A spool of thread lies on a horizontal plane. The reel is pulled by a thread. At what angles \alpha between the force and the horizontal will the coil accelerate towards the taut thread?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#2.7">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.7.20^*.$ A spool of thread lies on a horizontal plane. The reel is pulled by a thread. At what angles $\alpha$ between the force and the horizontal will the coil accelerate towards the taut thread?

    <h3>Solution</h3>
    <p>

The second Newton's law for the horizontal axis
$$ma=F\cos\alpha-F_{fr}\quad(1)$$
Let's write a moment of inertia with respect to the axis of rotation of the spool with angular acceleration $\varepsilon$
$$F_{fr}\cdot R - F\cdot r = \frac{mR^2}{2}\cdot \varepsilon\quad(2)$$
The moment of inertia of the spool in our model (a solid cylinder, symmetry axis)
$$I=\frac{mR^2}{2}$$
Because there's no slippage:
$$a=\varepsilon\cdot R\quad(3)$$
Let's substitute $(3)$ into $(1)$
$$F_{fr}=F\cos\alpha-m\varepsilon R\quad(4)$$
and put into $(2)$
$$\left(F\cos\alpha-m\varepsilon R\right)\cdot R - F\cdot r = \frac{mR^2}{2}\cdot \varepsilon$$
After algebraic transformations
$$F\left(R\cos\alpha-r\right)=\frac{3mR^2\varepsilon}{2} \Leftrightarrow \boxed{\varepsilon=\frac{2F\left(R\cos\alpha-r\right)}{3mR^2}} \quad(5)$$
For the angular acceleration to be clockwise, and consequently the motion to be co-directional with the direction of $\vec{F}$, the external force moment from equation $(2)$ must be positive, and consequently the expression $(5)$ must be greater than 0. The moment of external forces in equation $(2)$ must be positive $(\varepsilon = 0)$
$$\frac{2F\left(R\cos\alpha-r\right)}{3mR^2}>0$$
$$R\cos\alpha-r>0\Leftrightarrow \boxed{\cos\alpha>\frac{r}{R}}$$
The larger the angle $\alpha$, the smaller the value of the trigonometric function $\cos\alpha$, making the expression for the angle
$$\boxed{\alpha<\arccos\left(\frac{r}{R}\right)}$$

    <h4>Answer</h4>
    <p>
        $$\cos\alpha>\frac{r}{R}$$
    </p>


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