Torque angular momentum equation in case of pure rolling I was doing some problems in rotational mechanics. 
In one such problem a solid sphere was kept on an inclined plane and pure rolling was taking place. In the solutions they have applied:
$$\tau = \frac{dL}{dt}$$
Torque equal to rate of change of angular momentum equation about the bottommost point.
But I know that the above mentioned equation is valid only for inertial frame of reference and the bottommost point is accelerated towards the centre so it should be a non-inertial frame of reference and hence pseudo torque should be applied.
Any help will be appreciated.
PS

This is one of the given solutions in the book.
The question just asks the velocity of COM of sphere after it has descended through a height H.
I have already mentioned the necessary details.
But i am posting the whole question.
A uniform sphere of mass m and radius R rolls without slipping down an inclined plane set at an angle to the horizontal.
The question has three parts
(1)Magnitude of friction coefficient when slipping is absent
(2)kinetic energy of sphere t seconds after the beginning of motion and (3) the velocity of COM at the moment it has descended through height H.
It is an easy question.
 My book mentions five ways to solve the question of which i have posted the one that i did not understand.
 A: The following analysis is performed in an inertial "lab/ground" reference frame.
Let our system be a collection of point particles labelled by index $i$: $\{m_i,\mathbf{x}_i,\mathbf{v}_i,\mathbf{a}_i\}$ 
and let $P$ be a general point that may or may not be in motion with respect to the inertial frame.
  The angular momentum with respect to point $P$ is defined1 as follows.
$$\mathbf{L}_P=\sum_i m_i\; (\mathbf{x}_i -\mathbf{x}_P)\times(\mathbf{v}_i-\mathbf{v}_P)\tag{1}$$ Taking time derivative on both sides, we have:\begin{align}\frac{d\mathbf{L}_P}{dt}&=\sum_i m_i\;(\mathbf{x}_i -\mathbf{x}_P)\times(\mathbf{a}_i-\mathbf{a}_P) \\
&=\sum_i m_i\;(\mathbf{x}_i -\mathbf{x}_P)\times\mathbf{a}_i-\sum_im_i\;\mathbf{x}_i\times \mathbf{a}_P +\sum_im_i\;\mathbf{x}_p \times\mathbf{a}_P \\ &=\sum_i(\mathbf{x}_i -\mathbf{x}_P)\times \mathbf{F}_i+(\mathbf{x}_{cm}-\mathbf{x}_P)\times(-m\mathbf{a}_P) \\ &=\boldsymbol{\tau}_P+(\mathbf{x}_{cm}-\mathbf{x}_P)\times(-m\mathbf{a}_P) \tag{2}\end{align}
The second term on the RHS of $(2)$ is precisely the "fictitious/pseudo" torque associated with the fictitious force $-m\mathbf{a}_P$, which OP has mentioned.   
Let's get back to OP's original question and consider the point of contact of the sphere with the incline to be point $P$. In this case, we observe: 
$$\mathbf{a}_P \text{ is parallel to } \mathbf{x}_{cm}-\mathbf{x}_P \stackrel{\text{Using eq. $(2)$}}\Rightarrow \frac{d\mathbf{L}_P}{dt}=\boldsymbol{\tau}_P$$
Also, since the velocity of the point of contact is zero, we observe that $(1)$ can be rewritten as follows:
\begin{align}\mathbf{L}_P&=\sum_i m_i\;(\mathbf{x}_i-\mathbf{x}_{cm})\times (\mathbf{v}_i-\mathbf{v}_{cm})+\sum_i m_i \; (\mathbf{x}_i-\mathbf{x}_{P})\times \mathbf{v}_{cm} \\ &= \mathbf{L}_{cm}+m\;(\mathbf{x}_{cm}-\mathbf{x}_P)\times\mathbf{v}_{cm}\\ L_P &=\frac{2}{5}mR^2\omega+mRv_{cm}\end{align}
References 


*

*Kirk T. McDonald, Comments on Torque Analyses; eq. $(14)$-$(15)$

1 One may also define angular momentum with respect to point $P$ as follows.
$$\mathbf{L}_P=\sum_i m_i\;(\mathbf{x}_i-\mathbf{x}_P)\times\mathbf{v}_i$$   The differences between the two definitions and detailed discussions are present in $[1]$.
A: yes the situation is indeed a non inertial one, but if we consider a inertial frame such as
with respect to(wrt) inclined plane, we don't get any axis to apply torque (using ICOR) it comes out that point of center of mass(COM) is the axis of rotation, so we apply torque wrt COM frame which is indeed non inertial, torque is produced by friction in this case, which has same direction as in inertial and non inertial one, unless the direction of force causing torque is different in inertial and non inertial frame of reference, we need not to apply pseudo torque here
