Design a double dwell cam to move a follower from 0 to 50 mm in 75 degree in a C

Question

Design a double dwell cam to move a follower from 0 to 50 mm in 75 degree in a Cycloidal motion, dwell for 75 degree, fall 50 mm in 75 degree in a constant acceleration motion, and dwell for the remainder. The total cycle must take 5 seconds. Find the displacement (h), velocity, and acceleration of the cam for the following intervals. If the cam must have a base circle of 100 mm, rotates clockwise, and uses a knife-edge follower, determine x coordinate of the cam surface profile (Rx) and the y coordinate of cam surface profile (Ry) at the above times. If the Miller cutter for the knife-edge follower has a radius of 12 mm, find the x coordinate of the cutter center (Cx) and the y coordinate of the cutter center (Cy).

Explanation / Answer

solution:

1) here we have radial cam and knife edge follower with following motion

first cycloidal rise to 50 mm,then dwell which follow by constnt accelaration fall to 0 mm and the again dwell for remain time

2) here total cycle time=5 sec and rotation is 2* pi

hence w=2*pi/time=2*pi/5=1.2566 rad/s

cam angular velocity=1.566 rad/s

3) now i have drawn displacement diagram for this cam motion with displacement as ordinate and time as abscissa, for 1 s=2 cm on abscissa and 1 cm=10 mm on ordinate axis

4) from displacement diagram i get displacement at time mention above as

t=1 s ,s=49 mm

t=1.5 s,s=50 mm

t=2.4 s,s=43 mm

t=4 s,s=0 mm

5) cycloidal motion velocity and accelaration at given time is given by

for t=1 s,angle=72 degree=1.2566 radian

v=2*w*s/angle in radian=2*1.2566*49/1.2566=98 mm/s

and accelaration is given by

a=2*pi*w^2*s/(angle)^2=2*pi*1.2566*1.2566*49/1.2566^2=307.87 mm/s2

6) velocity and accelaration at

t=1.5 s and t=4 s

as it is dwell period hence there is no displacement,hence

v=0 mm/s

a=0 mm/s2

6) velocity and accelaration at t=2.4 s for uniform retardation as

s=43 mm

v=2*w*s/angle

here angle=172.8 degree or 3.015 rad

v=2*1.2566*43/3.015=35.83 mm/s

accelaration=a=4*w^2*s/(angle)^2=4*1.2566^2*43/(3.015)^2=29.85 mm/s2

7) here coordinate of point on cam at angle or given time can be given if angle measured with respect to follower in contact with cam at s=0 mm and it has motion in cc direction

s=displacement,r=base circle radius=100 mm

Rx=(s+r)sin(angle)

Ry=(s+r)cos(angle)

9) for time

t=1 s,angle=72 degree,s=49 mm,r=100 mm

Rx=149sin72=141.70 mm

ry=149cos72=46.04 mm

10) for t=1.5 s,angle=108 degree,s=50 mm

Rx=150sin108=142.65 mm

Ry=150cos108=-46.35 mm

for t=2.4 sec, angle=172.8,s=43 mm

Rx=143sin172.8=17.92 mm

Ry=143cos172.8=-141.87 mm

for t=4,angle=288 s=0

rx=100sin288=-95.10 mm

Ry=100cos288=30.90 mm

11) as miller cutter is placed on another end of folloer then it only has resiproprocating motion and hence it Cx remain constant only cy vary equal to lift of follower

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