Answer to Question #196813 in Mechanical Engineering for muhammed mudhasir

Question

.From the following data draw the profile of a cam in which the follower moves with simple harmonic motion (SHM) during

ascent while it moves with uniformly accelerated and decelerated motion during descent.

least radius of cam = 50 mm.,

angle of ascent = 48o

angle of dwell b/w ascent & descent= 48o

angle of descent = 60o

lift of follower = 40 mm.

diameter of roller= 30 mm

If the cam rotates at 360 r.p.m anti-clockwise, find the maximum velocity & acceleration of the follower

during descent. The follower axis passes through the axis of cam shaft.

Answer 1

$\theta _0=48^0=0.837 rad$

$\theta _D=42^0=0.73 rad$

$\theta _0=60^0=1.04 rad$

N=250 rpm

$\omega =\frac{2 \pi N}{60}=\frac{2 \pi *360}{60}=37.69911 rad/s$

$S=40 mm=0.04m$

Maximum velocity of the follower on the out stroke

$V_0 =\frac{ \pi \omega S}{2\theta}=\frac{ \pi *37.6991*0.04}{2*0.837}=2.82999m/s$

Maximum velocity of the follower on the return stroke

$V_r =\frac{ \pi \omega S}{2\theta}=\frac{ \pi *37.6991*0.04}{2*1.04}=2.27760m/s$

Maximum acceleration of the follower on the out stroke

$a_0 =\frac{ \pi^2 \omega^2 S}{2\theta ^2}=\frac{ \pi^2 *37.6991^2*0.04}{2*0.837^2}=259.37315m/s^2$

Maximum acceleration of the follower on the return stroke

$a_r=\frac{ \pi \omega S}{2\theta}=\frac{ \pi^2*37.6991^2*0.04}{2*1.04^2}=259.37315 m/s^2$