Finding velocity of a rolling disc down a slope When finding the velocity of a rolling disc using newton's second law method we need to take into account the friction. We use $$  mg \sin(\theta) - F = ma  $$ and $$ Fr = I \alpha$$ $$I = \dfrac{mr^2}{2}$$ 
However, if we are finding the velocity using the energy method. We use   $ T =  \dfrac{1}{2} mv^2 + \dfrac{1}{2} Iw^2 $  and  $ V = mgh$ .
After some manipulation of either set of equations we can achieve $v = \sqrt{\dfrac{4sfg \sin\theta}{3}}$
So why is it that there is no account for friction when using the energy method?
 A: We don't explicitly account for the work done by friction because friction actually does no net work. In terms of energy dissipation by friction, if there is rolling with no slipping, then there is no energy loss. However, it is the case that friction is the force that is responsible for allowing some of our potential energy to be converted into kinetic energy due to rotation. Since we are converting between types of energy with no loss in energy, then we can just consider the kinetic and potential energies. See below to see how we do account for this in the "rotational kinetic energy" term.

If you want to, you can consider the work done by friction in terms of the rotation (note, this is not the total work done by friction here, as that is $0$). The work done by a torque is given by
$$W=\int\tau\ \text d\theta$$
Assuming a constant friction force, we get a constant torque, and so
$$W_{f}=Fr\Delta\theta=I\alpha\Delta\theta$$
Now, under constant angular acceleration, and assuming the disk was released from rest, $\Delta\theta=\frac{\omega^2}{2\alpha}$, therefore
$$W_f=\frac12I\omega^2$$
So you see, we actually do account for this work by using this "rotational kinetic energy" term.
