1. Jan 4, 2009

### tachu101

1. The problem statement, all variables and given/known data
A dis of mass (m) and radius (r) with I=(1/2)(m)(r^2) is rolling down an incline dragging with it a mass (m) attached with a light rod to a bearing at the center of the disc. The friction coefficients are the same for both masses, Uk and Us.

Determine the linear acceleration of the mass (m) (not the disk)
Determine the friction force acting on the disc
Determine the tension in the rod.

2. Relevant equations
Torque=Ia ; a(linear)= a(rot)*radius ; friction= u*Normal

3. The attempt at a solution

Should I sum forces first? Or is there some thing i'm missing

2. Jan 4, 2009

### Staff: Mentor

Start out, as always, by drawing free body diagrams of both objects identifying all the forces acting.

3. Jan 4, 2009

### tachu101

$$\sum$$Fx= Wsin$$\Theta$$(disk)-Friction(disk)-Tension+Tension+Wsin$$\Theta$$(mass)-Friction(mass)= Ma

$$\sum$$Fy= N-Wcos$$\Theta$$=0

4. Jan 4, 2009

### Staff: Mentor

Rather than lump the forces together, treat each object separately. Applying Newton's 2nd law, you'll end up with three equations.

5. Jan 4, 2009

### tachu101

For Disk : Fx= Wsintheta-friction-T=Ma
For Mass : Fx= Wsintheta-friction+T=Ma
For both: Fy= N-Wcostheta=0

Now What?

6. Jan 4, 2009

### Staff: Mentor

So far, so good. Realize that the friction forces on each are not necessarily the same, so use different symbols. In the case of the sliding mass, you can express the friction in terms of the normal force.

Apply Newton's 2nd law to the rotation of the disk to get another equation. (I wasn't counting the vertical force equation when I said you'd end up with three equations.)

7. Jan 4, 2009

### tachu101

Net torque= Friction(disk)* Radius= Ia(rotational)= (1/2)(M)(R^2)(alinear/R)? Is this then ((mgcostheta)*Uk)*R= (1/2)(M)(R^2)(a linear/R). This will solve for the linear acceleration of the disk. And that is the same as the linear acceleration of the mass right? For the friction force acting on the disk would I just solve for the friction on the disk instead. To find tension would I put the values into Wsintheta-friction on disk-T = Ma , but then what value would Ma have?

8. Jan 4, 2009

### Staff: Mentor

Good.
No. Careful here: The friction against the disk is static friction, thus you can't simply equate it to μN. You'll have to solve for the friction.

Hint: The friction force on the dragged mass is kinetic friction, which can be equated to μN.
Yes, since they are attached they both have the same linear acceleration.

By combining your equations, you'll be able to solve for the friction force on the disk, the acceleration, and the tension.

9. Jan 4, 2009

### tachu101

Friction= (1/2)(M)(a(linear)) ? then I plug that in to solve for the values?

10. Jan 5, 2009

### Staff: Mentor

Yes. Combine that equation with your others and you'll be able to solve for the unknowns.