Find the gear ratio for maximum angular acceleration

[fahraynk]

Homework Statement

A constant torque is applied to a pinion which has a moment of inertia of I_m. The pinion(A) drives two gears, one (B) which is connected to a mass which has a moment of inertia = I_m and the other(C) is connected to a mass which has a moment of inertia = 2I_m. The gear ration R_1=A/B is fixed and is equal to 3. What should the gear ration R_2=A/C be to give the maximum angular acceleration of gear 4? Neglect the mass of the gears.

Homework Equations

The Attempt at a Solution

So... Torque out/ Torque in = gear ratio(GR), and I think angular acceleration should be proportional to torque:
$$\alpha_C*2I_m =T_C$$

So, this would mean that the higher the gear ratio, the higher the torque out, the higher the angular acceleration. Since torque in is fixed and the size of the driver gear A is fixed...
$$T_a*GR=T_C\\\\
T_a*A/C=T_C$$$$
So... the lower the number of teeth on gear C is, the higher the torque and the higher the angular acceleration of gear C. So if there is 0 teeth... should be infinite angular acceleration! So something is obviously wrong...

But the books answer is sqrt(1.8)=1.34.
I have no idea how they got this. Google is not helping. Please help.

 

[jbriggs444]

So... the lower the number of teeth on gear C is, the higher the torque and the higher the angular acceleration of gear C. So if there is 0 teeth... should be infinite angular acceleration! So something is obviously wrong...

You have three masses being driven by one input torque. You are in control of only one of the relevant gear ratios.

The connection between the driving torque and the pinion (A) is direct. That's a 1 to 1 gear ratio.
The connection between the driving torque and gear (B) is on a 1 to 2 gear ratio.
The connection between the driving torque and gear (C) is on a gear ratio that you are free to select.

If you make the gear ratio to C extremely high (so that gear C stays nearly stationary while A and B move rapidly) then the gear ratio works against you. The input torque will be able to accelerate A and B at a more or less fixed rate. The higher the gear ratio, the more slowly C will accelerate.

If you make the gear ratio to C extremely low (so that gear C moves very rapidly compared to A and B) then the gear ratio still works against you. The input torque is divided by the gear ratio. The lower the gear ratio, the more slowly C will accelerate.

Your job is to convert this into one or more equations and then solve those equations for the gear ratio that finds the best "sweet spot" between the extremes.

 

[fahraynk]

You have three masses being driven by one input torque. You are in control of only one of the relevant gear ratios.

The connection between the driving torque and the pinion (A) is direct. That's a 1 to 1 gear ratio.
The connection between the driving torque and gear (B) is on a 1 to 2 gear ratio.
The connection between the driving torque and gear (C) is on a gear ratio that you are free to select.

If you make the gear ratio to C extremely high (so that gear C stays nearly stationary while A and B move rapidly) then the gear ratio works against you. The input torque will be able to accelerate A and B at a more or less fixed rate. The higher the gear ratio, the more slowly C will accelerate.

If you make the gear ratio to C extremely low (so that gear C moves very rapidly compared to A and B) then the gear ratio still works against you. The input torque is divided by the gear ratio. The lower the gear ratio, the more slowly C will accelerate.

Your job is to convert this into one or more equations and then solve those equations for the gear ratio that finds the best "sweet spot" between the extremes.

With gear A in the middle, gear B and C adjacent to gear A on both sides :
$$F_AR_A=F_AR_B-F_AR_C\\\\
GR_{AB}=\frac{R_B}{R_A}=3\\\\
R_B=3R_A\\\\
$$
That simplifies to:
$$1=3-R_C/R_A\\\\
2R_A=R_C$$
This gives me 2 equations, 2 unknowns :
$$R_B=3R_A\\\\2R_A=R_C$$

If I take angular acceleration about each gear:$$
\alpha_AI_m=FR_B-FR_C\\\\
\alpha_BI_m=FR_A+FR_C\\\\
\alpha_C2I_m=FR_B-FR_A$$
which gives:
$$
\alpha_A=\alpha_B-2\alpha_C\\\\
\alpha_BI_m=\alpha_A+2\alpha_C\\\\
2\alpha_C=\alpha_B-\alpha_A$$

If I a not doing anything wrong, (But I think I probably am wrong)... the acceleration of gear B should be as high as possible to maximize gear C...
Anyway... 5 equations, 6 unknowns. I have no clue what I am doing wrong.

