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1. Coefficient of Speed Fluctuation It is the difference b/n the max. & min. speeds of the driven shaft expressed as a ratio of the driving shaft speed for constant angle α b/n the axes of the driving & driven shafts. For small angle angle α , sin α = α and tan α = α , hence

2. Angular Acceleration of Driven shaft Differentiating the above expression w.r.t. time will gives us the angular acceleration of the driven shaft

3. The negative sign does not show that there is always retardation but it depends upon the value of sin 2θ For angular acceleration to be maximum, differentiate dω b / dt with respect to θ and equate to zero. Then, the result is approximated as:

4. Note: For small values of α (less than 30°), then the above expression may approximately be written as From which the value of θ for the max. angular acceleration will be determined & can be substituted in the equation of acceleration, to obtain the max. angular acceleration.

5. Double Universal or Hook’s Joint We know that velocity of the driven shaft is not constant , but varies from maximum to minimum values. In order to have a constant velocity ratio of the driving and driven shafts, an intermediate shaft with a Hooke’s joint at each end as shown in Fig . is used. This type of joint is known as double Hooke s ’s joint .

6. Let the driving, intermediate & driven shafts, in the same time, rotate through angles θ , φ and γ respectively . Then from equation (iii) as discussed previously. we will have: From the above two equations one can see that This shows that the speed of the driving and driven shafts are equal & constant . 1. The axes of driving & driven shafts are in the same plane, & 2 The driving and driven shafts make equal angles with the intermediate shaft

7. Quiz 10% An epicyclic gear train is shown In fig. Find out the rpm of pinion D if arm A rotate at 60 rpm in anticlockwise direction. No of teeth on wheels are given below.