Question

For the planetary gear set,
W7
=0 rpm, $\omega 3$ = $\omega$ arm= 280rpm, and the number of teeth in the gears are, N2=27, N5=39, N6=73 and N7=70, what is the angular velocity of gear 2, $\omega 2$ in rpm.

Answer 1

If we take arm stationary and give angular velocity 'a' to gear 2, the 'angular velocity' row of the following table will give you the results for the angular velocities of other gears.

Also the 'Rotations' row gives the results when we consider each of the gears are not rotating about their own centroidal axis and arm rotates at angular velocity 'b'.

Gears	Arm (Gear3)	Gear 2	Gear 4	Gear 5	Gear 6	Gear 7
Angular velocity	0	ta	$-a\frac{N3}{N4}$	N2 N4 N2 ta = a- N4N5 N5	$+a\frac{N2}{N5}$	N2 N6 -a N5 N7
Rotations	b	b	b	b	b	b
Total Rotations	$0+b=b$		$-a\frac{N3}{N4}+b$	$a\frac{N2}{N5}+b$	$a\frac{N2}{N5}+b$	N2 N6 -a N5 N7 + 6

Now, given $\omega _{7}=0$

$\omega _{3}=200$

$N_{2}=27$

  $N_{5}=39$

  $N_{6}=73$

  $N_{7}=70$

Also, the distance between the centers of gear 6 and gear 7 is same as the the distance between the centers of gear 5 and gear 2. And gear 4 lies between gear 5 and gear 2.

Therefore, $m_{67}(\frac{N_{6}+N_{7}}{2})=m_{245(}\frac{N_{5}+N_{2}}{2}+N_{4})$

or, $\frac{m_{67}}{m_{245}}(\frac{73+70}{2})=\frac{39+27}{2}+N_{4}$

or, $\frac{m_{67}}{m_{245}}71.5=33+N_{4}$

where $m_{67}$ is the module of both the gear 6 and 7, and $m_{245}$ is the module of gear 2, gear 4 and gear 5.

For $N_{4}$ to be an integer the left hand side of the above equation should be an integer.

Therefore, the least acceptable value of $\frac{m_{67}}{m_{245}}$ is 2.

Therefore,

or, $143=33+N_{4}$

or, $110=N_{4}$

From the table now we get

$-a\frac{N2}{N5}\frac{N6}{N7}+b=0$ (given, $\omega 7=0$ )

$b=280$ (given, $\omega _{3}=280$ )

Now, we obtain

$-a\frac{27}{39}\frac{73}{70}+280=0$

$a=387.8$

Now, $\omega _{2}=a+b$ (from the table)

or, $\omega _{2}=387.8+280$

or, $\omega _{2}=667.8$

NOTES : DIAMETER OF A GEAR=MODULE OF THE GEAR X NUMBER OF TEETH OF THAT GEAR

MODULES OF THE MATING GEARS ARE ALWAYS HAVE TO BE THE SAME.

HERE WE CONSIDERED THE VALUE OF $\frac{m_{67}}{m_{245}}$ IS 2. GENERALLY IN ANY GEAR SET MODULES ARE SAME FOR EACH GEAR. BUT IF THIS WAS HAPPENED HERE WE COULD NOT GET AN INTEGRAL VALUE FOR THE NUMBER OF TEETH OF GEAR 4.THIS IS ONLY POSSIBLE IF WE CONSIDER THOSE TWO SETS OF GEARS 6 ,7 AND GEARS 2,4,5 HAVE DIFFERENT MODULES AS THEY ARE NOT MATING GEARS (GEAR 6 AND GEAR 5).