GATE Mechanical Engineering (ME) 2018 Shift 2 Solved Paper

Question : 48 of 65

Marks: +1, -0

In a cam-follower, the follower rises by

h

as the cam rotates by

δ

(radians) at constant angular velocity

ω

(radians\/s). The follower is uniformly accelerating during the first half of the rise period and it is uniformly decelerating in the latter half of the rise period.
Assuming that the magnitudes of the acceleration and deceleration are same, the maximum velocity of the follower is

$\;{4 h ω}/{δ}$
hω
$\;{2 h ω}/{δ}$
$2 h ω$

Validate

Solution:

time,

t=\;{δ {/} 2}/{ω}

Since the cam is having uniform velocity, first half of rise will be accelerating, other half will be decelerating. Hence velocity will be maximum at height

= {h} {/} 2

h {/} 2=0 × t+\;{1}/{2} a t^{2}

a=\;{h}/{t^{2}}

v=u+a t

v=a t=\;{h}/{t^{2}} × t=\;{h}/{t}=\;{2 h ω}/{δ}

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