Question

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 of the interval ?

(a) 188.6 m/s

(b) 235.5 m/s

(c) 300 m/s

(d) 400 m/s

The question was asked in an international level competition.

My enquiry is from Acceleration of a Particle along a Circular Path in section Kinematics & Kinetics of Motion of Machine Kinematics

Answer 1

The correct choice is (a) 188.6 m/s

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