 

[jbriggs444]

With gear A in the middle, gear B and C adjacent to gear A on both sides :
$$F_AR_A=F_AR_B-F_AR_C\\\\
GR_{AB}=\frac{R_B}{R_A}=3\\\\
R_B=3R_A\\\\
$$

Can you define the variables in those equations? I do not know what any of them are supposed to mean.

 

[fahraynk]

Can you define the variables in those equations? I do not know what any of them are supposed to mean.

sorry
Force on gear A = F_A
force on gear B = F_B
force on gear C = F_C
GR = gear ratio
R_B = radius of B
R_c = radius of C
R_A = radius of A

 

[jbriggs444]

sorry
Force on gear A = F_A
force on gear B = F_B
force on gear C = F_C
GR = gear ratio
R_B = radius of B
R_c = radius of C
R_A = radius of A

OK. So let's go down the equations you have written and see if you can justify them.

With gear A in the middle, gear B and C adjacent to gear A on both sides :
$$F_AR_A=F_AR_B-F_AR_C$$

Where does this one come from?

Let me suggest that you work with torques and angular accelerations instead of forces and radii.

Write down the equation of rotation for gear A. This is the rotational equivalent of Newton's second law: The sum of the external torques is equal to to the angular acceleration times the moment of inertia.

$$\sum \tau = \alpha I$$

Now repeat for the other gears. A kind of rotational equivalent of Newton's third law applies. The gearing system gives it a twist: The torque of A on B is multiplied by the gear ratio between them to compute the torque of B on A. The angular acceleration of B is the angular acceleration of A multiplied by [the negative of] their gear ratio.

 

[fahraynk]

OK. So let's go down the equations you have written and see if you can justify them.Where does this one come from?

Sorry reply took so long. To try to justify that equation I wrote :
the force of gear A * radius = force gear B* radiusB - Force gear C * radius C
Because gear A is in middle, gear B rotates 1 way and gear C rotates the other way so the force would be in the opposite direction.

I really appreciate your help. I have been googling for hours but can't find a solved problem that has a driver driving 2 separate gears for some reason.

Let me suggest that you work with torques and angular accelerations instead of forces and radii.

Write down the equation of rotation for gear A. This is the rotational equivalent of Newton's second law: The sum of the external torques is equal to to the angular acceleration times the moment of inertia.

$$\sum \tau = \alpha I$$

Now repeat for the other gears. A kind of rotational equivalent of Newton's third law applies. The gearing system gives it a twist: The torque of A on B is multiplied by the gear ratio between them to compute the torque of B on A. The angular acceleration of B is the angular acceleration of A multiplied by [the negative of] their gear ratio.

To try again with the torque=I*alpha like you suggest...
T_m = torque from motor driving pinion A, which drives gears B and C.
$$
I_A\alpha_A=T_m-T_B-T_C\\\\
I_B\alpha_B=T_B\\\\
I_C\alpha_C=T_C\\\\
I_A=I_B=I_m\\\\
I_C=2I_m
$$
I can't figure out how to connect these equations. They tell me the gear ratio btwn A and B is:
$$\frac{R_B}{R_A}=3\\\\ GR_{AB}=3$$
But... the gear ratio as I am aware is determined this way for torque:
$$
\frac{F_AR_B}{F_AR_A}=\frac{Torque_B}{Torque_A}\\\\
\frac{R_B}{R_A}=\frac{T_B}{T_A}\\\\
GR_{AB}=\frac{T_B}{T_A}$$
Above says that the torque at gear B should be the GR_AB*Torque_A
This is how I learned gear ratio... But I don't think this holds if gear A is driving 2 gears at the same time.

