Answer to Question #214707 in Mechanical Engineering for Prince

Question

The following particulars relate to a symmetrical tangent cam operating a roller follower :-

Least radius = 36 mm, nose radius = 30 mm, roller radius = 21 mm, distance between cam 

shaft and nose centre = 28 mm, angle of action of cam = 150°, cam shaft speed = 750 r.p.m. 

Assuming that there is no dwell between ascent and descent, determine the lift of the valve 

and the acceleration of the follower at a point where straight flank merges into the circular 

nose.

Answer 1

$Total \space lift \space+r_1=PO+r_2\\ 20+30-16=d-5$

Flank radius, R

$R= \frac{r^2_1-r_2^2+d^2-2r_1 d cos \alpha}{2(r_1-r_2-d cos \alpha)}\\ R= \frac{30^2-5^2+45^2-2*30*45 d cos75}{2(30-5-45 cos 75)}\\ R=82.42 mm$

Flank angle $\phi$

$\frac{PO}{sin \phi}= \frac{PQ}{sin(180- \phi)}\\ sin \phi = \frac{sin(180- 75)}{84.42-5} \implies \phi = 34.2 ^0$

Acceleration at the end of the contact with the flank when $\theta = \phi=29.6^0$

$a=\omega ^2(R-r_1) cos \phi \\ a=(\frac{2 \pi *600}{60}) ^2 (82.4-30)*10^{-3} cos 34.2\\ a=171.09 m/s^2$

Reterdation at the beginning of the contact with the nose

$a=-\omega ^2(R-r_1) cos \phi \\ a=-(\frac{2 \pi *600}{60}) ^2 (45)*10^{-3} cos 29.6\\ a=-146.92 m/s^2$