We have four forces: gravity, the normal force due to the incline, and friction. We will use a dot to represent the object as it is a point mass. Recall we also have to define a coordinate system, so we will choose a convenient set of basis vectors. The tangential/normal basis vectors $\hat{e}_t, \hat{e}_n$ work best for this problem. Therefore, our free-body diagram is the following:

Now, we can proceed with Newton's 2nd Law: \[\begin{aligned} \sum \vec{F}_{ext} &= m\vec{a} \\ m\vec{g} + \vec{F}_s + \vec{N} &= m\vec{a} \end{aligned}\]

Decomposing the vector equation into scalar equations in $\hat{e}_t, \hat{e}_n$ : \[\begin{aligned} \hat{e}_t: \qquad F_s + mg\sin\theta = ma_t \\ \hat{e}_n: \qquad N - mg\cos\theta = ma_n \end{aligned}\]

We will choose the polar basis to model this show basis

There are two forces present: Gravity, which is the force that allows for circular motion to be possible, and another force present. Think of this force as something that represents some kind of thrust on a rocket modeled as a point mass. Circular motion is still possible without it, however, the addition of it helps understand and model complex dynamical systems better. All in all, the free-body diagram is as follows:

Now, we can proceed with Newton's 2nd Law: \[\begin{aligned} \sum \vec{F}_{ext} &= m\vec{a} \\ m\vec{g} + \vec{F} &= m\vec{a} \end{aligned}\]

Decomposing the vector equation into scalar equations in $\hat{e}_t, \hat{e}_n$ : \[\begin{aligned} \hat{e}_r&: \qquad F_r - mg = ma_r \\ \hat{e}_\theta&: \qquad F_\theta = ma_\theta \end{aligned}\]

We can now substitute in the acceleration in polar found here: \[\begin{aligned} \vec{a} &= (\ddot{r} - r\dot\theta^2)\hat{e}_r + (r\ddot\theta + 2\dot{r}\dot\theta)\hat{e}_\theta \\ &= a_r \hat{e}_r + a_\theta \hat{e}_\theta \end{aligned}\]

Substituting: \[\begin{aligned} F_r - mg &= m\ddot{r} - mr\dot\theta^2 \\ F_\theta &= mr\ddot\theta + 2m\dot{r}\dot\theta \end{aligned}\]

Since $r$ is constant, $\dot r = 0$ and $\ddot r = 0$. Therefore: \[\begin{aligned} F_r - mg &= - mr\dot\theta^2 \\ F_\theta &= mr\ddot\theta \end{aligned}\]

There's a reason choosing a coordinate system is important when solving kinetics problems. We can verify that $F_\theta = mr\ddot\theta$ is true using Newton's 2nd Law in rotation. However, we have to choose an appropriate axis to apply it about, since the choice of an axis matters. We will choose point $O$ to be our reference point. Using that: \[\begin{aligned} \sum \vec{M}_O = I_O\vec{\alpha} \end{aligned}\]

where $I_O = mr^2$ (see here), and $\alpha = \ddot\theta$. Now the external moment about point $O$ is as follows: \[\begin{aligned} \vec{M}_O &= \vec{r} \times \vec{F} + \vec{r} \times m\vec{g} \\ &= r\hat{e}_r \times (F_r \hat{e}_r + F_\theta \hat{e}_\theta) + r\hat{e}_r \times (-mg \hat{e}_r) \\ &= F_r r (\hat{e}_r \times \hat{e}_r) + F_\theta r (\hat{e}_r \times \hat{e}_\theta) - mgr (\hat{e}_r \times \hat{e}_r) \\ &= F_\theta r \, \hat{k} \end{aligned}\]

Substituting everything: \[\begin{aligned} F_\theta r &= mr^2 \ddot\theta \\ F_\theta &= mr \ddot\theta \end{aligned}\]

Precisely as we obtained from linear Newton's 2nd Law.

We have four forces: gravity at point $C$, normal force from the ground at point $Q$, force $\vec{F}$ at point $P$, and friction at point $Q$. Therefore, the free-body diagram is the following:

Now, we can proceed with Newton's 2nd Law. Recall Newton's 2nd Law for rigid bodies is about the center of mass. Therefore: \[\begin{aligned} \sum \vec{F}_{ext} &= m\vec{a}_C \\ m\vec{g} + \vec{F}_s + \vec{N} + \vec{F} &= m\vec{a}_C \end{aligned}\]

Applying Newton's 2nd Law in rotation (Euler's 2nd Law). We will choose the center of mass as the axis of rotation: \[\begin{aligned} \sum \vec{M}_{ext} &= I_C \, \vec \alpha \\ \vec{r}_{CP} \times \vec{F} + \vec{r}_{CQ} \times \vec{F}_s &= I_C \, \vec \alpha \end{aligned}\]