Question

A certain machine requires a torque of (5000 + 500 sin θ) N-m to drive it, where θ is the angle of rotation of shaft measured from certain datum. The machine is directly coupled to an engine which produces a torque of (5000 + 600 sin 2 θ) N-m. The flywheel and the other rotating parts attached to the engine has a mass of 500 kg at a radius of gyration of 0.4 m. If the mean speed is 150 r.p.m., 

find : 

1.  the fluctuation of energy,

2.  the total percentage fluctuation of speed, and 

3.  the maximum and minimum angular acceleration of the flywheel and the corresponding shaft position.

Answer 1

Given : T₁ = (5000 + 500 sinθ) N-m; T₂ = (5000 + 600 sin 2θ) N-m; m = 500 kg; k = 0.4;N = 150 r. p. m. or ω = 2π x 150/60 = 15.71 rad/s

1.  the fluctuation of energy,

We know that change in torque

T₂ - T₁ = (5000 + 600 sin 2θ) - (5000 + 500 sinθ)

= 600 sin 2θ - 500 sinθ

This change is zero when

The turning moment diagram is shown in Fog . The maximum fluctuation of energy lies between C and D (i.e. between 180 and  249.6), as shown shaded in FIG,

Maximum fluctuation of energy,

2.  Total percentage fluctuation of speed, 

Let C_s =  Total percentage fluctuation of speed.

We know that maximum fluctuation of energy (ΔE),

3. Maximum and minimum angular acceleration of the flywheel and the corresponding shaft positions

Change in torque, T = T₂ - T₁ = (5000 + 600 sin 2θ) - (5000 + 500 sinθ)

= 6000 sin 2θ - 500 sinθ  ....(i)

Differentiating this expression with respect to θ and equating to zero for maximum or minimum values.

Substituting θ = 35° in equation (i)

T_max = 600 sin 70° - 500 sin 35° = 277 N-m

Substituting θ = 127.6° in equation (i) , we have minimum torque,

T_min = 600 sin 255.2° - 5000 sin 127.6° = 976 N-m

We known that maximum acceleration,

and minimum acceleration,