Gears and Gear Ratios

Hi Antony- this is a great question and the answer depends on how we fix the diagram!

The diagram shows a system that won't work as the teeth on the larger gear would not mesh properly with the teeth on the smaller gear (I suspect the artist has resized a stock image without a good understanding of how gears work!) and the system would likely either break or get stuck.

If the teeth on the gear A were made to mesh properly with the smaller gears whilst he radius of A is kept the same, then the larger gear would need more teeth. This would then mean that it would have a lower angular velocity (RPM) than B or C.

If gear A is scaled down so that the teeth mesh properly and the number of teeth is kept the same, then gear A would be the same size as B and C, and would have the same angular velocity (RPM).

Hope that helps!

Hi Antony

If you think about a bicycle, the rear wheel will turn with a different RPM than your pedals - so the size does matter. In fact it not really the size, but the number of teeth on each wheel (or cog).

Cog A will have the lowest RPM

- in terms of A level Physics, I recommended that my students always try to use the technical or scientific language in their answers, this gives examiners confident that the student 'knows their stuff'.

In this case we could use the term angular velocity, speed can be a bit woolly, and may lead to imprecise answers that may not hit all of the available marks on a mark scheme.

Coming back to the question, if cog A has twice as many teeth as cog B then cog B will have twice the angular velocity compared to cog A.

As in all good physics this can be justified with an equation:

Cog A number of teeth x Cog A angular velocity = Cog B number of teeth x Cog B angular velocity

If Cog A has 16 teeth and cog B has 8 teeth, to keep it simple, lets give cog A an angular velocity of 1 radian per second then:

so Cog B has an angular velocity of 2 rad/s

I hope this helps.

To Rpm of b and c would be higher than A as A has a larger radius.

Hi Anthony,

Both the diameter of the gears and the number of teeth of the gears must be considered but always the gear with the larger diameter (the driver) has more teeth and is slower and has lower RPM. If, the RPM of "A" (the driver) is known you must multiply that RPM by Gear Ratio to find the RPM of "B" (the driven gear).

A would move slower, because, the gear ratio depends on the number of teeth and angular velocity, omega. omega equals linear velocity, v divided by the radius of the gear. Since the radius of A is larger than B & C, its angular velocity will be small.

The answer to your question depends on the context that the question is set in.

From a basic mechanics / engineering point of view the gears would all rotate at the same RPM because they have the same number of teeth.

From a Y13 circular motion point of view they will all have the same angular velocity.

omega = delta theta / delta T

all of the gears will rotate through the same angle in the same time so the angular velocity is the same: r (the radius of the gears) does not appear in the equation do the angular velocity is independent of radius

The velocity that a point on the outside of the gear rotates at will be different

v = 2 pi r / T

Each of the gears will rotate once in the same time period (T) but the circumfrence of the larger gear is larger so a point on the outside of the larger gear will move faster than a point on the smaller gears: in the equation v is proportional to r

Gears will rotate at the same speed as they all have 20 teeth. (Speed A * Number of teeth A) = (Speed B * Number of teeth B).

Torque will be different due to size difference.

Torque = (Force’s Circumferential Component) * (Gear Radius)

Torque Ratio = (Radius A)/(Radius B)

As it is larger gear 'A' would have a lower RPM, even though the cogs have the same amount of teeth it can be seen from the diagram that it is unlikely that the teeth would lock evenly and so I believe we can ignore the teeth as a factor.

Interleaving gears need to have compatible teeth. Those sprockets pictured are typically used with chains and are not usually used for direct sprocket to sprocket torque transfer. Sprocket A appears to be an enlarged version of sprockets B and C. If those were used like that then since sprocket A doesn't have a perfectly matching tooth profile to that of sprocket B there will eventually be times when the teeth collide and the sprocket either locks, counter-rotates slightly or creates a jerk in the otherwise fairly smooth motion. It could even break or bend the axle or the sprocket, or fling the sprocket away. Assuming that sprocket A is the driving sprocket then you might be able to prevent that from happening if you keep a constant counter-rotational torsional load on sprockets B or C so that the teeth of sprocket B accelerate towards being directly in contact with the teeth of sprocket A as soon as the previous tooth of sprocket B is released from being in contact with sprocket A. That counter-torque and the resulting angular acceleration would have to be high enough compared with the rotational speed of sprocket A, otherwise it wouldn't be able to prevent the teeth of sprockets A and B from randomly colliding.

