Question

Two shafts with an included angle of 160° are connected by a Hooke’s joint. The driving shaft runs at a uniform speed of 1500 r.p.m. The driven shaft carries a flywheel of mass 12 kg and 100 mm radius of gyration. Find the maximum angular acceleration of the driven shaft.

(a) 3090 rad/s^2

(b) 4090 rad/s^2

(c) 5090 rad/s^2

(d) 6090 rad/s^2

This question was addressed to me in semester exam.

I need to ask this question from Double Hooke’s Joint topic in division Friction of Machine Kinematics

Answer 1

Correct answer is (a) 3090 rad/s^2

The explanation is: Given : α = 180° – 160° = 20°; N = 1500 r.p.m.; m = 12 kg ; k = 100 mm = 0.1 m

We know that angular speed of the driving shaft,

ω = 2 π × 1500 / 60 = 157 rad/s

and mass moment of inertia of the driven shaft,

I = m.k^2 = 12(0.1)^2 = 0.12 kg – m^2

Let dω1 / dt = Maximum angular acceleration of the driven shaft, and

θ = Angle through which the driving shaft turns.

We know that, for maximum angular acceleration of the driven shaft,

cos 2θ = 2sin^2α/2 – sin^2α = 2sin^220°/2 – sin^220° = 0.124

2θ   = 82.9° or θ = 41.45°

and dω1 / dt = ω^2cosα sin2θsin^2α/(1 – cos^2θsin^2α)^2

                        = 3090 rad/s^2.