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.
I already did the first part as shown in the picture below:
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.
Best Answer
One solution is the above. The other one is from the rotation at the point when the spool touching the ground. For simplicity, let $\alpha = 0$:
kinetic relation: $a_{||} = R \frac{d \omega}{dt}$
Newton law in rotation: $F(R-r) = (\gamma + 1) R^2\frac{d \omega}{dt}$
So that: $a_{||} = \frac{1- \frac{r}{R}}{\gamma + 1} \frac{F}{m}$