# 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 tangential component of the acceleration of the mid-point of the bar after 5 seconds after the acceleration begins ?

+1 vote
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 tangential component of the acceleration of the mid-point of the bar after 5 seconds after the acceleration begins ?

(a) 18287 m/s^2

(b) 18387 m/s^2

(c) 18487 m/s^2

(d) 18587 m/s^2

The question was posed to me by my college director while I was bunking the class.

Question is taken from Numericals On Kinematics Of Motion topic in division Kinematics & Kinetics of Motion of Machine Kinematics

by (106k points)
Right option is (c) 18487 m/s^2

Easiest 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

Let α = Constant angular acceleration.

We know that ω = ω0+ α.t

157 = 125.7 + α × 5 or α = (157 – 125.7) /5 = 6.26 rad/s^2

Radius corresponding to the middle point,

r = 1.5 /2 = 0.75 m

∴ Tangential acceleration = α. r = 6.26 × 0.75 = 4.7 m/s^2

Radial acceleration = ω^2 . r = (157)^2 0.75 = 18 487 m/s^2