Find the acceleration and the work done by a force pulling a spool?

[Davidllerenav]

Homework Statement

A spool with thread wound on it, of mass m, rests on a rough horizontal surface. Its moment of inertia relative to its own axis is equal to $I=\gamma mR^2$, where $\gamma$ is a numerical factor, and $R$ is the outside radius of the spool. The radius of the wound thread layer is equal to $r$. The spool is pulled without sliding by the thread with constant force F directed at an angle $\alpha$ to the horizontal. Find:

a) the projection of the acceleration vector of the spool axis on the x-axis.
b)the work performed by the force during the first $t$ seconds after the beginning of motio.

Relevant Equations

Laws of Newton.
$\tau=lF\sin\theta$
$\tau=l\times F$
$\sum\tau=I\alpha$

I already solved the first part, but according the my book, the answer isn't quite correct. This is what I did.

Finally, I ended up with $a=\frac{F(r-R\cos\alpha)}{Rm(\gamma+1)}$. But according to my book, the answer is $a=\frac{F(\cos\alpha-\frac{r}{R})}{m(1+\gamma)}$, what am I doing wrong?
For the second part, I don't know how to find the work done by the force $F$.

 

[haruspex]

Finally, I ended up with $a=\frac{F(r-R\cos\alpha)}{Rm(\gamma+1)}$. But according to my book, the answer is $a=\frac{F(\cos\alpha-\frac{r}{R})}{m(1+\gamma)}$, what am I doing wrong?

The only difference is the sign. Maybe you are defining the positive direction for the acceleration opposite to the way the book defines it.

 

[Davidllerenav]

The only difference is the sign. Maybe you are defining the positive direction for the acceleration opposite to the way the book defines it.

Yes, I think so. Does that mean that I'm not wrong? I was wondering, in what direction does the spool move?

 

[haruspex]

Yes, I think so. Does that mean that I'm not wrong? I was wondering, in what direction does the spool move?

Looking at your working (it is painful referencing lines in work posted as image; please don't do that, images are for diagrams and book extracts), you have an error after the line $F\cos(\alpha)-F_r=ma$.

 

[Davidllerenav]

(it is painful referencing lines in work posted as image; please don't do that, images are for diagrams and book extracts)

Sorry, I just did it so to avoid writing everything and taking indidual images for diagrams.

you have an error after the line Fcos(α)−Fr=maFcos⁡(α)−Fr=maF\cos(\alpha)-F_r=ma.

Oh, it should be $F_r=F\cos\alpha-ma$, right?

 

[haruspex]

Sorry, I just did it so to avoid writing everything and taking indidual images for diagrams.

Oh, it should be $F_r=F\cos\alpha-ma$, right?

Right, and that should give you the right sign in the answer.

 

[Davidllerenav]

Right, and that should give you the right sign in the answer.

But wouldn't that give me $m(\gamma-1)$ on the denominator?

 

[Davidllerenav]

But wouldn't that give me $m(\gamma-1)$ on the denominator?

I think I did it, I ended up with $a=\frac {F\cdot r-F\cos\alpha}{-Rm (\gamma+1)}$ thus I have $a=\frac {F\cos\alpha-F\cdot r}{Rm (\gamma+1)}$ , am I right?

How can I find the work done.by $F$?

 

[haruspex]

I think I did it, I ended up with $a=\frac {F\cdot r-F\cos\alpha}{-Rm (\gamma+1)}$ thus I have $a=\frac {F\cos\alpha-F\cdot r}{Rm (\gamma+1)}$ , am I right?

How can I find the work done.by $F$?

Hard to be sure whether your method is now entirely right without seeing all of your latest working. One thing to be careful of is consistency of signs in "a=αR" equations. In the present case, if you are defining a to be right-positive then that makes α clockwise-positive.

For the work you have two options. You can use force x distance or KE achieved. The former can be tricky where the point of application of the force is moving - which is the distance to use? So safer to find the velocity reached.

Edit: just noticed you have dropped an R in the top line in your final answer for a, making it dimensionally inconsistent. I assume that is a typo.