estion-2 A cam has a base radius of 4 cm and a roller follower. The prime circle

Question

estion-2 A cam has a base radius of 4 cm and a roller follower. The prime circle diameter is 11 cm and the am rotates at 240 rpm. he motion of the follower is given in table 1 Table 1 Initial angle of Final angle of Rise(+) or (cm) 0 Interval interval interval Type of motion return() Degrees)(Degrees) 0 200 210 210 360 Simple Harmonic Dwell Uniform Velocity Answer the following a) What is the diameter of the roller b) How much time it takes the cam to complete hal rotation (minut c) What will be the velocity and acceleration of the Tollower after the cam has rotated 40 degrees from the 0 degree position What will be the displacement and velocity of the follower after the cam has rotated 220 degrees from the 0 degree position d) e) In which interval follower will have minimum velocity 0 The time (minutes) it takes the follower to rise 40 em g) Determine displacement of follower from O degree position till 40 degrees position (graphically or analytically) and draw the cam profile for this period

Explanation / Answer

(a)

Radius of base circle = Rb = 4 cm

Radius of prime circle = Rp = 11 / 2 = 5.5 cm

Diameter of roller = 2 *( Rp - Rb ) = 2 * (5.5 - 4 ) = 3 cm

..............................................................................................................................................................

(b)

Speed of cam = 240 rpm

Time required for half revolution of Cam = 60 / (240 * 2) = 0.125 sec.

..............................................................................................................................................................

(c) During rising period follower perform simple harmonic motion (SHM).

let , s = follower displacement (instanteneous)

h = maximum follower displacement = 5 cm

omega = angular velocity of cam = 25.12 rad/ sec

phi = cam ratation angle for maximum displacement = 200 degree = 3.488 rad

theta = cam rotation angle (instantaneous) = 40 degree

s = h/2 {1 - cos( pi * theta / phi ) }   

velocity of follower = {( h/2 * pi * omega ) / phi } * sin ( pi * theta / phi )

= {( .05/2 * 3.14 * 25.12 ) / 3.488 } * sin( pi * 40/ 200 )

= 0.332 m/sec  

acceleration of follower = { h/2 * (pi * omega / phi )2 } * cos( pi * theta / phi )

= { .05/2 * (3.14 * 25.12 / 3.488)2 } * cos( pi * 40 / 200 )

= 10.34 m / sec2

................................................................................................................................................................

(d)

for constant velocity follower:

displacement of follower = h * angle between start and instantaneous position /  angle of return period

= 5 * (220 - 210) / (360 - 210)

= 0.33 cm ( from the top position ) OR - 0.33 cm

velocity of follower = h * omega / angle of return period = 0.05 * 25.12 / {(360-210) *3.14/ 180}

= 0.4801 m /sec

...................................................................................................................................................................

(e)

minimum velocity of follower ( in rising period) = {( h/2 * pi * omega ) / phi } * sin (0)

= 0 m/sec ( at begining of rising period ; theta = 0 degree)

Minimum velcotity of follower ( in returning period) =  0.4801 m /sec  

Therfore minimum velocity of follower occures during rising and dwelling period

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.