# Homework Help: Dynamics Physics - Acceleration Analysis / Gear Ratios /

1. Jun 17, 2013

### f2434427@rmqkr

First time posting.

1. The problem statement, all variables and given/known data

Gears A and B, of Masses 4kg and 10kg, respectively, are rotating about their mass centers. The radius of gyration about the axis of rotation is 100mm for A and 300mm for B. A constant couple of C=0.75Nm acts on gear A. Neglecting friction, compute:

- Angular Acceleration of each gear
- Tangential contact force between the gears at C

May want to look at attachment for image.
2. Relevant equations

3. The attempt at a solution

Started with the smaller gear.

Using T=I$\alpha$ , where I=mr^2

Rearrange the equation to get $\alpha$ = 0.75 / (4*0.1^2) = 18.75 rad/s^2

Not to sure how to draw a relationship with the 2nd larger gear. Maybe using w1/w2 = r2/r1 then $\alpha$ = w^2 * r

Thanks,

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2. Jun 17, 2013

### Simon Bridge

Welcome to PF;
The applied torque has to accelerate both gears doesn't it?
If this were a linear case, you'd use a free body diagram to couple the masses wouldn't you?

3. Jun 17, 2013

### f2434427@rmqkr

Ah yes, that would make sense. Forgot to account for the 2nd gears inertia

hmm. Dont quite understand what you mean by coupling the masses though

4. Jun 17, 2013

### Simon Bridge

When you push on one mass, it pushes on the other one .. the masses are said to be coupled ... the motion of one affects the motion of another.

5. Jun 17, 2013

### f2434427@rmqkr

Alrighty,

Well the force applied from gear A, will have an opposite reaction on Gear B.

T=F*d

F=0.75/0.15 = 5n

then applying this force to Gear B.

T=F*D = 5 * 0.45 = 2.25 Nm

alpha = T/I = 2.25/(10*0.3^2) =2.5 rad/s^2

Looks about right to me..... hopefully. Logically, first gear would have to spin faster than the 2nd.

6. Jun 17, 2013

### Simon Bridge

That's right - the smaller gear will reach a higher angular velocity in the same time, so must have a higher acceleration. You should be able to see from the relative sizes how the accelerations are related.