Dynamics rigid body homework question

  • #1
Pipsqueakalchemist
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Homework Statement
I have the problem below
Relevant Equations
V_b = V_a + (w)X(R_ab)
So I’m confused about a few things in the solution. Why is it that speed of the V_h appears on both gear a and b? Is it because both gears are both in contact so they have equal speeds just in opposite direction. Same confusion for V_e being on the top of gear b. So Just because gear e doesn’t move and the top of gear b is in contact with it, that means the point of contact has the same speed? Also How can I use the V_h = V_e + (w)X(R_h/e) if they are separate gears and not one rigid body.
 

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  • #2
Hi,

Pipsqueakalchemist said:
Why is it that speed of the V_h appears on both gear a and b? Is it because both gears are both in contact
Yes, there is a no slip condition. In this case the velocity ## v_H ## is in the same direction at the point of contact, but the angular velocities ##\omega ## will be in the opposite direction. Note, depending on how the gears mesh, the angular velocities may also be in the same direction (e.g. the way gear B meshes with E), but the velocity ## v ## will always be in the same direction

[EDIT]: Also note that the angular velocities are not necessarily equal, that depends on the radii & number of the gears in mesh

Pipsqueakalchemist said:
Same confusion for V_e being on the top of gear b. So Just because gear e doesn’t move and the top of gear b is in contact with it, that means the point of contact has the same speed?
It is because they are in contact that they have the same speed at that point. Gear B is is instantaneously stationary about its point of contact (i.e. the velocity at the point of contact is 0, but it can be thought of as rotating around that point).

Pipsqueakalchemist said:
Also How can I use the V_h = V_e + (w)X(R_h/e) if they are separate gears and not one rigid body.
See above, ## V_e ## also refers to the speed of gear B at the point of contact, so I believe this equation is only referring to one equation. This equation can be used as we are using the stationary point of contact as the reference.

I hope that made some sense. If not, let me know.
 
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  • #3
Pipsqueakalchemist said:
Homework Statement:: I have the problem below
Relevant Equations:: V_b = V_a + (w)X(R_ab)

So I’m confused about a few things in the solution. Why is it that speed of the V_h appears on both gear a and b? Is it because both gears are both in contact so they have equal speeds just in opposite direction. Same confusion for V_e being on the top of gear b. So Just because gear e doesn’t move and the top of gear b is in contact with it, that means the point of contact has the same speed? Also How can I use the V_h = V_e + (w)X(R_h/e) if they are separate gears and not one rigid body.
In order to better understand these relationships among meshing gears, you could visualize them as instantaneous levers, as shown in attached diagram.

Please, see:
https://www.tec-science.com/mechani...ary-gear/willis-equation-for-planetary-gears/

image.jpeg
 
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  • #4
Yeah thanks guys, that helps and makes sense to me
 
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