Gears Problem -- tray for holding the disk in a tabletop Blu-ray player

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SUMMARY

The discussion focuses on calculating the motor speed required to operate the load/unload tray of a tabletop Blu-ray player, utilizing nylon spur gears with a module of 2.5 mm. The tray must move at a speed of 0.1 m/s, leading to the conclusion that the motor must turn at approximately 171.43 rpm, which translates to an angular velocity of about 17.95 rad/s. The participants clarify the relationship between gear teeth, module, and the necessary calculations to determine motor speed.

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Homework Statement


The mechanism that operates the load/unload tray for holding the disk in a tabletop Blu-ray player uses nylon spur gears, a rack, and a belt drive. The gear that meshes with the rack has a module of 2.5 mm. The gears have the numbers of teeth indicated, and the two sheaves have diameters of 7 mm and 17 mm. The rack is connected to the tray that holds the disk. For the tray to move at 0.1 m/s, how fast must the motor turn?
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Homework Equations

The Attempt at a Solution


##
\omega_{4}=\omega_{5}\\
\omega_{2}=\omega_{3}\\
d_{1} \omega_{1} = d_{2} \omega_{2} \\
\omega_{2} = \frac{\omega_{1}d_{1}}{d_{2}} \\
N_{4} \omega_{4} = N_{3} \omega_{3} \\
N_{4} \omega_{5} = N_{3} \omega_{2} \\
N_{4} \omega_{5} = N_{3} \frac{\omega_{1}d_{1}}{d_{2}} \\
\omega_{1}=\frac{N_{4}\omega_{5}d_{2}}{N_{3}d_{1}} \\
##

I know that the module of the gear that meshes with the rack is 2.5[mm] so the module of the rack must also be 2.5[mm].
##
v=0.1[m/s]=0.1[\frac{m}{s}] \times 60[\frac{s}{min}]= 6[\frac{m}{min}] \\
v=6 [\frac{m}{rev}][\frac{rev}{min}] = 6[\frac{m}{rev}][rpm] \\
##

I need help from here.

Also, why is the units of the module [mm] and not [mm/teeth]? The formula for it is 2r/N
 
Last edited:
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##
v=0.1[m/s]=0.1[\frac{m}{s}] \times 60[\frac{s}{min}]= 6[\frac{m}{min}]\\
v=6 [\frac{m}{rev}][\frac{rev}{min}] = x\omega_{5} \\
x=14[\frac{tth}{rev} \times 2.5 \frac{mm}{tth}=0.035 \frac{m}{rev} \\
v=6 [\frac{m}{rev}][\frac{rev}{min}] = 0.035 \frac{m}{rev} \times \omega_{5} \\
\omega_{5}=171.4285714 [rpm]= 171.4285714 [rpm] \times 2\pi \frac{rad}{rev} \times \frac{1}{60} \frac{min}{s}= 17.95195802 [rad/s]
##

Did I get the angular velocity correctly for the gear meshed with the rack/tray?
 

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