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A cam is designed to be knife edge follower with the following data:
1. Cam lift =40 mm during 90 deg of cam rotation with simple harmonic motion.
2. Dwell for next 30 deg.
3. During the next 60 deg of cam rotation , the follower returns to its original position with simple harmonic motion.
4. Dwell during the remaining 180 deg.

Draw the profile of the cam when,

a) The line of stroke of the follower passes through the axis of the cam shaft, and

b) The line of stroke is offset 20 mm from the axis of the shaft.

The radius of the base circle of the cam is 40 mm. determine the maximum velocity and acceleration of the follower during its ascent and descent, if the cam rotates at 240 r.p.m.

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Given:- S = 40 mm=0.4m; $\theta_O$= 900$^{\circ}$ = $\frac{\pi}{2}$rad=1.571 rad; $\theta_R$= 600$^{\circ}$ = $\frac{\pi}{3}$rad = 1.047 rad; N= 240 rpm.

(a) Profile of the cam when the line of stroke of the follower passes through the axis of the cam shaft.

(b) Profile of the cam when the line of stroke of the follower is offset 20 mm from the axis of the cam shaft.

Maximum velocity of the follower during its ascent and descent

We know that angular velocity of the cam

$\omega$ = $\frac{2\pi N}{60}$ =$\frac{2\pi \times 240}{60}$ = 25.14 rad/s.

We also know that the maximum velocity of the follower during its ascent

$v_{O}$=$\frac{\pi\omega S}{2\theta_{O}}$ =$\frac{\pi\times 25.14\times 0.04 }{2\times 1.571}$= 1 m/s.

and maximum velocity of the follower during its descent

$v_{R}$=$\frac{\pi\omega S}{2\theta_{R}}$ =$\frac{\pi\times 25.14\times 0.04 }{2\times 1.047}$ = 1.51m/s.

Maximum acceleration of the follower during its ascent and descent

We know that the maximum acceleration of the follower during its ascent,

$a_{O}$= $\frac{\pi^{2}\omega^{2}}{2(\theta_{o})^{2}}$ = $\frac{\pi^{2}(25.14)^{2}\times0.04}{2(1.571)^{2}}$ = 50.6 m/s.

and maximum acceleration of the follower during its descent,

$a_{R}$= $\frac{\pi^{2}\omega^{2}}{2(\theta_{R})^{2}}$ = $\frac{\pi^{2}(25.14)^{2}\times0.04}{2(1.047)^{2}}$= 113.8 m/s.