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1989 Paper 3 Q14
D: 1700.0 B: 1472.2
moments of inertia variable density disc pulley system impulse rough pulley rotational dynamics inelastic string Atwood machine

  1. 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 \(\frac{3}{5}ma^{2}.\)
  2. 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 (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.{]}

Solution:

  1. TikZ diagram
    \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*}
  2. TikZ diagram
    \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 \(\displaystyle \frac{(2M-m)g}{2M +\frac{17}{10}m} Mt\)

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