1. To move outwards through 40 mm during 100 deg rotation of the cam.

2. To dwell for next 80 deg.

3. To return to its starting position during next 90 deg, and

4. To dwell for the rest period of a revolution i.e.90 deg.

Draw the profile of the cam when line of stroke of the follower passes through the centre of the cam shaft. the displacement of follower is to take place with uniform acceleration and uniform retardation. Determine the maximum velocity and acceleration of the follower when the cam shaft rotates at 900 rpm.

Draw the displacement, velocity and acceleration diagrams for one complete revolution of the cam.

Given :-S =40 mm,= 0.4m; $\theta_{0}=100 ^{\circ}=100\times\frac{\pi}{180}$ =1.745 rad; $\theta_{R}=90 ^{\circ}=\frac{\pi}{2}$ = 1.571 rad; N=900 rpm.

Maximum velocity of the follower during out stroke and return stroke.

We know that angular velocity of the cam shaft,

$\omega=\frac{2\pi N}{60}=\frac{2\pi \times 900}{60}$ = 94.26 rad/s.

$v_{o}=\frac{2\omega S}{\theta_{0}}=\frac{2\times94.26\times0.04}{1.745}$ = 4.32 m/s.

and maximum velocity of the follower during return stroke,

$v_{R}=\frac{2\omega S}{\theta_{R}}=\frac{2\times94.26\times0.04}{1.571}$ = 4.8 m/s.

The velocity diagram as shown in fig. (b)

Maximum acceleration of the follower during out stroke and return stroke.

We know that the maximum acceleration of the during out stroke,

$a_{o}=\frac{4\omega^2 S}{\theta_{0}^2}=4(94.26)^{2}\times0.04/(1.145)^{2}$ = 467 $m/s^{2}$

and maximum acceleration of the follower during return stroke,

$a_{R}=\frac{4\omega^2 S}{\theta_{R}^2}=4(94.26)^{2}\times0.04/(1.571)^{2}$ = 576 $m/s^{2}$.

The acceleration diagram is shown in fig. (c)