Problem source

Two identical particles $P$ and $Q$, each of 
mass $m$, are attached to the ends of a diameter
of a light thin circular hoop of radius $a$. 
The hoop rolls without slipping
 along a straight line
on a horizontal table with  the plane of the hoop vertical.
Initially, $P$ is in contact with 
the table. At time $t$, the hoop has rotated through an angle
$\theta$. Write down the position at time $t$ of $P$, relative to its starting
point, in cartesian coordinates, and  determine its speed in
terms
of $a$, $\theta$ and $\dot\theta$. 
Show that the total 
kinetic energy of the two particles is $2ma^2\dot\theta^2$.
Given that the only external forces on the system are gravity
and the vertical reaction of the table on the hoop,
show that the hoop rolls with constant speed.

Solution source

\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.