Problem source

\begin{questionparts} \item A solid circular disc has radius $a$ and mass $m.$ The density is proportional to the distance from the centre $O$. Show that the moment of inertia about an axis through $C$ perpendicular to the plane of the disc is $\frac{3}{5}ma^{2}.$
\item A light inelastic string has one end fixed at $A$. It passes under and supports a smooth pulley $B$ of mass $m.$ It then passes over a rough pulley $C$ which is a disc of the type described in \textbf{(i)}, free to turn about its axis which is fixed and horizontal. The string carries a particle $D$ of mass $M$ at its other end. The sections of the string which are not in contact with the pulleys are vertical. The system is released from rest and moves under gravity for $t$ seconds. At the end of this interval the pulley $B$ is suddenly stopped. Given that $m<2M$, find the resulting impulse on $D$ in terms of $m,M,g$ and $t$. 
{[}You may assume that the string is long enough for there to be no collisions between the elements of the system, and that the pulley $C$ is rough enough to prevent slipping throughout.{]} \end{questionparts}

Solution source

\begin{questionparts}
\item 
\begin{center}
    \begin{tikzpicture}
        \def\r{3};
        \coordinate (O) at (0,0);

        \draw[thick, black, solid, domain=0:360, samples=100]
            plot ({3*cos(\x)},{sin(\x)});
        \draw[thick, black, dashed, domain=0:360, samples=100]
            plot ({1.5*cos(\x)},{0.5*sin(\x)});
        \draw[thick, black, dashed, domain=0:360, samples=100]
            plot ({1.6*cos(\x)},{0.55*sin(\x)});

        \draw[dashed] (0,-2) -- (0, 2);
        \filldraw (O) circle (1.5pt) node[right] {$O$};
    \end{tikzpicture}
\end{center}

\begin{align*}
m &= \int_0^a \underbrace{(\rho r)}_{\text{mass per area}} \underbrace{\pi r^2}_{\text{area}} \d r \\
&= \rho \pi \frac{a^3}{3} \\
\\
I &= \sum m r^2 \\
&= \sum (\rho r) \pi r^2 \cdot r^2 \\
&\to \int_0^a \rho \pi r^4 \\
&=  \frac15 \rho \pi a^5 \\
&= \frac35 m a^2
\end{align*}

\item 
\begin{center}
    \begin{tikzpicture}
        \draw (0, 5) -- (5,5);

        \draw (1,5) -- (1,1);
        \draw (1.5, 1) circle (0.5);
        \draw (2,1) -- (2,3);
        \draw (3, 3) circle (1);
        \draw (4,3) -- (4,-1);

        \node[above] at (1,5) {$A$};
        \node at (1.5,1) {$B$};
        \node at (3,3) {$C$};
        \node[right] at (4,-1) {$D$};

        \filldraw (4,-1) circle (1.5pt);

        \draw[-latex, blue, ultra thick] (4,-1) -- ++(0,-1.25) node[right] {$Mg$};
        \draw[-latex, blue, ultra thick] (1.5,1) -- ++(0,-.6) node[right] {$mg$};
        \draw[-latex, blue, ultra thick] (4,-1) -- ++(0,1) node[right] {$T_C$};
        \draw[-latex, blue, ultra thick] (4,3) -- ++(0,-1) node[right] {$T_C$};
        \draw[-latex, blue, ultra thick] (2,3) -- ++(0,-0.75) node[left] {$T_B$};
        \draw[-latex, blue, ultra thick] (2,1) -- ++(0,0.75) node[left] {$T_B$};
        \draw[-latex, blue, ultra thick] (1,1) -- ++(0,0.75) node[left] {$T_B$};
    \end{tikzpicture}
\end{center}

\begin{align*}
\text{N2}(\downarrow, D): &&  Mg -T_C &= Mf \\
\overset{\curvearrowright}{C} && (T_C - T_B)a &= I \frac{f}{a} \\
&&&= \frac35 m a f \\
\text{N2}(\uparrow, B): && 2T_B-mg &= \frac12 m f \\
\\
\Rightarrow && Mg-T_B &= \left (M + \frac35 m \right)f \\
\Rightarrow && Mg - \frac12 mg &= \left (M + \frac35 m + \frac14 m \right)f \\
\Rightarrow && f &= \frac{(M-\frac12 m)g}{M + \frac{17}{20} m} \\
&&&= \frac{(2M-m)g}{2M +\frac{17}{10}m}
\end{align*}

Therefore the speed after a time $t$ is $\displaystyle \frac{(2M-m)g}{2M +\frac{17}{10}m} t$ and the impulse will be the change in momentum, ie $\displaystyle \frac{(2M-m)g}{2M +\frac{17}{10}m} Mt$

\end{questionparts}