Page 584 - Dynamics of Mechanical Systems

0593_C16_fm Page 565 Tuesday, May 7, 2002 7:06 AM

Mechanical Components: Cams 565

Hence, h(θ)becomes:

h θ () = h +  h − h      2 π  ( θθ ) − sin   2 π  ( θθ )   (16.13.9)
−
−
1
2
1
 2 π    θ − θ  1  θ − θ  1 
2
1
1
2
or
−
 h − h   h − h    θθ  
h θ () = h + 2 1 ( θθ ) − 2 1 sin 2 π 1 (16.13.10)
−
1   θ − θ  1  2 π      θ − θ   
1
1
2
2
Then, by differentiation, we obtain the follower velocity as:
v = dh = dh dθ = ω dh (16.13.11)
θ
dt d dt dθ
where ω is the (constant) angular speed of the cam. Hence,
  h − h   h − h   θθ  
−
v = ω   2 1  −  2 1  cos 2 π  1   (16.13.12)
  θ 2 − θ 1   θ 2 − θ 1   θ 2 − θ 1  
Similarly, for the follower acceleration we have:

−
2
2
a = dh = ω 2 dh = ω 2     h − h   2π   sin 2π  θθ 1   (16.13.13)
1
2

 
 
dt 2 dθ 2   θ 2 − θ 1   θ 2 − θ 1   θ 2 − θ 1  
Finally, for the jerk we obtain:
2
−
3
3
jerk = dh = ω 3 dh = ω 3     h − h   2π   2 cos 2π  θθ 1     (16.13.14)
1
2
 

dt 3 dθ 3    θ 2 − θ 1   θ 2 − θ 1   θ 2 − θ 1   
Observe that at the transition points (θ = θ and θ = θ ) the jerk has the finite value:
2
1
ω 3   h − h 1   2 π   2
 
2
 θ − θ   θ − θ 
2
2
1
1
Observe further from Eq. (16.13.13) that the maximum values of the follower acceleration
occur at cam rotation angles where the sine argument 2π(θ – θ )/(θ – θ ) has values π/4
1
1
2
and 3π/4. At these angles, the magnitude of the maximum acceleration a max is:
θ
a = 2πω 2 h θ − ) 2 (16.13.15)
h − ) ( 2
max ( 2 1 1
From Eqs. (16.12.11) and (16.12.12) we see that for the sinusoidal rise function the maxi-
2
2
2
mum follower acceleration is (π /2)(h – h )/(θ – θ ) . Because 2π is greater than π /2, we
1
2
1
2
see that the maximum follower acceleration of the cycloidal rise function is larger than