# Momentum of this rotating gear?

1. Mar 18, 2013

### risecolt

I assume that the equation that applies for this problem is M = mg * r.
The large gear is rigid while the small gear rotates about the axis which passes through the center of the large gear by rotating a motor. What is the momentum which is required to turn the motor?

I'm confused because I'm used to calculating the momentum of an object rotating about an axis, when the point of rotation is at the same location as the axis. But in my problem, the center of rotation and the center what is making it rotate is different. The gears are weigthless.

http://cognitivenetwork.yolasite.com/resources/Gear.png [Broken]

Last edited by a moderator: May 6, 2017
2. Mar 18, 2013

### etudiant

If the gears are weightless, they will have 0 momentum.
The problem needs a bit more exact specification.

3. Mar 18, 2013

### risecolt

The gears are weightless, but not the motor which makes it rotate. The gear carries the motor around the large gear, as it rotates about it.

4. Mar 18, 2013

### tiny-tim

welcome to pf!

hi risecolt! welcome to pf!
yes, but if there's no resistance, no force is needed

doesn't the question mention any resistance?

(the mass of the motor isn't resistance, it's just mass)

5. Mar 18, 2013

### etudiant

The problem is not well defined.
Unless the small gear and motor are constrained, they will simply slip off the large gear rather than rotating around it. One might get some momentum estimate from the mass of the motor and the rpm of the small motor, which would give a tip speed of the gear and hence an approximate forward speed of the motor/gear assembly.

6. Mar 19, 2013

### risecolt

There is a force, that force is equal to gravity multiplied by the mass of the motor.
That force is located at the center of the motor.
The arm is equal to the radius of the small gear.
Are you saying that I can't use the formula M = F x r, because the momentum required depends on the friction coefficient between the contact surfaces of the two gears?

Those two gears are held in place by a mechanical structure. Just imagine that they do not simply slip off.

7. Mar 19, 2013

### sophiecentaur

I don't see where gravity comes into this. It all seems to be in a horizontal plane.??
You would need to draw in, explicitly, any framework that is holding this arrangement together - else the problem is not defined. What is fixed and what is allowed to move?

8. Mar 19, 2013

### risecolt

Yes, it's all on the horizontal plane, or more technically the top plane.
You don't see how the gravity would pull the mass of the motor downwards?!
The large gear in the center is rigid. The small gear is attached to the motor.
The motor turns, and the small gear rolls around the rigid gear in the center.
But the gears are still held in place, from falling apart. Meaning I have eliminated one of their three degrees of freedom. The plane it moves in is in the x and z coordinate, while it is locked from moving in the y coordinate.

9. Mar 19, 2013

### etudiant

The mass of this lashup would be centered at the axis of the small wheel, at distance
r1+r2 from the large wheel axis.
Assume the gears are held in contact by some frame linking the two axes, then the angular momentum of the small gear/motor combo will be set by the rotation speed of the small gear translated into angular velocity around the large gear.

10. Mar 20, 2013

### sophiecentaur

I see that but what has that got to do with Momentum, which involved Mass and not Weight? I thought this was an 'ideal' system, with no friction - and it would only be friction that could be affected by gravity here.

11. Mar 20, 2013

### risecolt

This is starting to make sense to me. So that means that I should set up the equation:

Angular Momentum = Force x arm = angular velocity x moment of inertia?

F x r = w x I

12. Mar 20, 2013

### risecolt

The angular momentum is what determines the torque required to make the motor turn the mass of the motor about the axis of the large gear. The weight is determined by the mass multiplied by the gravity generated by the earth's magnetic field. You can exclude friction in this example.

I was thinking if I could perhaps use the following equation: M = angular velocity x moment of inertia. So that if the maximum torque which the motor can generate by the motor is known, and if I could calculate the moment of inertia, I could calculate the fastest speed the gear could rotate around the axis?

