The diagram in Figure 1 shows a 4-bar linkage system. The links are initially at
ID: 1841879 • Letter: T
Question
The diagram in Figure 1 shows a 4-bar linkage system. The links are initially at rest at theta_1 = -90 degree when a motor at A applies a torque, tau. Each link is 1 m long and each has a mass of 1 kg. Determine the work done by the motor to raise the 4-bar linkage to theta_1 = 90 degree and an angular velocity of 10 rad/s. If the motor applied a constant torque, tau what is the value of tau? For an electric motor with 95% efficiency, what is the input power given parts a and b when theta_1 = 90 degree ?Explanation / Answer
solution:
for a
1)when body rotate at constant velocity means angular accelaration is zero and hence torque due to inertia of rod is zero,hence torque is sum of torque of weight of rod AB at its C.G. and torque due to system weigth at B.
weigth at B=weight of BC and weight CD
as this weight has to raise by motor
hence torque at pivot A is
T=Wab*l1/2+(Wbc+Wcd)l1=1*9.81*.5+2*9.81*1=24.525 N m
where displacement is angle s=pi radians
hence work done=T*S=24.525*pi=77.047 N m
where input power is,w=10 rad/s
P=T*w=24.525*10=245.25 J
2)when mechanism is rotating with constant torque means at constant accelaration,then we have to consider inertia due to rotating masses as it is acting at point B for link BC and CD as point mass and induce torque around A.as follows
inertia=Iab+I(bc,cd)
I=ml1^2/3+(mbc+mcd)l1^2
I=2.333 kg m2
where for s=pi and w=10 angular accelaration is given by
w^2=wo^2+2*a*s
wo=0
a=w^2/2*s=15.91 rad/s2
hence torque is
T=I*a=37.1356 N m
and power is
P=T*w=371.356 J