Answer to Question #260153 in Mechanical Engineering for Suman

Question

Draw the profile of a cam in which the follower moves with simple harmonic motion during ascent while it moves with uniform velocity motion during descent. The other data is as follows.

Least radius of cam = 50mm; Angle of ascent = 480; Angle of dwell between ascent and descent = 420; Angle of descent = 600; Lift of follower = 40mm; Diameter of roller = 30mm; If the cam rotates at 360 r.p.m. anticlockwise, find the maximum velocity and acceleration of the follower during descent.

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

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