Proof that torque on a gyroscope equals derivative of its spin angular momentum Definitions:
$\vec{\tau}$ = torque
$\vec{L}$ = angular momentum
$\vec{R}$ = radius
$\vec{F}$ = force
$M$ = mass
$\vec{A}$ = acceleration
$\vec{V}$ = velocity
I am familiar with the proof that $\vec{\tau} = \vec{R} \times \vec{F} = \vec{R} \times M\vec{A} = \vec{R} \times M\frac{\mathrm{d}}{\mathrm{dt}}\vec{V} = \frac{\mathrm{d}}{\mathrm{dt}} (\vec{R} \times M\vec{V}) = \frac{\mathrm{d}}{\mathrm{dt}}\vec{L}$ , and in my research, this is the proof that is cited to justify the claim that the torque on a gyroscope due to gravity is equal to the derivative of its spin angular momentum.  But in the case of the gyroscope, the torque radius (extending from the pivot to the center of the gyro wheel) and the angular momentum radius (the radius of the gyro wheel) are not the same, that is, the $\vec{R}$ in the torque equation is not, according to my understanding, the same as the $\vec{R}$ in the angular momentum equation.  So why does this proof apparently suffice to make this claim?
For example, the following diagram:

where
$\vec{d}$ = the arm extending from the pivot to the gyro wheel
$\vec{r}$ = radius of the gyro wheel
$\vec{g}$ = acceleration due to gravity
$\vec{V} = \vec{\omega} \times \vec{r}$ = spin velocity of gyro
What is the proof that $\vec{d} \times M\vec{g} = \frac{\mathrm{d}}{\mathrm{dt}}(\vec{r} \times M\vec{V})$
 A: 
What is the proof that $\vec{d} \times M\vec{g} = \frac{\mathrm{d}}{\mathrm{dt}}(\vec{r} \times M\vec{V})$
I don't think that this is correct ?.
with:
$$\vec{d} \times M\vec{g}=\vec\tau=\frac{d}{dt}\vec L\tag 1$$
where $~\vec L~$ is the angular momentum of system.
but only where the components of the torque $~\vec\tau~$ are  zero ,the
components of the angular momentum $~\vec L~$ are conserved.
with
$$\vec L=\vec d\times M\vec v+\Theta\,\vec\omega$$
and
$$\vec d= d\left[ \begin {array}{c} -\sin \left( \psi \right) \cos \left( 
\varphi  \right) \\ \cos \left( \psi \right) \cos
 \left( \varphi  \right) \\ \sin \left( \varphi 
 \right) \end {array} \right] 
$$
$$\vec v=d\,
 \left[ \begin {array}{c} \sin \left( \psi \right) \sin \left( 
\varphi  \right) \dot\varphi -\cos \left( \psi \right) \cos \left( 
\varphi  \right) \dot\psi \\  -\cos \left( \psi \right) 
\sin \left( \varphi  \right) \dot\varphi -\sin \left( \psi \right) \cos
 \left( \varphi  \right) p\psi \\ \cos \left( 
\varphi  \right) \dot\varphi \end {array} \right] 
$$
$$\vec\omega=
 \left[ \begin {array}{c} \dot\varphi +\sin \left( \psi \right) \omega
\\  \sin \left( \varphi  \right) \dot\psi +\cos \left( 
\psi \right) \omega\,\cos \left( \varphi  \right) 
\\  \cos \left( \varphi  \right) \dot\psi -\cos \left( 
\psi \right) \omega\,\sin \left( \varphi  \right) \end {array}
 \right]
$$
$$\mathbf\Theta=
\left[ \begin {array}{ccc} I_{{x}}&0&0\\ 0&I_{{y}}+
I_{{w}}&0\\ 0&0&I_{{z}}\end {array} \right]
$$
thus equation (1)
$$\vec\tau=M\,g\,d\,\left[ \begin {array}{c} \cos \left( \psi \right) \cos \left( 
\varphi  \right) \\ \sin \left( \psi \right) \cos
 \left( \varphi  \right) \\ 0\end {array} \right]
$$
from here
$$(\vec L)_z=\left(\vec d\times M\vec v+\Theta\,\vec\omega\right)_z=\left( M \left( \cos \left( \varphi  \right)  \right) ^{2}{d}^{2}+{
\it I_z}\,\cos \left( \varphi  \right)  \right) \dot\psi -{\it I_z}\,\cos
 \left( \psi \right) \omega\,\sin \left( \varphi  \right) 
=\text{constant}$$
