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\begin{center}
    
\begin{tikzpicture}
    % Draw the horizontal line
    \draw[thick] (-2,-2) -- (4,-2);

\coordinate (O) at (0,0);
    \coordinate (P) at (-{sqrt(3)},-1);
    \coordinate (Q) at ({sqrt(3)},1);
    \coordinate (T) at (0,-2);
    
    % Draw the circle
    \draw[thick] (O) circle (2cm);
    
    % Draw the radius to point O
    \draw[thick] (O) -- (P);
    
    % Draw the dashed line
    \draw[dashed] (O) -- (0,-2);
    
    % Add points
    \node[left] at (P) {$P$};
    \node[right] at (Q) {$Q$};
    \node[above] at (O) {$O$};

\pic [draw, angle radius=0.8cm, "$\theta$"] {angle = P--O--T};
    
    % Add the arrow at the top
    \draw[-{Stealth[length=3mm]}, thick] (-0.5,2.5) -- (0.5,2.5);
\end{tikzpicture}

\end{center}

We can see that the position of $O$ is $\begin{pmatrix} a \theta \\ a \end{pmatrix}$ since the hoop is not slipping. $P$'s position relative to $O$ is $\begin{pmatrix}  -a\sin\theta\\a(1-\cos \theta) \end{pmatrix}$, therefore the position of $P$ is $\begin{pmatrix}  a(\theta-\sin\theta) \\ a(1-\cos \theta) \end{pmatrix}$.

We can now calculate $\mathbf{v}_P = a \begin{pmatrix}  (\dot{\theta}-\dot{\theta}\cos\theta) \\ \dot{\theta}\sin \theta \end{pmatrix} = a \dot{\theta} \begin{pmatrix}  (1-\cos\theta) \\ \sin \theta \end{pmatrix}$

We can also see that

\begin{align*}
    && |\mathbf{v}_P|^2 &= a^2\dot{\theta}^2 \l \l 1 - \cos \theta \r^2 + \sin^2 \theta \r \\
    && &= a^2\dot{\theta}^2 ( 2 - 2\cos \theta) \\
    && &= 2a^2\dot{\theta}^2 ( 1 - \cos \theta) \\
    && &= a^2\dot{\theta}^2 4 \sin^2 \frac{\theta}{2} \\
    \Rightarrow |\mathbf{v}_P| &= 2a \dot{\theta} \left | \sin \frac{\theta}2 \right |
\end{align*}

Not that the position of $Q$ is $\begin{pmatrix}  a(\theta+\sin\theta) \\ a(1+\cos \theta) \end{pmatrix}$

Therefore

\begin{align*}
    && |\mathbf{v}_Q|^2 &= a^2\dot{\theta}^2 \l \l 1 + \sin \theta \r^2 + \l 1 + \cos \theta \r^2 \r \\
    &&  &= a^2\dot{\theta}^2 \l \l 1 + \sin \theta \r^2 + \cos^2 \theta  \r \\
    &&  &= 2a^2\dot{\theta}^2 \l 1 + \cos \theta \r \\
\end{align*}

Therefore

\[ \text{K.E.} = \frac12m|\mathbf{v}_P|^2 + |\mathbf{v}_Q|^2 = \frac12m2a^2 \dot{\theta}^2 (1 - \cos \theta + 1-\cos \theta) = 2ma^2 \dot{\theta}^2\]

Since there are no external forces acting conservation of energy tells us that kinetic energy is constant, ie $4ma^2 \dot{\theta}\ddot{\theta} = 0 \Rightarrow \ddot{\theta} = 0$, ie the hoop is rolling with constant speed.

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