 

[jbriggs444]

Sorry reply took so long. To try to justify that equation I wrote :
the force of gear A * radius = force gear B* radiusB - Force gear C * radius C
Because gear A is in middle, gear B rotates 1 way and gear C rotates the other way so the force would be in the opposite direction.

Sorry, that makes no sense to me. If you are trying to apply Newton's 2nd law (F=ma) to a gear on a fixed axle, the "a" will always be zero because the axle is not moving. It will apply whatever force is needed to enforce that constraint.

To try again with the torque=I*alpha like you suggest...
T_m = torque from motor driving pinion A, which drives gears B and C.
$$
I_A\alpha_A=T_m-T_B-T_C\\\\
I_B\alpha_B=T_B\\\\
I_C\alpha_C=T_C\\\\
I_A=I_B=I_m\\\\
I_C=2I_m
$$
I can't figure out how to connect these equations.

Is the torque of A on B [about the axis for B] equal to the torque of B on A [about the axis for A]? Even when B and A have different radii?

 

[fahraynk]

Is the torque of A on B [about the axis for B] equal to the torque of B on A [about the axis for A]? Even when B and A have different radii?

I don't follow you here. But I tried adding this to the equations for clarity (T_AonB = torque of A on B)

If alpha_A is 0, because I guess A just moves at a constant velocity to give torque to B and C? Then :
$$T_M=I_M\alpha_B+2I_m\alpha_C\\\\
I_M\alpha_B=T_{AonB}=GR_{AB}*T_M=3T_M\\\\
2I_M\alpha_C=T_{AonC}=GR_{AC}*T_M$$
Cant be right, because then you would want an infinite gear ratio

 

[jbriggs444]

Suppose that you have two gears that are meshing. Call them A and B. The gear ratio between them is three to one.

Would you agree that the radius of the two gears are also in the radius of three to one?
Would you agree that the force of the teeth of A on B is equal to the force of the teeth of B on A?
Would you agree that the moment arm for the force on A is different from the moment arm for the force on B?
Would you agree that the torques must therefore be different -- and in a ratio of three to one?

 

[fahraynk]

Suppose that you have two gears that are meshing. Call them A and B. The gear ratio between them is three to one.

Would you agree that the radius of the two gears are also in the radius of three to one?
Would you agree that the force of the teeth of A on B is equal to the force of the teeth of B on A?
Would you agree that the moment arm for the force on A is different from the moment arm for the force on B?
Would you agree that the torques must therefore be different -- and in a ratio of three to one?

If I have 2 gears I think its easy.
Force = Torque/radius
$$F_1=\frac{T_1}{R_1}=F_2=\frac{T_2}{R_2}\\\\
\frac{T_1R_2}{R_1}=T_2$$

This is 1 equation, 1 unknown and I think it agrees with all 4 points you asked (plus it agrees with formula in books)

But if there are 2 gears, I need an extra equation. Say middle gear is gear 1, gears to left and right are gear 2 and gear 3, respectivly, with radius's R1,R2,R3. T1 is known and given from the motor driving it.
$$\frac{T_1}{R_1}=\frac{T_2}{R_2}+\frac{T_3}{R_3}\\\\
R_2=3R_1\\\\
\frac{T_1}{R_1}-\frac{T_2}{3R_1}=\frac{T_3}{R_3}\\\\
\frac{1}{R_1}[T_1-\frac{T_2}{3}]=\frac{T_3}{R_3}$$

So, I feel like I need a second equation to relate T2 to T1. Unless something is fundamentally wrong with my sum of forces. the force on the edge of gear 1 should equal the force on the edge of gear 2 plus the force from gear 3
All I can think of is
$$I_M\alpha_2+2I_m\alpha_3 = 0\\\\
\frac{1}{R_1}[T_1-\frac{I_m\alpha2}{3}]=\frac{2I_m\alpha_3}{R_3}\\\\
\frac{1}{R_1}[T_1-\frac{I_m(-2\alpha_3)}{3}]=\frac{2I_m\alpha_3}{R_3}$$
Am I on the right track here?

