93 A cam drive is required for a shaker platform This platfo
Solution
solution:
1)here total cycle time is t=.7+.2+.56=1.4 sec and angle turned is 2*pi
hence required angular velocity
w=angle/time=2*pi/1.4=4.4879 rad/s
2)here equation of motion for constant accelaration are
for process angle turned =m
angle of cam roation=n where 0<n< 2*pi
3)displacemnt ,velocity and accelartion is given by
H=1 \'\', for 0<n<m/2
S=2*H(n/m)^2
V=4*H*w(n/m^2)
a=4*H(w/m)^2
for m/2<n<m
S=H[1-2(1-(n/m))^2]
V=4H(w/m)[1-(n/m)]
a=4*H(w/m)^2
this is equation for rise and fall they avelocity and accelaration relation are same but displacemnt has to subtract from rise present at start of cycle
5)for given operation of cam in given question equation of motion are
for 0<n<pi/2 or 0<t<.7
S=.2026n^2
V=1.818n
a=8.1629 rad/s2
for pi/2<n<pi
S=[1-.2026(pi-n)^2]
V=1.818(pi-n)
a=-8.1629 rad/s2
6) for pi<n<4.039 or for dwell .7<t<.9
s=1\'\'
v=a=0
7)for fall equation of motion are
for 4.039<n<2.2439/2
S=1-.3972n^2
V=3.5653n
a=16 rad/s2
for 2.2439/2<n<2pi
S=1-.3972(2.2439-n)^2
V=3.5653(2.2439-n)
a=-16 rad/s2
7)here using this equation we can easily find displacement ,velocity and accelaration of cam at any instant
here matlab code is
n=[0:.001:pi/2]
S=.2026n^2;
V=1.818n;
a=8.1629;
n=[pi/2:.001:pi]
S=[1-.2026(pi-n)^2];
V=1.818(pi-n);
a=-8.1629;
n=[pi:.001:4.039]
s=1;
v=0;
a=0;
n=[4.039:.001:5.1610]
S=1-.3972n^2;
V=3.5653n;
a=16;
n=[5.1610:.001:2*pi]
S=1-.3972(2.2439-n)^2;
V=3.5653(2.2439-n);
a=-16;
plot(S,n);
title(\'displacement curve\')
plot(V,n);
title(\'velocity curve\')
plot(a,n);
title(\'accelaration curve\')
7)in this way using matlab code we can find displacement,velocity and accelaration curve more easily



