I'm getting tripped up on something which probably has a simple answer but is presently escaping me at the moment.

Let's say that a smaller gear is geared up with a larger gear. If I apply a force perpendicular to the first gear's axis, The gear will undergo an angular acceleration $\alpha$ via this relationship:

$$\tau_1 = I_1 \alpha_1$$

If the wheel has a radius $r_1$ then the torque on the wheel is $r_1F_1$, so:

$$r_1F_1 = I_1 \alpha_1$$

If I apply this force for a time $t$, then the wheel will achieve an angular speed $\omega_1 = \alpha_1 t$, so:

$$\omega_1 = \alpha_1t = \frac{r_1F_1 }{I_1}t$$

Now consider the second gear. I can follow the same logic with the second wheel and also say:

$$\omega_2 = \alpha_2t = \frac{r_2F_2 }{I_2}t$$

Now, let's make two simplifying assumptions: let's assume that $I_1 = I_2$ (I think this is reasonable -- maybe we achieve this with different materials and mass distributions). Now comes another assumption that I can't quite justify quite as easily but I think is okay: let's assume that $F_1 = F_2$. I realize that with regular gears, some of the force on the second gear will be transmitted along the directional component pointing toward the wheel rather than perpendicular to it, but I imagine maybe with some ingenuity, we could minimize this to the point where it's negligible for our purposes.

Assuming we can get away with my two assumptions, we can get rid of subscripts on $I$ and $F$:

$$\omega_1 = \frac{r_1F }{I}t, \qquad \omega_2 = \frac{r_2F }{I}t$$

Combining equations we get:

$$ \frac {\omega_1}{r_1} = \frac{Ft }{I}, \qquad \frac {\omega_2}{r_2} = \frac{Ft }{I}$$

$$ \frac {\omega_1}{r_1} = \frac {\omega_2}{r_2} $$

But this has to be wrong: what it says is that increasing the radius of a gear also increases it's angular speed. But clearly the larger gear turns more slowly than the smaller gear.

So what did I do wrong? Was it a bad assumption to assume that $F_1=F_2$? Is my model too simple? Do I need to account for the second gear's torque on the first gear? Is it some other silly thing I've forgotten?


1 Answer 1


The force you have postulated must act like a static friction force. If no other force is acting to make one of the gears accelerate, that force will be zero, and the gears cannot accelerate. There must be a 2nd torque-producing force which is not acting on one of the gears, causing the friction force to be non-zero.

That extra force, coupled with the fact the the tangential velocities must be equal will give you the correct relationship for both the $\alpha$ and $\omega$ terms.

If the gears are moving at constant $\omega$s then the interaction force must be zero or the gears will accelerate, contrary to the constant $\omega$ claim.

Let some other force $F_T$ produce a torque on one of the gears, also. You can do the maths.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.