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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 end of the interval ?

(a) 235.5 m/s

(b) 335.5 m/s

(c) 435.5 m/s

(d) 535.5 m/s

I have been asked this question during an internship interview.

My doubt is from Numericals On Kinematics Of Motion topic in division Kinematics & Kinetics of Motion of Machine Kinematics

1 Answer

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by (106k points)
Right answer is (a) 235.5 m/s

The explanation: 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

Linear velocity at the end of 5 seconds

We also know that linear velocity after 5 seconds,

v5 = r . ω = 1.5 × 157 = 235.5 m/s

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