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2014 Paper 1 Q11
D: 1500.0 B: 1500.0
pulleys Atwood machine Newton's second law tension acceleration constraint equations inequality AM-GM inequality

The diagrams below show two separate systems of particles, strings and pulleys.In both systems, the pulleys are smooth and light, the strings are light and inextensible, the particles move vertically and the pulleys labelled with \(P\) are fixed. The masses of the particles are as indicated on the diagrams.

TikZ diagram
  1. For system I show that the acceleration, \(a_1\), of the particle of mass \(M\), measured in the downwards direction, is given by \[ a_1= \frac{M-m}{M+m} \, g \,, \] where \(g\) is the acceleration due to gravity. Give an expression for the force on the pulley due to the tension in the string.
  2. For system II show that the acceleration, \(a_2\), of the particle of mass \(M\), measured in the downwards direction, is given by \[ a_2= \frac{ M - 4\mu}{M+4\mu}\,g \,, \] where \(\mu = \dfrac{m_1m_2}{m_1+m_2}\). In the case \(m= m_1+m_2\), show that \(a_1= a_2\) if and only if \(m_1=m_2\).

Solution:

  1. \(\,\)
    TikZ diagram
    \begin{align*} \text{N2}(\uparrow, m): && T - mg &= ma_1 \\ \text{N2}(\uparrow, M): && T-Mg &= -Ma_1 \\ \Rightarrow && (M-m)g &= a_1(m+M) \\ \Rightarrow && a_1 &= \frac{M-m}{M+m}g \\ && T &= mg + ma_1 \\ &&&= \frac{2mM}{M+m}g \end{align*}
  2. System II is the same as system I, but with \(m\) replaced with \(2\frac{T}{g} = \frac{4mM}{M+m}\). In particular, this means that: \begin{align*} && a_2 &= \frac{M - \frac{4m_1m_2}{m_1+m_2}}{M + \frac{4m_1m_2}{m_1+m_2}} g \\ &&&= \frac{M-4\mu}{M+4\mu}g \end{align*} If \(m = m_1 + m_2\) then \begin{align*} && a_1 &= a_2 \\ \Leftrightarrow && \frac{M-m_1-m_2}{M+m_1+m_2} &= \frac{M - \frac{4m_1m_2}{m_1+m_2}}{M + \frac{4m_1m_2}{m_1+m_2}} \\ \Leftrightarrow && \frac{M-m_1-m_2}{M+m_1+m_2} &= \frac{M(m_1+m_2) -4m_1m_2}{M(m_1+m_2) + 4m_1m_2} \\ \Leftrightarrow && M^2(m_1+m_2)+4m_1m_2M &- M(m_1+m_2)^2 - 4m_1m_2(m_1+m_2) \\ &&\quad \quad = M^2(m_1+m_2) - 4m_1m_2M &+M(m_1+m_2)^2-4m_1m_2(m_1+m_2) \\ \Leftrightarrow && 8m_1m_2M&= 2M(m_1+m_2)^2 \\ \Leftrightarrow && 0 &= (m_1-m_2)^2 \\ \Leftrightarrow && m_1 &= m_2 \end{align*}

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2012 Paper 3 Q9
D: 1700.0 B: 1500.0
moments of inertia pulley rotational dynamics Newton's second law friction couple Atwood machine inequality rigid body

A pulley consists of a disc of radius \(r\) with centre \(O\) and a light thin axle through \(O\) perpendicular to the plane of the disc. The disc is non-uniform, its mass is \(M\) and its centre of mass is at \(O\). The axle is fixed and horizontal. Two particles, of masses \(m_1\) and \(m_2\) where \(m_1>m_2\), are connected by a light inextensible string which passes over the pulley. The contact between the string and the pulley is rough enough to prevent the string sliding. The pulley turns and the vertical force on the axle is found, by measurement, to be~\(P+Mg\).

  1. The moment of inertia of the pulley about its axle is calculated assuming that the pulley rotates without friction about its axle. Show that the calculated value is \[ \frac{((m_1 + m_2)P - 4m_1m_2g)r^2} {(m_1 + m_2)g - P}\,. \tag{\(*\)}\]
  2. Instead, the moment of inertia of the pulley about its axle is calculated assuming that a couple of magnitude \(C\) due to friction acts on the axle of the pulley. Determine whether this calculated value is greater or smaller than \((*)\). Show that \(C<(m_1-m_2)rg\).

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2000 Paper 2 Q10
D: 1600.0 B: 1500.0
pulley systems Newton's second law constraint equations inextensible string acceleration Atwood machine compound pulley kinematics two-phase motion

A long light inextensible string passes over a fixed smooth light pulley. A particle of mass 4~kg is attached to one end \(A\) of this string and the other end is attached to a second smooth light pulley. A long light inextensible string \(BC\) passes over the second pulley and has a particle of mass 2 kg attached at \(B\) and a particle of mass of 1 kg attached at \(C\). The system is held in equilibrium in a vertical plane. The string \(BC\) is then released from rest. Find the accelerations of the two moving particles. After \(T\) seconds, the end \(A\) is released so that all three particles are now moving in a vertical plane. Find the accelerations of \(A\), \(B\) and \(C\) in this second phase of the motion. Find also, in terms of \(g\) and \(T\), the speed of \(A\) when \(B\) has moved through a total distance of \(0.6gT^{2}\)~metres.

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1992 Paper 2 Q14
D: 1600.0 B: 1500.0
pulley systems Newton's second law tension constraint equations equilibrium condition Atwood machine acceleration inextensible string

\noindent

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\noindent In the diagram \(P_{1}\) and \(P_{2}\) are smooth light pulleys fixed at the same height, and \(P_{3}\) is a third smooth light pulley, freely suspended. A smooth light inextensible string runs over \(P_{1},\) under \(P_{3}\) and over \(P_{2},\) as shown: the parts of the string not in contact with any pulley are vertical. A particle of mass \(m_{3}\) is attached to \(P_{3}.\) There is a particle of mass \(m_{1}\) attached to the end of the string below \(P_{1}\) and a particle of mass \(m_{2}\) attached to the other end, below \(P_{2}.\) The system is released from rest. Find the tension in the string, and show that the pulley \(P_{3}\) will remain at rest if \[ 4m_{1}m_{2}=m_{3}(m_{1}+m_{2}). \]

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1991 Paper 1 Q11
D: 1500.0 B: 1500.1
momentum energy conservation Atwood machine inelastic collision geometric progression piledriver string becomes taut pulley system

A piledriver consists of a weight of mass \(M\) connected to a lighter counterweight of mass \(m\) by a light inextensible string passing over a smooth light fixed pulley. By considerations of energy or otherwise, show that if the weights are released from rest, and move vertically, then as long as the string remains taut and no collisions occur, the weights experience a constant acceleration of magnitude \[ g\left(\frac{M-m}{M+m}\right). \] Initially the weight is held vertically above the pile, and is released from rest. During the subsequent motion both weights move vertically and the only collisions are between the weight and the pile. Treating the pile as fixed and the collisions as completely inelastic, show that, if just before a collision the counterweight is moving with speed \(v\), then just before the next collision it will be moving with speed \(mv/\left(M+m\right)\). {[}You may assume that when the string becomes taut, the momentum lost by one weight equals that gained by the other.{]} Further show that the times between successive collisions with the pile form a geometric progression. Show that the total time before the weight finally comes to rest is three times the time from the start to the first impact.

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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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