Confusion regarding rest mass of system in relativistic conditions In Morin's Mechanics there is this interesting question where we are required to find the mass of a system under relativistic conditions:

12.8. System  of  masses    **
Consider a dumbbell made of two equal masses, $m$. The dumbbell  spins  around, with its center pivoted at the end of a stick (see Fig. 12.15). If the  speed of the masses is $v$, then the energy of the system is $2γm$. Treated  as a whole, the system is at rest. Therefore, the mass of the system must  be $2γm$. (Imagine enclosing it in a box, so that you can’t see what’s  going on inside.) Convince yourself that the system does indeed behave  like a mass of $M = 2γm$, by pushing on the stick (when the dumbbell  is in the “transverse” position shown in the figure) and showing that $F ≡  dp/dt =  Ma$.


The mass of the system is quite clearly $2γm$, considering that
$$M=\sqrt{(\sum_i E_i)^2-(\sum_i p_i)^2}=\sqrt{(2*\gamma m)^2}=2 \gamma m$$
However, when I attempted to confirm the result by Morin's argument, I came across a contradiction.
The force on each particle, F, is given by
$$F=\frac{dp}{dt}=\frac{d}{dt}(\gamma m v)=mv\frac{d\gamma}{dt}+ m\gamma\frac{dv}{dt}=mv^2\gamma^3 a+m \gamma a = m \gamma^3 a$$
Total force required to accelerate the particles is therefore
$$F_{total}=2m\gamma^3 a = \gamma^2 M a \neq M a$$
What is wrong in my chain of reasoning?
 A: 
Consider a dumbbell made of two equal masses, [...]

It may be advantageous to denote the two dumbbell weights nevertheless distinctly, say as $A$ and $B$; with equal masses $m_A = m_B := m$.

Convince yourself that the system does indeed behave like a mass of $M = 2 \, \gamma \, m$, by pushing on the stick (when the dumbbell is in the “transverse” position shown in the figure) [...]
The force on each particle,

