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## Q. 8.8

Planetary Geartrain Speeds

The planetary geartrain has two inputs (the sun gear and carrier) and one output (the ring gear). (See Figure 8.41.) When viewed from the right-hand side, the hollow carrier shaft is driven at 3600 rpm clockwise, and the shaft for the sun gear turns at 2400 rpm counterclockwise. (a) Determine the speed and direction of the ring gear. (b) The carrier’s shaft is driven by an electric motor capable of reversing its direction. Repeat the calculation for the case in which it is instead driven at 3600 rpm counterclockwise. (c) For what speed and direction of the carrier’s shaft will the ring gear not rotate?

Approach
To fi nd the speed and direction of the appropriate gear in each operating condition, we apply Equations (8.24)

$\omega _{s}+n\omega _{r} -(1+n)\omega_{c}=0$                          (8.24)

and (8.25),

$n=\frac{N_{r}}{N_{s}}$                        (8.25)

together with the sign convention shown in Figure 8.36. With clockwise rotations being positive, the known speeds in part (a) are  $\omega _{c} = 3600$rpm  and $\omega _{s} = - 2400$rpm. In part (b), when the carrier’s shaft reverses direction,$\omega _{c} = - 3600$ rpm.

## Verified Solution

(a) Based on the numbers of teeth shown in the diagram, the form factor is

$n = \frac{100 \ teeth}{20 \ teeth} \longleftarrow [n=\frac{N_{r}}{N_{s}}]$

$= 5 \ \frac{\cancel {teeth}}{\cancel {teeth}}$

= 5

Substituting the given quantities into the planetary geartrain design
equation, [ $\omega _{s} + n\omega _{r} -(1+n)\omega_{c}=0$]

$(-2400 rpm) + 5\omega _{r} – 6(3600 rpm) = 0$

and $\omega _{r} = 4800$ rpm. Because $\omega _{r}$ is positive, the ring gear rotates clockwise.

(b) When the carrier’s shaft reverses, $ω_{c} = -3600$ rpm. The speed of the ring gear using the planetary geartrain design equation [ $\omega _{s} + n\omega _{r} -(1+n)\omega_{c}=0$] becomes

$(-2400 rpm) + 5\omega _{r} – 6(-3600 rpm) = 0$

and$ω_{r} = – 3840$ rpm. The ring gear rotates in the direction opposite that in part (a) and at a lower speed.

(c) The speeds of the sun gear and carrier are related by the planetary geartrain design equation [$\omega _{s} + n\omega _{r} -(1+n) \omega_{c}=0$ ], and at that special condition

$(-2400 rpm) – 6ω_{c} = 0$

and $ω_{c} = – 400$ rpm. The carrier’s shaft should be driven counterclockwise at 400 rpm for the ring gear to be stationary.

Discussion
The form factor is a dimensionless number because it is the ratio of the numbers of teeth on the ring and sun gears. Because this calculation involves only rotational speeds, not the velocity of a point as in Equation (8.2),

$v=rω$            (8.2)

it is acceptable to use the dimensions of rpm rather than convert the units for angular velocity into radians per unit time. These results demonstrate the complicated yet flexible nature of planetary geartrains to produce a wide range of output motions.

Clockwise carrier:  $ω_{r}$  = 4800 rpm (clockwise)

Counterclockwise carrier:  $ω_{r}$= 3840 rpm  (counterclockwise)

Stationary ring: $ω_{c}$ =  400 rpm (counterclockwise)