Answer to Question #196811 in Mechanical Engineering for muhammed mudhasir

Question

.A cam is to be designed for a knife edge follower with the following data:

1. Cam lift = 60 mm during 90° of cam rotation with simple harmonic motion.

2. Dwell for the next 30°.

3. During the next 60° of cam rotation, the follower returns to its original position with simple harmonic motion.

4. Dwell during the remaining 180°.Draw the profile of the cam when the line of stroke is

offset 20 mm from the axis of the cam shaft. The radius of the base circle of the cam is 50 mm. Determine the maximum velocity and acceleration of the follower during its ascent and descent, if the cam rotates at 250 r.p.m.

Answer 1

Maximum velocity of the follower during its ascent and descent

We know that angular velocity of the cam

$ω = \frac{2πN}{60} =\frac{2π*240}{60} = 25.14 rad/s.$

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

$v_O=\frac{πωS}{2θ_O} =\frac{π×25.14×0.04}{2×1.571}= 1 m/s.$

and maximum velocity of the follower during its descent

$v_R =\frac{πωS}{2θ_R} =\frac{π×25.14×0.04}{2×1.047} = 1.51m/s.$

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

$a_O=\frac{π^2ω^2}{2(θ_O)^2} =\frac{π^2×25.14^2}{2×1.571^2}= 50.6 m/s^2.$

and maximum acceleration of the follower during its descent

$a_R =\frac{π^2ω^2}{2(θ_R)^2} =\frac{π^2×25.14^2}{2×1.047^2} = 113.8m/s^2$