 

[jbriggs444]

If I have 2 gears I think its easy.

This is 1 equation, 1 unknown and I think it agrees with all 4 points you asked (plus it agrees with formula in books)

But if there are 2 gears, I need an extra equation. Say middle gear is gear 1, gears to left and right are gear 2 and gear 3, respectivly, with radius's R1,R2,R3. T1 is known and given from the motor driving it.
$$\frac{T_1}{R_1}=\frac{T_2}{R_2}+\frac{T_3}{R_3}$$

As I understand this equation, you are taking a torque balance on gear one (after dividing through by $R_1$) and assuming that the net torque (motor plus the other two external torques) totals to zero. But it does not. Gear 1 is accelerating. It is subject to an unbalanced net torque.

 

[fahraynk]

As I understand this equation, you are taking a torque balance on gear one (after dividing through by $R_1$) and assuming that the net torque (motor plus the other two external torques) totals to zero. But it does not. Gear 1 is accelerating. It is subject to an unbalanced net torque.

Oh... yeah, you said this earlier : "Sorry, that makes no sense to me. If you are trying to apply Newton's 2nd law (F=ma) to a gear on a fixed axle, the "a" will always be zero because the axle is not moving. It will apply whatever force is needed to enforce that constraint."
So I figured that meant their is no acceleration on gear 1 (I used to call gear 1 gear A)
I was working on the equation from my last post, and got this :
$$I_mT_1\frac{3R_3}{6R_1-2R_3}=\alpha_3$$
But now that you said gear 1 should also have an acceleration... does that mean I should start over and the sum of forces should be this :

$$\frac{T_1}{R_1}=\frac{T_2}{R_2}+\frac{T_3}{R_3}+\alpha_1R_1$$
Should I work with this equation?
Thanks for your continued help

 

[jbriggs444]

Oh... yeah, you said this earlier : "Sorry, that makes no sense to me. If you are trying to apply Newton's 2nd law (F=ma) to a gear on a fixed axle, the "a" will always be zero because the axle is not moving. It will apply whatever force is needed to enforce that constraint."
So I figured that meant their is no acceleration on gear 1 (I used to call gear 1 gear A)
I was working on the equation from my last post, and got this :
$$I_mT_1\frac{3R_3}{6R_1-2R_3}=\alpha_3$$
But now that you said gear 1 should also have an acceleration...

A gear has no linear acceleration. It is pinned to the axle in its center.
A gear has angular acceleration. It is free to rotate about that axis.

does that mean I should start over and the sum of forces should be this :

Sum torques, not forces.

 

[fahraynk]

A gear has no linear acceleration. It is pinned to the axle in its center.
A gear has angular acceleration. It is free to rotate about that axis.

Sum torques, not forces.

moment of inertia gear 1, gear 2, gear 3 = I_m, I_m, 2I_m
T_m is torque of motor on gear 1.
$$T=\frac{dL}{dt}=I\frac{dW}{dt}=I*\alpha\\\\
T_m=I_m\alpha_1+I_m\alpha_2+I_m\alpha3\\\\
\alpha_2=3\alpha_1\\\\
T_m=I_m\alpha_1+3I_m\alpha_1+I_m\alpha3=4I_m\alpha_1+I_m\alpha3$$

Now what is wrong, lol...

 

[jbriggs444]

moment of inertia gear 1, gear 2, gear 3 = I_m, I_m, 2I_m
T_m is torque of motor on gear 1.
$$T=\frac{dL}{dt}=I\frac{dW}{dt}=I*\alpha\\\\
T_m=I_m\alpha_1+I_m\alpha_2+I_m\alpha3\\\\
\alpha_2=3\alpha_1\\\\
T_m=I_m\alpha_1+3I_m\alpha_1+I_m\alpha3=4I_m\alpha_1+I_m\alpha3$$

Now what is wrong, lol...

