Inertia change through a gearbox question

  • Thread starter mech_rocks
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  • #1

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Hello,

This is my first post on the webpage I hope you guys can help me out. I'm trying to understand the concept of the change of mass moment of inerita through a gearbox. I know the formula, but why does the gear ratio affect the mass moment of inertia [kg m^2]. Lets say you have a input shaft rotating at 540 rpm, a gearbox with a 2:1 ratio, and a disk on the output shaft so why is the inertia "seen" at the beginning of the input shaft equal to the inertia produced by the disk multiplied by 4 considering (effective inertia = inertia x gear ratio^2 ) ???

Maybe you could describe the physics behind this... I sure would appreciate it. Thanks!
 

Answers and Replies

  • #2
Filip Larsen
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I'm not sure this is the text book answer, but try consider the rotational energy with and without the gearbox. First, lets apply an 10 W engine to the shaft without the gearbox for 10 seconds. The rotational energy of the disc must now be 100 J, disregarding all friction. Now insert the gear box and again apply the 10 W engine for 10 sec. The rotational energy of the disc must still be 100 J since the gearbox neither absorb nor introduce energy, and since it is the same disc with the same moment of inertia, its rotational speed must also be the same. But this means the shaft on the engine side only rotates with half angular speed (or double, depending on the "orientation" of the gearbox) compared to before and so from the engines perspective the shaft seems to have four times the moment of inertia.

So in general, if you equate the rotation energy in the two situations now using the general gearbox ratio r, you get

[tex]1/2 I_e (\omega/r)^2 = 1/2 I_d \omega^2[/tex]

which after cancelling [itex]\omega[/itex] implies that
[tex]I_e = r^2 I_d [/tex]
 
Last edited:
  • #3
Ok I kind of understand what your saying. I appreciate your time. Thanks!
 

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