Gears

Use

I don’t think we actually ever do anything with gears other than velocity transfer lol

Pitch:
$p=\frac{2\pi R_1}{N_1}=\frac{2\pi R_2}{N_2}$ Where N: number of teeth
Velocity:
$v=R{\theta }_1^{\prime }=R{\theta }_2^{\prime }$ Where no slipping is also assumed.
Where theta’ = w (angular acceleration)

No power loss: No heat inertia or no acceleration and no friction loss:
$T_1{\theta }_1^{\prime }=T_2{\theta }_2^{\prime }$
In terms of work:
$T_1{\theta }_1=T_2{\theta }_2$
where
$\frac{{\theta }_2}{{\theta }_1}=\frac{T_1}{T_2}=\frac{N_2}{N_1}$ where N2/N1 is the gear ratio.

Note on gear ratios

If N2/N1 > 1 then w2 < w1 and T2 > T1 (speed reducer)

Equivalent Inertia

See Equivalent Mass and Inertia

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Examining the effect of the system (shaft and gear) we get T1 and T2 which are not the same…
but
$n_2{T}_{1G}=n_1{T}_{2g}$ Since there is no moment of inertia here:
$T_2-T_{2G}=0\ast {\theta }^{{}^{\prime \prime }}$ $T_2=T_{2G}$
and by using a previous equation
$T_{1G}=\frac{n_1}{n_2}{T}_2$

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Solving Gear Questions

See my practice problems for a more detailed explanation

Key Points

When analyzing gear ratios:

• If you have a motor driving a gear train, the gear ratios applied to a gear are baed off of what the gear needs to drive from the motors perspective. For example, the gears on the same axel as a motor have the total equivalent gear ratio (product of N) because they still need to drive the entire system. The first intermediate gear however, will only have subsequent gear ratios not including the first one because the first gear ratio is accounted for by the gear it interacts with.
• You also need to find an equivalent torque by the whole gear train which is also done using gear ratios
• Use the axis of the output shaft as the neutral axis you take moments about

In terms of gears, Inertia is: $I^{\prime }=\frac{I^{}}{N}$ where I’ is the output inertia and I is in the input inertia