Free body diagram. Sum the torques on one object to get the angular acceleration of one object.

Edit: I am going to preach a bit at you. [I am a product of a computer science education in the 70's. The mantra then as now is document, document, document. You are not writing for the computer. You are not writing for yourself. You are writing for the person trying to make sense of what you have written]

1. If you are going to put a variable name into an equation, first stop and define what that variable name denotes. Do not make us guess what $T_1$, $T_2$ and $T_3$ or $I_m$ mean.

2. If you are going to write down an equation, justify it.

Your first equation above is incorrect and unjustified. We are not dealing with a single rigid object with moment of inertia I and angular acceleration $\alpha$. We have three objects, each subject to a net torque and each with different (albeit related) angular accelerations.

Your second equation above is incorrect and unjustified. Same reason.

Your third equation is justifiable. Yes, we are given the gear ratio between gears 1 and 2.

Your fourth equation above is certainly unjustified. I have no idea what the fourth equation is trying to say.

None of the equations you have written down use the gear ratio that you are trying to optimize.

Please... write down ONE EQUATION for the angular acceleration for gear 1 in terms of the torques on it and its moment of inertia.

 

[fahraynk]

Please... write down ONE EQUATION for the angular acceleration for gear 1 in terms of the torques on it and its moment of inertia.

I_1, I_2,I_3, alpha_1,alpha_2,alpha_3, R_1,R_2,R_3 are moments of inertia, angular accelerations, radius of gears 1, 2, 3
T_m is the torque the motor applies to gear 1.
$$I_1\alpha_1=T_m-I_2\alpha_2\frac{R_2}{R_1}-I_3\alpha_3\frac{R_3}{R_1}$$

First I did a sum of forces on gear 1 at its radius = torque from motor * radius of gear 1- minus the force it exerts on gear 2 and gear 3 which is I_2*alpha_2*R_2 and I_3*alpha_3*R_3
Then I divided by radius of gear 1 to get it in a form of angular accelerations in terms of torque like you asked.

 

[jbriggs444]

I_1, I_2,I_3, alpha_1,alpha_2,alpha_3, R_1,R_2,R_3 are moments of inertia, angular accelerations, radius of gears 1, 2, 3
T_m is the torque the motor applies to gear 1.
$$I_1\alpha_1=T_m-I_2\alpha_2\frac{R_2}{R_1}-I_3\alpha_3\frac{R_3}{R_1}$$

First I did a sum of forces on gear 1 at its radius = torque from motor * radius of gear 1- minus the force it exerts on gear 2 and gear 3 which is I_2*alpha_2*R_2 and I_3*alpha_3*R_3
Then I divided by radius of gear 1 to get it in a form of angular accelerations in terms of torque like you asked.

Thank you. Now let me translate so that it fits the way I am thinking about the situation...

You are trying to write something in the form $I \alpha = \sum \tau$ (moment of inertia times angular acceleration equals the sum of the torques).

You have the torque from the motor accounted for. That's the $T_m$ term.

You proceed to work on the term for the torque coming from gear 2. There is only one torque on gear 2 -- the torque from its interface with gear 1. Using $I \alpha = \sum \tau$, you determine that the torque of gear 1 on gear 2 must be the moment of inertia $I_2$ multiplied by the angular acceleration, $\alpha_2$. You multiply this by the gear ratio, $\frac{R_2}{R_1}$ to get the corresponding torque of gear 2 on gear 1.

But let us pause for a moment and think.

If $R_2$ is large and $R_1$ is small, the fraction $\frac{R_2}{R_1}$ is large. A large gear has mechanical advantage to impose a larger torque on a smaller gear?! That's backward. If gear 2 is large and gear 1 is small, a large torque on gear 2 should correspond to a small torque on gear 1.

Proceeding... You take this resulting term, corrected to read $I_2 \alpha_2 \frac{R_1}{R_2}$ and negate it before adding it to the sum. You are adopting a sign convention where all of the angular accelerations and torques are positive. Gear 1 acts as a drag on gear 2, so its contribution is negative. That's fine. I have no problem with that sign convention.

