The motion of any point on the rolling ring can be described as a superposition of the translational motion of the center O and rotational motion about O. Let's define a coordinate system with the origin on the horizontal surface, $\hat{\imath}$ pointing horizontally to the right, and $\hat{\jmath}$ pointing vertically upwards.

The center of the ring O has velocity $\vec{v}_O = v_0 \hat{\imath}$. Since $v_0$ is constant, the acceleration of the center is $\vec{a}_O = \vec{0}$.

For pure rolling, the velocity of the center $v_0$ and the angular velocity $\omega$ are related by $v_0 = \omega R$. For the ring to move forward, the rotation must be clockwise, so the angular velocity vector is $\vec{\omega} = -\omega \hat{k} = -\frac{v_0}{R} \hat{k}$. Since $v_0$ is constant, the angular acceleration $\vec{\alpha} = \vec{0}$.

We consider the point P at the same height as the center. Let's choose P to be at the front of the ring. The position vector of P relative to O is $\vec{r}_{PO} = R \hat{\imath}$.

[Q1] Instantaneous velocity of point P The velocity of P is the vector sum of the velocity of the center and the velocity of P relative to the center.

$$\vec{v}_P = \vec{v}_O + \vec{v}_{PO} = \vec{v}_O + \vec{\omega} \times \vec{r}_{PO}$$

Substituting the known quantities:

$$\vec{v}_{PO} = \left(-\frac{v_0}{R} \hat{k}\right) \times (R \hat{\imath}) = -v_0 (\hat{k} \times \hat{\imath}) = -v_0 \hat{\jmath}$$

Thus, the total velocity of P is:

$$\vec{v}_P = v_0 \hat{\imath} - v_0 \hat{\jmath}$$

The magnitude of the velocity is $|\vec{v}_P| = \sqrt{v_0^2 + (-v_0)^2} = v_0\sqrt{2}$.

[Q2] & [Q3] Tangential and Normal Acceleration of point P First, we find the total acceleration of P.

$$\vec{a}_P = \vec{a}_O + \vec{a}_{PO} = \vec{a}_O + \vec{\alpha} \times \vec{r}_{PO} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{PO})$$

Since $\vec{a}_O = \vec{0}$ and $\vec{\alpha} = \vec{0}$:

$$\vec{a}_P = \vec{\omega} \times (\vec{\omega} \times \vec{r}_{PO}) = \vec{\omega} \times \vec{v}_{PO}$$ $$\vec{a}_P = \left(-\frac{v_0}{R} \hat{k}\right) \times (-v_0 \hat{\jmath}) = \frac{v_0^2}{R} (\hat{k} \times \hat{\jmath}) = -\frac{v_0^2}{R} \hat{\imath}$$

This total acceleration is the centripetal acceleration of P relative to O.

The tangential and normal components of acceleration are projections of the total acceleration $\vec{a}_P$ onto directions parallel and perpendicular to the velocity vector $\vec{v}_P$. The unit vector in the direction of velocity is:

$$\hat{u}_t = \frac{\vec{v}_P}{|\vec{v}_P|} = \frac{v_0 \hat{\imath} - v_0 \hat{\jmath}}{v_0\sqrt{2}} = \frac{1}{\sqrt{2}}(\hat{\imath} - \hat{\jmath})$$

Tangential Acceleration [Q2] The tangential acceleration vector is the projection of $\vec{a}_P$ onto $\hat{u}_t$.

$$a_{P,t} = \vec{a}_P \cdot \hat{u}_t = \left(-\frac{v_0^2}{R} \hat{\imath}\right) \cdot \left(\frac{1}{\sqrt{2}}(\hat{\imath} - \hat{\jmath})\right) = -\frac{v_0^2}{R\sqrt{2}}$$ $$\vec{a}_{P,t} = a_{P,t} \hat{u}_t = \left(-\frac{v_0^2}{R\sqrt{2}}\right) \frac{1}{\sqrt{2}}(\hat{\imath} - \hat{\jmath}) = -\frac{v_0^2}{2R}(\hat{\imath} - \hat{\jmath})$$

Normal Acceleration [Q3] The normal acceleration vector is the remaining component of the total acceleration.

$$\vec{a}_{P,n} = \vec{a}_P - \vec{a}_{P,t}$$ $$\vec{a}_{P,n} = \left(-\frac{v_0^2}{R} \hat{\imath}\right) - \left(-\frac{v_0^2}{2R}(\hat{\imath} - \hat{\jmath})\right)$$ $$\vec{a}_{P,n} = \left(-\frac{v_0^2}{R} + \frac{v_0^2}{2R}\right)\hat{\imath} - \frac{v_0^2}{2R}\hat{\jmath}$$ $$\vec{a}_{P,n} = -\frac{v_0^2}{2R}\hat{\imath} - \frac{v_0^2}{2R}\hat{\jmath} = -\frac{v_0^2}{2R}(\hat{\imath} + \hat{\jmath})$$