ch where la no. of links; ja no. of joints, th- no of higher pair In our case, l= 4; ;=4; h=0 Therefore , F = 3 (4-1)-2 (4) -0 = 3(3) - 2 (4) 1 degree of freedonn as per The following mechanism has Kutzbach Equation.