Angular momentum for gear

[risecolt]

In the image showed in the link below I've illustrated a mechanism in which a small gear rotates about a large gear when the motor is spinning. The large gear is rigid and both of the gears are masless, except for the motor. The small gear has a radius of 25mm and the large gear has a radius of 45mm. I need to calculate what angular momentum (torque) is required when the mass of the motor and the angular velocity is known. I have eliminated one degree of freedom, the gears are held in place from getting off.
I'm using the following equation:

M = w x I (angular velocity x moment of inertia).
By simplifying the shape of the motor into a cylinder:
I = m/2 * (r1^2 + (2*r1 + 2*r2)^2
The motor weighs 1 kg and the angular speed is 60 RPM (2pi rad/second).
M = (2pi) * 1/2 * (0.025^2 + (2*0.025 + 2*0.045)^2)) = 0.0635387 Nm

But when I try to compare this with a more classic example:
M = F x r = 1kg*9.81m/s^2 * 0.025 m (radius of the small gear) = 0.24525.
These two values are significantly different.

Which one of these calculations are correct?

http://cognitivenetwork.yolasite.com/resources/Gear.png

Diagram for moment of inertia:
http://www.caddisegni.com/upload/calcoli/1-2.jpg

I realize I've made a small mistake.
So if I once have calculated desired angular momentum, I define torque as the rate of change of angular momentum.
So if I want the angular momentum to occur within 1 second, I can set up the equation:
torque / second = angular momentum. And once the speed has been achieved, there is no more torque.

So my next problem should be: What should the rate of change of angular momentum be, in order to overcome the friction preventing the motor from spinning? In other words, if the angular velocity is too low, the motor will stop. How do I go about to do this?

 

[sci-phy]

And once the speed has been achieved, there is no more torque.

Sorry, but that's not true.. You always have to supply power (and hence torque) to overcome the friction.

 

[sci-phy]

Hi risecolt,

I don't think you need to go too deep into the physics of it. The power rating of the motor is enough. If the motor has rating P and the small gear is spinning with an angular velocity $ω_{1}$, the torque on the small gear is $τ_{1}$ = $P / ω_{1}$ . So, considering the gear ratio, the angular velocity of larger gear is $ω_{2}$ = $ω_{1}$ * $d_{1}$ / $d_{2}$. So the torque on this gear will be $τ_{2}$ = $P / ω_{2}$

Hope it helps

 

[risecolt]

Sorry, but that's not true.. You always have to supply power (and hence torque) to overcome the friction.

I agree. But I am thinking within a certain timeframe, and ideal system.
But we both know what we mean.

 

[risecolt]

Hi risecolt,

I don't think you need to go too deep into the physics of it. The power rating of the motor is enough. If the motor has rating P and the small gear is spinning with an angular velocity $ω_{1}$, the torque on the small gear is $τ_{1}$ = $P / ω_{1}$ . So, considering the gear ratio, the angular velocity of larger gear is $ω_{2}$ = $ω_{1}$ * $d_{1}$ / $d_{2}$. So the torque on this gear will be $τ_{2}$ = $P / ω_{2}$

Hope it helps

That's a bit of a problem. Because the large gear doesn't spin. It is rigid.
Only the small gear rotate. So the small gear has to do all the work. Rotate itself, and translate itself about the axis of the large gear. As a bachelor degree student I must include as much physics as possible. But perhaps you can tell me the relationship between this equation and the previous? How about:

Torque = power / angular velocity = moment of inertia * angular acceleration?