1. Dec 6, 2013

### wr8899

1. The problem statement, all variables and given/known data
A yoyo of mass m= 2 kg and moment of inertia I_cm= 0.0625 kg m^2 consists of two solid disks of radius R=0.25 m, connected by a central spindle of radius r=0.1875 m and negligible mass. A light string is coiled around the central spindle. The yoyo is placed upright on a flat rough surface and the string is pulled with a horizontal force F= 22 N, and the yoyo rolls without slipping.

(a) What is the x-component of the acceleration of the center of mass of the yoyo? (in m/s^2 )
(b)What is the x-component of the friction force? (in N)

2. Relevant equations
F = ma, torque = I*alpha, a = R*alpha

3. The attempt at a solution

I tried to solve the problem this way:

Using F = ma, and the given, I had F - friction force = ma, with friction force = 22 - 2a. With torque = I*alpha, I got 0.25*friction force - 0.1875*22 = 0.0625*alpha.

To put them all together, I got this equation, R(22-2a) - 0.1875*22 = (0.0625/0.25)*a
And then plugging the given, I got a = 1.83 m/s^2, but I don't think I'm doing this correctly.

Can someone please help me figure out where I went wrong right away. Thanks so much, sorry to be a bother here.

2. Dec 6, 2013

### haruspex

There's an ambiguity in the question. Is the horizontal string tangential to the spindle above or below the centre of mass? Taking it to be below, I get the same answer as you did. (The easiest way is to take moments about the point of contact with the ground - then you don't need to consider the frictional force.)

3. Dec 7, 2013

### Ruitker

What difference does it make if it is above the centre of mass?

4. Dec 7, 2013

### TSny

It affects the direction of the torque produced by the string about the center of mass.

5. Dec 7, 2013

### Ruitker

So in this instance, how would the formulae change?

6. Dec 7, 2013

### TSny

The signs of the torque terms depend on your conventions for positive and negative torque and whether or not the string is being pulled to the right or to the left as viewed from the side. You haven't specified these details. Once you are clear on the setup, you should be able to decide if a particular torque is positive or negative when setting up the equation.

7. Dec 7, 2013

### haruspex

If you use my approach, taking moments about the point of contact with the ground, you will see that the moment of the tension in the string changes because the distance is either R+r or R-r.