We could simplify this term since the gear ratio between 1 and 2 is known. But it is good to at least start in symbolic form. Well done.

The third term is the same again. We do not know the gear ratio this time, so we are forced to leave things in symbolic form. This matches the form of the second term. Pretty. Again, well done.

So the only problem so far is that the gear ratios are reversed. No big deal. We have an equation we can work with.

Next step...

You have three angular accelerations listed. Can you express $\alpha_2$ in terms of $\alpha_1$, $R_1$ and $R_2$? Can you do the same for $\alpha_3$ in terms of $\alpha_1$, $R_1$ and $R_3$?

 

[fahraynk]

You proceed to work on the term for the torque coming from gear 2. There is only one torque on gear 2 -- the torque from its interface with gear 1. Using $I \alpha = \sum \tau$, you determine that the torque of gear 1 on gear 2 must be the moment of inertia $I_2$ multiplied by the angular acceleration, $\alpha_2$. You multiply this by the gear ratio, $\frac{R_2}{R_1}$ to get the corresponding torque of gear 2 on gear 1.

From the general equation of torque between 2 gears $T_2=\frac{R_2}{R_1}T_1$ :
$$T_2=\frac{R_2}{R_1}T_1\\\\
I_2\alpha_2=\frac{R_2}{R_1}I_1\alpha_1\\\\
I_2=I_1\\\\
I_3=2I_1\\\\
\alpha_2=\frac{R_2}{R_1}\alpha_1\\\\
2\alpha_3=\frac{R_3}{R_1}\alpha_1$$
Does this make sense?

 

[jbriggs444]

From the general equation of torque between 2 gears $T_2=\frac{R_2}{R_1}T_1$

You know I just got done preaching at you about defining your variable names. What is $T_1$? What is $T_2$?

What I think they are is: $T_1$ is the torque of Gear 2 on Gear 1 and $T_2$ is the torque of Gear 1 on Gear 2. With that interpretation...

$T_2=\frac{R_2}{R_1}T_1$ is correct
$I_2\alpha_2=\frac{R_2}{R_1}I_1\alpha_1$ is not correct.

There is more than one torque acting on gear 1. It angular acceleration will not match that which would be due to the influence of gear 2 alone.

Again, I ask: Can you write an equation for $\alpha_2$ in terms of $\alpha_1$, $R_1$ and $R_2$. Hint: It's dead easy.

 

[fahraynk]

You know I just got done preaching at you about defining your variable names. What is $T_1$? What is $T_2$?

What I think they are is: $T_1$ is the torque of Gear 2 on Gear 1 and $T_2$ is the torque of Gear 1 on Gear 2. With that interpretation...

$T_2=\frac{R_2}{R_1}T_1$ is correct
$I_2\alpha_2=\frac{R_2}{R_1}I_1\alpha_1$ is not correct.

There is more than one torque acting on gear 1. It angular acceleration will not match that which would be due to the influence of gear 2 alone.

Again, I ask: Can you write an equation for $\alpha_2$ in terms of $\alpha_1$, $R_1$ and $R_2$. Hint: It's dead easy.

With that interpretation : $T_1$ is the torque of Gear 2 on Gear 1, you said the first equation is correct. A few posts up you confirmed the torque of gear 2 on gear 1 is $T1=I_2\alpha_2\frac{R_1}{R_2}$
So I GUESS the torque from gear 1 on gear 2 is $T_2=T_m\frac{R_2}{R_1}$ by the same logic
So... Maybe $T_m\frac{R_2}{R_1}=I_2\alpha_2\frac{R_1}{R_2}\frac{R_2}{R_1}\\\\ T_m\frac{R_2}{R_1}=I_2\alpha_2$
I''m really not sure. Been lost on this problem for days... I can write the equations for regular systems, like spring and dashpod systems easily... but there is something I am missing in these problems and after 4 days I still don't see it... Already spent hours googling can't find a similar problem of 1 gear driving 2.

