I need some help with this problem:

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 $$ 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:

  • 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 motion.

enter image description here I already did the first part as shown in the picture below: enter image description here

I ended up with $a=\frac{F(r-R\cos\alpha)}{Rm(\gamma+1)}$, but according to my book the answer is $a=\frac{F(\cos\alpha-\frac{r}{R})}{m(1+\gamma)}$. I don't understand why are the signs different, what am I doing wrong?

I don't have too much of a clue for the second part, maybe can you give me a hint? Hope you can help me.


You need to be careful in applying the right hand rule to determine the direction of the torques and the angular acceleration. Notice that since the spool is rolling without slipping, it will have angular acceleration in the clockwise direction. By the right hand rule, this means that the angular acceleration is into the page. If we take both the angular acceleration and horizontal acceleration to be positive, which you've done with $a = R\alpha$, then the torque from $F$ is directed out of the page (i.e. negative) while the torque from friction is into the page (i.e. positive). Thus you should have the following equations: $$F\cos\alpha-F_f = m a_{||} \\ a_{||} = R\alpha \\ F_f R - Fr = I \alpha.$$ Solving those for $a_{||}$ yields the answer from the book.

The second part is definitely trickier. My suggestion is the following. We know that $$\Delta K = W_{ext}.$$ We also know that the force of friction does no work since the spool doesn't slip. Therefore if we compute $\Delta K$ then we know how much work was done by the force. Remember that the kinetic energy has two contributions: one from the horizontal motion and one from the rotational motion. I found a final answer of $$W_{ext} = \frac{1}{2} \frac{F^2(\cos\alpha-\frac{r}{R})^2}{m(1+\gamma)}t^2.$$

Good luck!

  • $\begingroup$ I'm confused with the sign of torque. So it doesn't matter the direction or the rotation that the torque creates, the sign of the torque depends on whether the torque is in the same direction as angular velocity and acceleration? $\endgroup$ – davidllerenav Apr 25 at 15:56
  • $\begingroup$ You need to decide whether into the page is positive or negative. Since you've chosen to take $a_{||}=R\alpha$, you've chosen $\alpha>0$. So you've chosen into the page as positive. $\endgroup$ – WAH Apr 25 at 15:57
  • $\begingroup$ So, if I would have decided that into the page was negative, the torque caused by $F$ would be positive while the torque caused by $F_r$ would be negative, right? $\endgroup$ – davidllerenav Apr 25 at 16:00
  • $\begingroup$ Yes. And then you've have that $a_{||}=-R\alpha$. $\endgroup$ – WAH Apr 25 at 16:02
  • $\begingroup$ Is a bit confusing to be honest. If I understood coorectly, if the torque has the same direction as the angular velocity, both of them have the same sign and therefore if I choose that direction as positive, both will be positive, correct? $\endgroup$ – davidllerenav Apr 25 at 16:06

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