L1 = 6 mm L2 = 1.5 mm L3 = 7 mm L4 = 3 mm theta2 = 40 degree For the above four
ID: 2995513 • Letter: L
Question
L1 = 6 mm L2 = 1.5 mm L3 = 7 mm L4 = 3 mm theta2 = 40 degree For the above four bar mechanism for crank angle 40degree, determine the angular velocity of CD, omega 4 using purely analytical method. Determine omega 4 using the empirical formula in the textbook and compare the answer in part (a) Determine the angular acceleration of BC, alpha using purely analytical method. Determine alpha3 using the empirical formula in the textbook and compare the answer in part (c) BD = - 2(L1)(L2)cos(theta2) (4.9) y = cos - 1( - BD2/2(L3)(L4)) (4.10) theta3 = 2tan - 1[ - L2sintheta2 + L4siny/L1 + L3 - L2costheta2 - L4cosy] (4.11) theta4 = 2tan - 1[L2sintheta2 - L3siny/L2costheta2 + L4 - L1 - L3cosy] (4.12) As presented in Chapter 6, the velocity equation are omega3 = - opmega2[L2sin(theta4 - theta2/L3siny] (6.14) omega4 = - omega2[L2sin(theta3 - theta2)/L4siny] (6.15) The acceleration equation can be presented as alpha 3 = alpha2L2sin(theta2 - theta4) + omega L2cos(theta2 - theta4) - omega L4 + omegaL3cos(theta4 - theta3)/L3sin(theta4 - theta3) (7.23) alpha4 = alpha2L2sin(theta2 - theta3) + omega L2cos(theta2 - theta3) - omega L4cos(theta4 - theta3) + omega L3/L4sin(theta4 - theta3) (7.24) BD = - 2(L1)(L2)cos(theta2) (4.9) y = cos - 1( - BD2/2(L3)(L4)) (4.10) theta3 = 2tan - 1[ - L2sintheta2 + L4siny/L1 + L3 - L2costheta2 - L4cosy] (4.11) theta4 = 2tan - 1[L2sintheta2 - L3siny/L2costheta2 + L4 - L1 - L3cosy] (4.12) The velocity equations are as follows [Refs. 10,11,12,14]: omega3 = - omega2[L2sin(theta4 - theta2)/L3siny] (6.14) omega 4 = - omega2[L2sin(theta3 - theta2)/L4siny] (6.15)Explanation / Answer
solution for angular velocity of CD
omega3=?
omega3=-omega2*(L2*sin(theta3-theta2)/sin(gamma))
omega2=100;
BD=sqroot(L1^2+L2^2-2*L1*L2*cos(theta2))=4.95 mm
theta3=using equation 4.11=10.17 degrees
similarly
theta4=using equation 4.12=47.17 degrees
gamma=cos-1((L32+L42-BD2)/(2*L3*L4))=37 degrees
on putting those values in equation 6.14 omega3 =-10.37 rad/s
on putting those values in equation 6.15 omega4 =41.34 rad/s
to find alpha3
use equation 7.23 or differentiate the equation 6.14
alpha3=11854.49
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