If sprocket A had teeth that matched the profile of sprocket B then it would have several more teeth than those that are shown and they would be smaller.

If we assume that the teeth of sprocket A are exactly as shown, and if A is the driving sprocket, then since sprockets A, B and C all have the same number of teeth, they would all rotate at the same average rate, but, if there isn't a constant torsional load on sprocket B or C then sprocket A would move without contacting sprocket B for a small angle until it comes back into contact with the next tooth of sprocket B and continues to rotate sprocket B. That process would repeat for every tooth. If sprocket A rotates smoothly and slowly then sprockets B and C would start and stop their rotation as they await each next tooth of sprocket A. If sprocket A rotates smoothly but quickly then sprockets B and C may develop significant rotational momentum and will then oscillate between experiencing no contact with sprocket A and experiencing contact with sprocket A: Assuming no collisions between the teeth from incorrect intermeshing, when sprockets B and C rotate at their maximum RPM (within a very narrow permitted range of RPMs) after having just been nudged by a perfectly positioned tooth on sprocket A, then it may rotate in close synchrony with sprocket A without touching it, until friction slows it down enough for it to then come back into contact with one of the teeth of sprocket A and be nudged again. If the teeth of sprockets A and B collide then sprocket B may temporarily rotate at a higher rotational rate than sprocket A and have its next tooth catch up with the current tooth of sprocket A and bounce around like that until it settles into a smoother rotation again. If the collision coincides more with the other side of a tooth on sprocket A then it could make sprockets B and C suddenly decelerate or even counter rotate until they come into contact with the next tooth on sprocket A and resume their driven rotation, but perhaps with some bouncing around again until the motion settles back into smooth rotation.

In summary, they would each rotate with the same RPM on average because they have the same number of teeth and because sprocket B would continually be made to wait for sprocket A to come back into contact with it, but there would be some chaos because the profiles of the teeth don't properly interleave.

All three have the same number of teeth, so they have the same angular velocity. A has the largest radius, so its velocity is also the largest, because velocity is the product of angular velocity and radius.

As you've correctly pointed out because the gears have the same number of teeth, they would all move at the same RPM. It is probably easier to see with just two gears, where it is quite clear that the same number of teeth would the gears making the same number of rotations. Hence all RPMs will be the same, just the linear velocities at the boundaries of the gears would be different.

Hi Antony!

A fun question with gears!

When you’re asking if they all move at the same “speed” I’m going to presume you mean rotational speed, such as measured in rpm or rad/s.

I find the best way to think about is is that all the gears are meshed together, and therefore the same number of teeth will “move” on each gear.

Now, for a small gear (such an gear B or C) to fully rotate it takes less “teeth” than the big gear (such as A). Therefore, for every turn of the small gear, the big gear won’t quite complete a turn.

The big gear must therefore be turn slower!

Imagine if gear B had 10 teeth, and gear A had 30 (because it’s bigger). Every time gear A turned 1 full turn, gear B would have to do 3 full turns, and be going much faster!

I hope that helps. Let me know if you have any further questions.

Ben

The relationship of gears and their teeth is that the greater the radius of the gear the lower the RPM.

Also the number of teeth on a gear is directly proportional to it's speed.

Therefore their speeds will depend on the size of the gear and the number of teeth they have.

They all have the same number of teeth so A would rotate at the same rpm as B and C.

Speed of the gear A (if it is driven gear) is depend upon the gear ration of gear A and B. Similarly speed of the Gear B depend upon the gear ration of gear A and C (Driving gear supposed). Gear ratio is

(speed of driven/speed of driving gear)=(No of teeth of driving /No of teeth of driven gear)

so, If number of teeth in all the gears are same so the angular speed will remain same for all. However direction of the rotation will be different due to opposite tangential force on the periphery of the gear tooth. Also the linear velocity of any point on the gear circumference will differ depending upon the radius of the gear.