Force, momentum, velocity and acceleration are of course quantities with direction as well as magnitude. Accordingly, as determined initially by members of the reference frame of the dumbbell system at the “transverse” position:
$$\vec v_B = -\vec v_A$$
with
$$(\vec v_A \cdot \vec v_A) = (\vec v_B \cdot \vec v_B) := v^2 \gt 0,$$
and
$$\vec p_A = -\vec p_B = \frac{m \, \vec v_A}{\sqrt{1 - (\vec v_A \cdot \vec v_A) }} = \frac{-m \, \vec v_B}{\sqrt{1 - (\vec v_A \cdot \vec v_A) }} := \frac{m \, \vec v_A}{\sqrt{1 - v^2 }}.$$
Also, importantly, at the “transverse” position initial to pushing on the stick:
$$\frac{d}{dt} \! \left[ \phantom{\frac{d}{dt}} \! \! \! \! \! \! \! \vec v_A \right] =: \underset{\Delta t \longrightarrow 0}{\text{lim}}\left[ \left( \frac{\vec v_A + \Delta t \, \vec a}{1 + (\vec v_A \cdot \vec a) \, \Delta t} - \vec v_A \right) / \Delta t \right] = \vec a - \vec v_A (\vec v_A \cdot \vec a) = \vec a \, (1 - v^2)$$
and
$$\frac{d}{dt} \! \left[ \phantom{\frac{d}{dt}} \! \! \! \! \! \! \! \vec v_B \right] =: \underset{\Delta t \longrightarrow 0}{\text{lim}}\left[ \left( \frac{\vec v_B + \Delta t \, \vec a}{1 + (\vec v_B \cdot \vec a) \, \Delta t} - \vec v_B \right) / \Delta t \right] = \vec a - \vec v_B (\vec v_B \cdot \vec a) = \vec a \, (1 - v^2).$$
(This seems to be the origin of "the factor $1 / \gamma^2$" you had been missing.)
Then let's evaluate $$F_{\text{total}} := \frac{d}{dt} \! \left[ \phantom{\frac{d}{dt}} \! \! \! \! \! \! \! \vec p_{\text{total}} \right] = \frac{d}{dt} \! \left[ \phantom{\frac{d}{dt}} \! \! \! \! \! \! \! \vec p_A\left[ \, m, \vec v_A \, \right] + \vec p_B\left[ \, m, \vec v_B \, \right]  \right] = \frac{d}{dt} \! \left[ \phantom{\frac{d}{dt}} \! \! \! \! \! \! \! \vec p_A\left[ \, m, \vec v_A \, \right] \right] + \frac{d}{dt} \! \left[ \phantom{\frac{d}{dt}} \! \! \! \! \! \! \! \vec p_B\left[ \, m, \vec v_B \, \right] \right] $$
with
$$\! \frac{d}{dt} \! \left[ \phantom{\frac{d}{dt}} \! \! \! \! \! \! \! \vec p_A\left[ \, m, \vec v_A \, \right] \right] := \frac{d}{dt} \! \! \left[ \frac{m \, \vec v_A}{\sqrt{1 - (\vec v_A \cdot \vec v_A) }} \right]  = \frac{m}{\sqrt{1 - v^2 }} \, \frac{d}{dt} \! \! \left[ \phantom{\frac{d}{dt}} \! \! \! \! \! \! \! \vec v_A \! \right] \, + \,  m \, \vec v_A \, \frac{d}{dt} \! \! \left[ \phantom{\frac{d}{dt}} \! \! \! \! \! \! \! \frac{1}{\sqrt{1 - (\vec v_A \cdot \vec v_A) }} \right] \! \! =\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{m \, \vec a \, (1 - v^2)}{\sqrt{1 - v^2 }} \qquad + \qquad m \, \vec v_A \, \frac{(\vec v_A \cdot \vec a \, (1 - v^2))}{\left(\sqrt{1 - v^2 } \right)^3} = \\ \qquad \qquad \qquad \qquad \qquad \qquad m \, \vec a \, \sqrt{1 - v^2 } \qquad + \qquad  \frac{m \, v^2 \, \vec a}{\sqrt{1 - v^2 }} \qquad \qquad = \frac{m \, \vec a}{\sqrt{1 - v^2 }}, $$
while
$$\! \frac{d}{dt} \! \left[ \phantom{\frac{d}{dt}} \! \! \! \! \! \! \! \vec p_B\left[ \, m, \vec v_B \, \right] \right] := \frac{d}{dt} \! \! \left[ \frac{m \, \vec v_B}{\sqrt{1 - (\vec v_B \cdot \vec v_B) }} \! \right] \! = \! \frac{m}{\sqrt{1 - v^2 }} \, \frac{d}{dt} \! \! \left[ \phantom{\frac{d}{dt}} \! \! \! \! \! \! \! \vec v_B \right] \, + \,  m \, \vec v_B \, \frac{d}{dt} \! \! \left[ \phantom{\frac{d}{dt}} \! \! \! \! \! \! \! \frac{1}{\sqrt{1 - (\vec v_B \cdot \vec v_B) }} \right] \! = \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{m \, \vec a \, (1 - v^2)}{\sqrt{1 - v^2 }} \qquad + \qquad m \, \vec v_B \, \frac{(\vec v_B \cdot \vec a \, (1 - v^2))}{\left(\sqrt{1 - v^2 } \right)^3} = \\ \qquad \qquad \qquad \qquad \qquad \qquad m \, \vec a \, \sqrt{1 - v^2 } \qquad + \qquad \frac{m \, v^2 \, \vec a}{\sqrt{1 - v^2 }} \qquad \qquad = \frac{m \, \vec a}{\sqrt{1 - v^2 }}.$$
Together therefore
$$F_{\text{total}} := \frac{2 \, m \, \vec a}{\sqrt{1 - v^2 }} = M \, \vec a.$$