 

[Dr.D]

Actually, working with tangential forces and pitch radii is entirely correct. It is the forces along the line of action that actually cause the action of one gear on another. The tangential force is simply the tangential component of the contact force.

Eventually, it does come down to writing torque equations, and those torques can easily be written in terms of tangential forces and pitch radii.

 

[jbriggs444]

With that interpretation : $T_1$ is the torque of Gear 2 on Gear 1, you said the first equation is correct. A few posts up you confirmed the torque of gear 2 on gear 1 is $T1=I_2\alpha_2\frac{R_1}{R_2}$
So I GUESS the torque from gear 1 on gear 2 is $T_2=T_m\frac{R_2}{R_1}$ by the same logic

No. In the context of gears, torques come in pairs. The torque of gear 1 on gear 2 -- paired with the torque of gear 2 on gear 1. This comes from Newton's third law. The tangential forces are equal and opposite. The moment arms are in the ratio of the radii of the respective gears.
$$T_1=T_2\frac{R1}{R2}$$
$$T_2=T_1\frac{R2}{R2}$$

@Dr.D is correct -- you can choose to work with tangential forces. But the same logic still applies -- the tangential forces come in 3rd law pairs.

If you have a steam locomotive pulling a long train, the full force of the engine on the first car is not the same as the force of the last car on the caboose.

Now, I ask you for a third time, if you have the angular acceleration of gear 1: $\alpha_1$...
and if you know that the $\frac{R2}{R1}$ is equal to 2...
what is the angular acceleration of gear 2?

Hint: With a gear ratio of 2 to 1, the angular displacement, $\theta_2 = \frac{\theta_1}{3}$
Hint: With a gear ratio of 2 to 1, the angular velocity, $\omega_2 = \frac{\omega_1}{3}$

Edit: I hope I have the sense of the gear ratio right -- that $R_2$ is three times larger than $R_1$. Not dead sure about that part of the problem statement.

 

[Dr.D]

No. In the context of gears, torques come in pairs. The torque of gear 1 on gear 2 -- paired with the torque of gear 2 on gear 1. This comes from Newton's third law.

It seems to me that this is where some of the confusion lies. It is the contact forces that come in pairs, not the torques. The torque on each gear depends upon the radius at which the contact force acts, and that may not be the same for both gears. The contact forces are always equal and opposite; the torques not necessarily.

 

[jbriggs444]

It seems to me that this is where some of the confusion lies. It is the contact forces that come in pairs, not the torques. The torque on each gear depends upon the radius at which the contact force acts, and that may not be the same for both gears. The contact forces are always equal and opposite; the torques not necessarily.

You are, of course, correct. The relationship follows from Newton's third. As I go on to point out, it's not "equal and opposite" but, instead...

The tangential forces are equal and opposite. The moment arms are in the ratio of the radii of the respective gears.

 

[jack action]

A constant torque is applied to a pinion which has a moment of inertia of I_m. The pinion(A) drives two gears, one (B) which is connected to a mass which has a moment of inertia = I_m and the other(C) is connected to a mass which has a moment of inertia = 2I_m. The gear ration R_1=A/B is fixed and is equal to 3. What should the gear ration R_2=A/C be to give the maximum angular acceleration of gear 4? Neglect the mass of the gears.

Free body diagram of pinion (A):

• A constant torque $T$ is applied;
• A force $F_B$ is reacted at radius $A$ from the gear (B);
• A force $F_C$ is reacted at radius $A$ from the gear (C);
• Pinion (A) has moment of inertia $I_m$ and an angular acceleration $\alpha_A$.

Thus:
$$T = F_B A + F_C A + I_m\alpha_A$$
You have 3 unknowns ($F_B$, $F_C$ & $\alpha_A$) therefore you need 3 equations to solve the problem.

Free body diagram of gear (B):

• A force $F_B$ is applied at radius $B$ from the pinion (A);
• Gear (B) has moment of inertia $I_m$ and an angular acceleration $\alpha_B$.

