For a follower moving with cycloidal motion, the velocity of the
A. Beginning of the outstroke
B. End of outstroutstroke
C. Middle of the outstroke
D. None of the above
Please scroll down to see the correct answer and solution guide.
Right Answer is: C
SOLUTION
Explanation:
Cycloidal motion:
The displacement equation of a cycloidal motion of follower is given by:
\(s = \frac{h}{\pi }\;\left( {\frac{{\pi \theta }}{\phi } - \frac{1}{2}\sin \frac{{2\pi \theta }}{\phi }} \right)\)
Velocity equation for cycloidal motion is given by:
\(v = \frac{{ds}}{{d\theta }} \times \frac{{d\theta }}{{dt}}\)
\(v = \frac{{hω }}{{{ϕ }}}\;\left[ {1\; - cos\left( {\frac{{2\pi θ }}{{{ϕ }}}} \right)} \right]\)
where h = displacement of follower, θ = Angular displacement of Cam, ϕ = Angle of ascent, ω = Angular speed of cam.
∴ maximum velocity attainable by the follower will be at \(\theta = \frac{{{\phi }}}{2}\)
and maximum velocity is given by
\({v_{max}} = \frac{{2h\omega }}{{{\phi }}}\;\)
Now, from the displacement and velocity profile of follower as shown in fig. below
∴ we can conclude from the figure that the velocity of the follower is zero at the beginning and at the end of its stroke and increases gradually to a maximum at mid-stroke.
