+1 vote
in Machine Kinematics by (38.0k points)
A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the linear velocity at the beginning?

(a) 288.6 m/s

(b) 388.6 m/s

(c) 488.6 m/s

(d) 188.6 m/s

This question was posed to me at a job interview.

Asked question is from Numericals On Kinematics Of Motion in chapter Kinematics & Kinetics of Motion of Machine Kinematics

1 Answer

0 votes
by (106k points)
Correct choice is (d) 188.6 m/s

The explanation is: Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;

N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s

Linear velocity at the beginning

We know that linear velocity at the beginning,

v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s

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