Answer to Question #196814 in Mechanical Engineering for muhammed mudhasir

Question

A symmetrical tangent cam operating a roller follower has the following particulars: Radius of the

base circle of cam = 40 mm, roller radius = 20 mm, angle of ascent= 75o

, total lift = 20 mm, speed of

the cam shaft = 300 rpm. Determine: 1.the principle dimensions of the cam; 2.acceleration of the

follower at the beginning of the lift; and 3. acceleration of the follower where straight flank merges

into a circular nose.

Answer 1

r₁ = 40 mm

r₃ = 20 mm

Lift = x = 20 mm

$\omega$ = 300 rpm = 31.42 rad/s

α = 75°

OP + PT = OC + CT

OP = OC + CT- PT

OP = r₁ + x - r₂

OP = 60 - r₂

From the figure above;

OQ + QA = OA

OQ = OA - QA = r₁ - r₂ = 40 - r₂

From the below;

Cos a = OQ/OP

Substitute OQ & OP

Cos 75 = 40-r₂/ 60-r₂

r₂ = 33.02 mm

OP = d = 40 + 20 - 33.02 = 26.98 mm

From the triangles, GOB and POQ,

tanΦ = GB/OB = PQ/OB

= OP Sina/ OB

= d Sin 75/ (r1 + r3)

= 33.02 Sin 75/60

Φ = 31.34°

Acceleration of the follower,

(i) At the beginning of the lift, i.e., when θ=0

$a = ω²(r_1 + r_3) \dfrac{2-cos²\theta}{cos³\theta}$

= (31.42)² (40 + 20)(2-1)

= 59.23 m/s²

(ii) At the end of flank, i.e., when θ=Φ=25.6

$a = ω²(r_1 + r_3) \dfrac{2-cos²\theta}{cos³\theta}$

a = 59.23 × $(\dfrac{2-cos ² 25.6}{cos³ 25.6})$ = 100.12 m/s²