# Rotational Velocity with 2 Gears

1. Mar 29, 2016

### PrettyLights

1. The problem statement, all variables and given/known data
A small gear with mass m and radius r rotate around a central axis. A force is applied to an interior hub at a radial distance r/2 from the axis.
A. If a force of 4N is applied for 10s, what is the angular velocity of the small gear, assuming it starts at rest?
B. Once it reaches this speed, the small gear is engaged with a larger gear next to it, of mass 4m and radius 4r. If the angular speed of the small gear is maintained, what is the angular velocity of the larger gear, assuming no slipping?

2. Relevant equations
F=MAt
R x alpha= At
omega= omega initial + alpha x T
I=1/2 MR^2

3. The attempt at a solution
For the first piece, I tried to answer it by setting up F=MAt as 4=MAt and At=r x alpha, and then setting them equal and solving for omega, and then plugging them in to omega= omega initial + alpha x T. I feel like this isn't correct and that it is leaving out some key pieces. Any help is appreciated.

2. Mar 30, 2016

### haruspex

I assume At stands for tangential acceleration. Fnet=ma is a standard enough equation, but the applied force is not the net force in this case. There is also a force from the axle, holding the centre of the gear in place. Besides, not all of the gear's mass has the same linear acceleration. Think about moments and torque instead.

3. Mar 30, 2016

### PrettyLights

So, I can relate the moment of inertia to torque through the following:
T= R x F= 4 x R/2 = 2R
I= 1/2mR^2
So, since T=I x alpha --> 2R=1/2mR^2 x alpha

Is this in the right direction? I'm not exactly sure how to get to velocity from here. Should I isolate for alpha and use one of the rotational kinematic equations?

4. Mar 30, 2016

### haruspex

Yes, that's right so far.
You get angular velocity and displacement from angular acceleration and time in just the same way that you do for linear velocities, accelerations and displacements.

5. Mar 30, 2016

### PrettyLights

Gotcha- so, alpha=4/mR and (4/mR)x10= 40/mR

Thank you!