Question 11.11: The cotton reel problem A cotton reel is at rest on a rough ...

Suppose the ends of the reel have radius a, the axle wound with thread has radius b (with b < a), and the whole reel together with its thread has mass M. (We will neglect the mass of any thread pulled from the reel.) Let the reel have horizontal velocity V and angular velocity ω in the directions shown at time t. Note that we are not presuming the variables V, ω (or F) must take positive values. The signs of these variables will be deduced in the course of solving the problem.
The external forces on the reel are the string tension T and the fricton force F at the table. (The weight force and the normal reaction at the table cancel.) Hence, the equations of motion for the reel are

M \frac{d V}{d t}=T-F,                        (11.23)

M k^{2} \frac{d \omega}{d t}=a F-b T,                       (11.24)

where Mk² is the moment of inertia of the reel about {G, j}. We are not making any prior assumption about whether the reel slides or rolls. We will simply presume that the friction force F is bounded in magnitude by some maximum F^{\max } , that is,

-F^{\max } \leq F \leq F^{\max },                            (11.25)

and that F=+F^{\max }\left(\text { or }-F^{\max }\right) when the reel is sliding forwards (or backwards).
Whether the reel slides or rolls depends on v^{C} , the velocity of the contact particle C. Since v^{C} = V −ωa, it follows by manipulating the equations (11.23), (11.24) that v^{C}  satisfies the equation

\left(\frac{M k^{2}}{k^{2}+a b}\right) \frac{d v^{C}}{d t}=T-\gamma F                             (11.26)

where the constant γ is given by

\gamma=\frac{k^{2}+a^{2}}{k^{2}+a b}                       (11.27)

Different cases arise depending on how hard one pulls on the thread.

Strong pull T>\gamma F^{\max }

In this case, the right side of equation (11.26) is certain to be positive so that d v^{C} / d t>0 for all t. Since the system starts from rest, v^{C} = 0 initially and so v^{C} > 0 for all t > 0. In other words, the reel slides forwards. This in turn implies that F=F^{\max } so that the equations of motion (11.23), (11.24) become

\begin{aligned}\frac{d V}{d t} &=\frac{T-F^{\max }}{M}, \\\frac{d \omega}{d t} &=\frac{a F^{\max }-b T}{M k^{2}}.\end{aligned}

These equations imply that the reel slides forwards with constant acceleration and constant angular acceleration. Note that ω is positive for \gamma F^{\max }<T<(a / b) F^{\max } and negative for T>(a / b) F^{\max }.

Gentle pull T<\gamma F^{\max }
In this case, the reel must roll. The proof of this is by contradiction, as follows.

Suppose that the reel were to slide forwards at any time in the subsequent motion. Then there must be a time τ , at which v^{C} and d v^{C}/dt are both positive. The condition v^{C} > 0 implies that F=F^{\max } when t = τ , and the condition d v^{C}/dt > 0 then implies that T>\gamma F^{\max } when t = τ . This is contrary to asumption and so forward sliding can never take place. A similar argument excludes backward sliding and so the only possibility is that the reel must roll.

The reel must therefore satisfy the rolling condition V = ωa and this, together with the equations of motion (11.23), (11.24), implies that the reel must roll forwards with constant acceleration

\frac{d V}{d t}=\frac{a(a-b) T}{M\left(k^{2}+a^{2}\right)}.