13. Mar 20, 2013

### risecolt

I think I have finally found the solution because of your help.
I need to know what the angular velocity and the moment of inertia is, because calculating the force acting on the gear/motor alone is not enough, because torque increases by the increase of radius or speed. So if I consider the gear/motor as a cylinder to simplify things, I = 0.5 * (r1^2 + (r1+r2)^2), where r1 is the radius of the small gear, while the length at which what is rotating about the axis, is dislocated from its center, is the radius of the small gear + the radius of the large gear. So in an ideal system, if the max torque is known, the maximum allowed angular speed would be determined by the moment of inertia.

14. Mar 20, 2013

### sophiecentaur

BTW, are you assuming that the motor has no moment of inertia but just consists of an 'ideal' point mass?

15. Mar 20, 2013

### risecolt

Of course the motor has a moment of inertia, everything has. Only friction should be excluded, just like air resistance, because they can be very difficult to calculate. What is ideal about the mass is that it is centered in the motor.

16. Mar 20, 2013

### risecolt

Would you be so kind and see my last post, it is more relevant.

I had made a mistake previously. Torque is the change of rate of angular moment.
So if you know the angular moment, divide it by s (second).

17. Mar 21, 2013

### sophiecentaur

@risecolt
This has been muddled from the start. To get an answer, you have to start with a meaningful question
In the OP, you wrote:
This means nothing. In a frictionless system the torque (!) needed is infinitessimal.

You are still bringing "weight" into the argument (I assume you mean 'gravitational field'). Weight is irrelevant in the horizontal plane. It's all about Mass and Moment of Inertia.

There is no 'maximum speed' involved unless the motor itself is speed limited. The total angular momentum starts at zero and will remain at zero as the system is not fixed to the Earth (rotationally) - I have to assume. The casing of the motor will rotate one way and the whole motor will travel in a circle around the central pivot. The relevant thing to find out would be the relative angular acceleration of the motor (around its own axis to the angular acceleration of the motor about the central pivot. The relationship between the two angular rotation speeds will be given by the gear ratio and you can easily set up an equation involving this and equating angular momentum to zero.
It's just a more complicated form of the equivalent linear situation of rocket propulsion (reaction drive).

18. Mar 21, 2013

### risecolt

Thank you for your reply. I have previously suggested in a new thread that the torque is the change of rate in angular momentum. So if I divided the angular momentum by a second, I get the torque. But now I realize, it is more practical to just use Torque = angular acceleration * moment of inertia. So far so good. I know that the gear ratio is 0,55555555555555555555555555555556. The larger gear is 1,8 times larger.

However, when I've calculated this torque, it would be located on the (vertical) y axis of the motor. But is this torque simply transferred from the teeth of the small gear, to the motor, or do I still have to consider that for the small gear, there is rotation and (radial) translation (about the axis of the large gear).

Should following be the solution to my problem?

$\tau$ = $\alpha$ * $I$

By the way, I do realize, gravity is not taken into account, and I do find it a bit strange.
So I imagine myself that the system was confined within empty space with no gravity, I do agree only its own mass acts on the system, where moment of inertia is relevant.

19. Mar 21, 2013

### jbriggs444

If you divide angular momentum by time then you get torque.

If you divide angular momentum by one second you get a number of no particular relevance.
[You get the torque that would be required to achieve the given angular momentum if that torque were applied for a duration of one second]

You may have misread what risecolt wrote. If you have angular momentum expressed in kilogram meter2 per second and you divide by time expressed in seconds then you get torque expressed in kilogram meters2 per second2. The "second" enters in as a unit of measurement, not as a physical parameter of the problem.

That's a lot of 5's. Call the gear ratio 5 to 9 and save yourself some typing.

20. Mar 21, 2013

### risecolt

Thank you for replying. I am getting to the finish line now. So far I understand and accept that I must use the equation: torque = angular acceleration * moment of inertia.
This calculated torque would cross the axis of the center of the motor.
I know what the gear ratio is, but (1) do I use the gear ratio to calculate the angular acceleration of the small gear, and the torque remains the same, it's just transfered to the pin of the motor, or (2) I use the gear ratio to calculate the torque?