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- Thread starter Alistair McCheyne
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- #2

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[tex]\sum T_i = I_i \alpha_i[/tex]

And for every gear [itex]i[/itex] connected to gear [itex]j[/itex], you get:

[tex]\frac{T_{i\ out}}{T_{j\ in}} = \frac{N_i}{N_j}[/tex]

[tex]\frac{\alpha_i}{\alpha_j} = \frac{N_j}{N_i}[/tex]

This gives you 7 equations with 7 unknowns ([itex]T_{A\ out}[/itex], [itex]T_{B\ in}[/itex], [itex]T_{B\ out}[/itex], [itex]T_{C\ in}[/itex], [itex]\alpha_A[/itex], [itex]\alpha_B[/itex], [itex]\alpha_C[/itex]).

- #3

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Select an arbitrary rotation rate for shaft A (say 10 rad/sec)

Calculate the rotational KE of each shaft at this rate

The effective moment of inertia of shafts B and C you calculate by comparing the KE values to that of shaft A

For example :

The KE of shaft A at 10 rad / sec = 75 joules

The KE of shaft B at 5 rad / sec = 0.625 joules, so its effective moment of inertia = 1.5 * ( 0.625 / 75 ) = 0.0125 kg . m^2

Repeat for shaft C then add A, B and C for total effective moment of inertia of the system

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