Angular speed ratio and gear teeth ratio

[kidsasd987]

http://www.tech.plym.ac.uk/sme/mech226/gearsys/gearaccel.htm

Hello, I tried to derive n=N1/N2=omega2/omega1

with using T1=T2 but it is a wrong assumption.
I thought they should balance to satisfy Newton's 3rd law, but it is not.

Could you explain me why.

 

[SteamKing]

http://www.tech.plym.ac.uk/sme/mech226/gearsys/gearaccel.htm

Hello, I tried to derive n=N1/N2=omega2/omega1

with using T1=T2 but it is a wrong assumption.
I thought they should balance to satisfy Newton's 3rd law, but it is not.

Could you explain me why.

In the diagrams, it's not the torques which are equal but the energy of each gear,

E = T ⋅ ω,

such that T_a ⋅ ω_a = T_b ⋅ ω_b

T_a ≠ T_b because of the ratio between gears a and b.

 

[kidsasd987]

equal

In the diagrams, it's not the torques which are equal but the energy of each gear,

E = T ⋅ ω,

such that T_a ⋅ ω_a = T_b ⋅ ω_b

T_a ≠ T_b because of the ratio between gears a and b.

Hello, yes. It makes sense that work of each gear must be equal.

But I thought T1=T2 because torque is somewhat symmetrical to force. It changes moment of inertia, and I thought at the contact point of two gears, T1 and T2 must be the same just because of intuition that force and torque (angular momentum and momentum) are symmetrical.

If I push a box with a force F, by distance D.
Ive done FD amount of work to the box and vice versa, the box has done -FD amount of work to me.
Therefore, Torque*theata is symmetrical to Force*Distance.

However, unlike Force, T1=/=T2
Could you tell me why Newton's 3rd law is not valid for toruqe?

 

[jbriggs444]

Could you tell me why Newton's 3rd law is not valid for toruqe?

Newton's third law is valid for torque. But when you compute torque T₁ based on one axis of rotation and torque T₂ based on a different axis of rotation, that's comparing apples and oranges. If you want the torques from two third-law partner forces to be equal, the moment arms should be equal as well.

 

[kidsasd987]

Newton's third law is valid for torque. But when you compute torque T₁ based on one axis of rotation and torque T₂ based on a different axis of rotation, that's comparing apples and oranges. If you want the torques from two third-law partner forces to be equal, the moment arms should be equal as well.

Thanks. I like your analogy of apples and oranges. I must go grab some.