Thus:
$$F_B B = I_m\alpha_B$$
We introduce a 4th unknown ($\alpha_B$) thus we will need 4 equations.

Free body diagram of gear (C):

• A force $F_C$ is applied at radius $C$ from the pinion (A);
• Gear (C) has moment of inertia $2I_m$ and an angular acceleration $\alpha_C$.

Thus:
$$F_C C = 2I_m\alpha_C$$
We introduce a 5th unknown ($\alpha_C$) thus we will need 5 equations.

We also know that linear accelerations are the same for each gear at the contact point between pinion (A) and gear (B), thus:
$$a_A = a_B$$
$$\alpha_A A = \alpha_B B$$
Similarly for the linear accelerations between pinion (A) and gear (C):
$$\alpha_A A = \alpha_C C$$
There, we have 5 equations with 5 unknowns. So:
$$T = F_B A + F_C A + I_m\alpha_A$$
$$T = \left(\frac{I_m}{B} \alpha_B\right) A + \left(\frac{2I_m}{C} \alpha_C \right) A + I_m\alpha_A$$
$$T = \frac{A}{B} I_m\alpha_B + \frac{A}{C} 2I_m \alpha_C + I_m\alpha_A$$
$$T = \frac{A}{B} I_m\left(\frac{A}{B}\alpha_A\right) + \frac{A}{C} 2I_m \left(\frac{A}{C}\alpha_A\right) + I_m\alpha_A$$
$$T = \left(\frac{A}{B}\right)^2 I_m \alpha_A +\left(\frac{A}{C}\right)^2 2I_m \alpha_A + I_m\alpha_A$$
$$\alpha_A = \frac{1}{\left(\frac{A}{B}\right)^2 + 2\left(\frac{A}{C}\right)^2 + 1 }\frac{T}{I_m}$$
And then from the last 2 equations:
$$\alpha_B = \frac{A}{B}\alpha_A = \frac{^A/_B}{\left(\frac{A}{B}\right)^2 + 2\left(\frac{A}{C}\right)^2 + 1 }\frac{T}{I_m}$$
$$\alpha_C = \frac{A}{C}\alpha_A = \frac{^A/_C}{\left(\frac{A}{B}\right)^2 + 2\left(\frac{A}{C}\right)^2 + 1 }\frac{T}{I_m}$$
Now we can find where $\alpha_C$ will be max when its derivative will be zero, i.e.:
$$\frac{d\alpha_C}{d^A/_C} = \frac{\frac{\left(^A/_B\right)^2 + 1}{\left(^A/_C\right)^2} - 2}{\left( \frac{\left(^A/_B\right)^2 + 1}{^A/_C} + 2\ ^A/_C \right)^2}\frac{T}{I_m} = 0$$
Which means that:
$$^A/_C = \sqrt{\frac{\left(^A/_B\right)^2 + 1}{2}}$$
Or $^A/_C = \sqrt{5} = 2.236$. Now, I know you said the answer was supposed $\sqrt{1.8}$, but I'm pretty sure of my answer with the problem as stated (although I cannot be sure of what you meant by «gear 4»).

 

[haruspex]

Or A/C=√5=2.236.

That's what I get too.

 

[fahraynk]

We also know that linear accelerations are the same for each gear at the contact point between pinion (A) and gear (B), thus:
$$a_A = a_B$$
$$\alpha_A A = \alpha_B B$$
Similarly for the linear accelerations between pinion (A) and gear (C):
$$\alpha_A A = \alpha_C C$$
There, we have 5 equations with 5 unknowns.

AH! Nice. I didn't realize this condition existed... duh.
This really makes me chill now to... because I was going nuts over why gear ratios worked and couldn't find a derivation... Thanks!

 

[fahraynk]

You are, of course, correct. The relationship follows from Newton's third. As I go on to point out, it's not "equal and opposite" but, instead...

Thanks for your help jbriggs444! also haruspex jack action and Dr.D