Question #260153

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.

1
Expert's answer
2021-11-03T04:30:02-0400

θ0=480=0.837rad\theta _0=48^0=0.837 rad

θD=420=0.73rad\theta _D=42^0=0.73 rad

θ0=600=1.04rad\theta _0=60^0=1.04 rad

N=250 rpm

ω=2πN60=2π36060=37.69911rad/s\omega =\frac{2 \pi N}{60}=\frac{2 \pi *360}{60}=37.69911 rad/s

S=40mm=0.04mS=40 mm=0.04m

Maximum velocity of the follower on the out stroke

V0=πωS2θ=π37.69910.0420.837=2.82999m/sV_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

Vr=πωS2θ=π37.69910.0421.04=2.27760m/sV_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

a0=π2ω2S2θ2=π237.699120.0420.8372=259.37315m/s2a_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

ar=πωS2θ=π237.699120.0421.042=259.37315m/s2a_r=\frac{ \pi \omega S}{2\theta}=\frac{ \pi^2*37.6991^2*0.04}{2*1.04^2}=259.37315 m/s^2

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