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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 later half of the rise period. Assuming that the magnitudes of the acceleration and deceleration are same, the maximum velocity of the follower is
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4hω δ
- hω
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2hω δ
- 2hω
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Correct Option: C
Here, out stroke angle θ0 = δ
and stroke length = h
Angular velocity = ω
V = u + at
(Vo)max = o + a × | = | ......(1) | ||
2 | 2ω |
= O + | × a × | 2 | |||||
2 | 2 | 2 |
h = | |
4ω2 |
a = | .......(2) | |
δ2 |
From (1) & (2), we get
(Vo)max = | = | |||
2ω | δ |