1. Follower to move outwards through 50 mm during 120 deg of cam rotation,
2. Follower to dwell for next 60o of cam rotation,
3. Follower to returns its starting position during next 90o of cam rotation,
4. Follower to dwell for for the rest of cam rotation.

The minimum radius of cam is 50 mm and the diameter of roller is 10 mm. the line of stroke of the follower is off-set by 20 mm from the axis of the cam shaft. If the displacement of the follower place with uniform and equal acceleration and retardation on both the outward and return strokes, draw profile of the cam and find the maximum velocity and acceleration during out stroke and return stroke.

Subject: Kinematics of Machinery

Topic: CAM Mechanism

Difficulty: High

Given:- N =1000 rpm.; S = 50mm= 0.05m; $\theta_{0}$ = 120$^{\circ}$ = $\frac{2\pi}{3}$ = 2.1 rad: $\theta_{R}$ =90$^{\circ}$ = $\frac{\pi}{rad}$ =1.571 rad.

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

We know that angular velocity of the cam

$\omega$ =$\frac{2\pi N}{60}$ = $\frac{2π\times 1000}{60}$ = 104.7 rad/s

We also know that the maximum velocity of the follower during outstroke,

$v_{o}$= $\frac{2\omega S}{\theta_0}$ =$\frac{2\times 104.7\times 0.05}{2.1}$ = 5 m/s.

and maximum velocity of the follower during return stroke,

$v_{R}$=$\frac{2\omega S}{\theta_R}$ =$\frac{2\times 104.7\times 0.05}{1.571}$ = 6.66 m/s.

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

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

$a_{O}$ = $\frac{4\omega^2 S}{(\theta_O)^2}$ =$\frac{4(104.7)^2\times0.05}{(2.1)^2}$ = 497.2 $m/s^2$

$a_{R}$=$\frac{4\omega^2 S}{(\theta_R)^2}$ = $\frac{4(104.7)^2\times0.05}{(1.571)^2}$ = 888 $m/s^2$ .