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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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1989 Paper 3 Q15
D: 1700.0 B: 1503.8
probability cumulative distribution function uniform distribution order statistics median probability density function integration expectation variance

The continuous random variable \(X\) is uniformly distributed over the interval \([-c,c].\) Write down expressions for the probabilities that:

  1. \(n\) independently selected values of \(X\) are all greater than \(k\),
  2. \(n\) independently selected values of \(X\) are all less than \(k\),
where \(k\) lies in \([-c,c]\). A sample of \(2n+1\) values of \(X\) is selected at random and \(Z\) is the median of the sample. Show that \(Z\) is distributed over \([-c,c]\) with probability density function \[ \frac{(2n+1)!}{(n!)^{2}(2c)^{2n+1}}(c^{2}-z^{2})^{n}. \] Deduce the value of \({\displaystyle \int_{-c}^{c}(c^{2}-z^{2})^{n}\,\mathrm{d}z.}\) Evaluate \(\mathrm{E}(Z)\) and \(\mathrm{var}(Z).\)

Solution:

  1. \begin{align*} \mathbb{P}(n\text{ independent values of }X > k) &= \prod_{i=1}^n \mathbb{P}(X > k) \\ &= \left ( \frac{c-k}{2c}\right)^n \end{align*}
  2. \begin{align*} \mathbb{P}(n\text{ independent values of }X < k) &= \prod_{i=1}^n \mathbb{P}(X < k) \\ &= \left ( \frac{k+c}{2c}\right)^n \end{align*}
\begin{align*} &&\mathbb{P}(\text{median} < z+\delta \text{ and median} > z - \delta) &= \mathbb{P}(n\text{ values } < z - \delta \text{ and } n \text{ values} > z + \delta) \\ &&&= \binom{2n+1}{n,n,1} \left ( \frac{c-(z+\delta)}{2c}\right)^n\left ( \frac{(z-\delta)+c}{2c}\right)^n \frac{2 \delta}{2 c} \\ &&&= \frac{(2n+1)!}{n! n!} \frac{((c-(z+\delta))(c+(z-\delta)))^n 2\delta}{2^n c^n} \\ &&&= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}((c-(z+\delta))(c+(z-\delta)))^n 2\delta \\ \Rightarrow && \lim_{\delta \to 0} \frac{\mathbb{P}(\text{median} < z+\delta \text{ and median} > z - \delta)}{2 \delta} &= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}((c-z)(c+z))^n \\ &&&= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2) \\ \end{align*} \begin{align*} && 1 &= \int_{-c}^c \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\ \Rightarrow && \frac{(n!)^2 (2c)^{2n+1}}{(2n+1)!} &= \int_{-c}^c (c^2-z^2)^n \d z \end{align*} \begin{align*} \mathbb{E}(Z) &= \int_{-c}^c z \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\ &=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-c}^c z (c^2-z^2)^n \d z \\ &= 0 \end{align*} \begin{align*} \mathrm{Var}(Z) &= \mathbb{E}(Z^2) - \mathbb{E}(Z)^2 \\ &= \mathbb{E}(Z^2) \\ &= \int_{-c}^c z^2 \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\ &=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-c}^c z^2 (c^2-z^2)^n \d z \\ &=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \left ( \left [ -\frac{1}{2(n+1)}z(c^2-z^2)^{n+1} \right]_{-c}^c + \frac{1}{2(n+1)}\int_{-c}^c (c^2-z^2)^{n+1} \d z \right) \\ &= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \frac{1}{2(n+1)} \frac{((n+1)!)^2 (2c)^{2n+3}}{(2n+3)!} \\ &= \frac{(n+1)^2(2c)^2}{(n+1)(2n+2)(2n+3)} \\ &= \frac{2c^2}{2n+3} \end{align*}

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1989 Paper 3 Q16
D: 1700.0 B: 1484.0
probability conditional probability approximating binomial to normal distribution normal approximation hypothesis testing population model joint probability table continuity correction

It is believed that the population of Ruritania can be described as follows:

  1. \(25\%\) are fair-haired and the rest are dark-haired;
  2. \(20\%\) are green-eyed and the rest hazel-eyed;
  3. the population can also be divided into narrow-headed and broad-headed;
  4. no narrow-headed person has green eyes and fair hair;
  5. those who are green-eyed are as likely to be narrow-headed as broad-headed;
  6. those who are green-eyed and broad-headed are as likely to be fair-headed as dark-haired;
  7. half of the population is broad-headed and dark-haired;
  8. a hazel-eyed person is as likely to be fair-haired and broad-headed as dark-haired and narrow-headed.
Find the proportion believed to be narrow-headed. I am acquainted with only six Ruritanians, all of whom are broad-headed. Comment on this observation as evidence for or against the given model. A random sample of 200 Ruritanians is taken and is found to contain 50 narrow-heads. On the basis of the given model, calculate (to a reasonable approximation) the probability of getting 50 or fewer narrow-heads. Comment on the result.

Solution:

TikZ diagram
Conditions tell us: \begin{align*} && a+b+d+e &= 0.25 \\ && b+c+e+f &= 0.2 \\ && e &= 0 \\ && b+c &= e + f \\ && b &= c \\ && c+h &= 0.5 \\ && a &= g \\ \end{align*}
TikZ diagram
So \(4b = 0.2 \Rightarrow b = 0.05\)
TikZ diagram
And \begin{align*} && 0.25 &= a + d + 0.05 \\ && 1 &= 2a + d + 0.65 \\ \Rightarrow && a &= 0.15 \\ && d &= 0.05 \end{align*}
TikZ diagram
So the proportion who are narrow-headed is \(30\%\). It's obviously relatively unlikely for your six Ruritanian friends to all be broad-headed if it's a random sample, but friendship groups are are likely to be biased so it's not too surprising. Assuming there is a sufficiently large number of Ruritanians, we might model the number of narrow-headed Ruritanians from a sample of \(200\) as \(X \sim B(200, 0.3)\). Computing \(\mathbb{P}(X \leq 50)\) by hand is tricky, so let's use a binomial approximation to obtain: \(X \approx N(60, 42)\) and \begin{align*} \mathbb{P}(X \leq 50) &\approx \mathbb{P} \left (Z \leq \frac{50 - 60+0.5}{\sqrt{42}} \right) \\ &\approx \mathbb{P} \left (Z \leq -\frac{9.5}{6.5} \right) \\ &\approx \mathbb{P} \left (Z \leq -\frac{3}{2} \right) \\ &\approx 5\% \end{align*} (actually this approximation gives \(7.1\%\) and the binomial value gives \(7.0\%\)). This also seems somewhat surprising

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