Problems

Filters
Clear Filters

2 problems found

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

\psset{xunit=1.0cm,yunit=1.0cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-3.36,-3.71)(5.32,4.49) \pspolygon[linewidth=0pt,linecolor=white,hatchcolor=black,fillstyle=hlines,hatchangle=45.0,hatchsep=0.19](-3,4.22)(-3,4)(5,4)(5,4.22) \pscircle(-1,2){1} \pscircle(3,2){1} \pscircle(1,-1){1} \psline(0,2)(0,-1) \psline(2,2)(2,-1) \psline(-2,2)(-2,-1) \psline(4,2)(4,-1) \psline{->}(-2,-1.44)(-2,-2) \rput[tl](-2.25,-2.31){\(m_1g\)} \psline{->}(4,-1.44)(4,-2) \rput[tl](3.74,-2.25){\(m_2g\)} \psline{->}(1,-1)(1.02,-2.78) \rput[tl](0.72,-3.06){\(m_3g\)} \psline(-1,2)(-1,4) \psline(3,2)(3,4) \psline(-3,4)(5,4) \rput[tl](-1.19,1.67){\(P_1\)} \rput[tl](2.83,1.64){\(P_2\)} \rput[tl](0.83,-0.5){\(P_3\)} \begin{scriptsize} \psdots[dotstyle=*](-1,2) \psdots[dotstyle=*](3,2) \psdots[dotstyle=*](1,-1) \psdots[dotstyle=*](-2,-1) \psdots[dotstyle=*](4,-1) \end{scriptsize} \end{pspicture*} \par
\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 \[ 4m_{1}m_{2}=m_{3}(m_{1}+m_{2}). \]

View
1988 Paper 2 Q14
D: 1600.0 B: 1488.9
circular motion conical motion rotating rod rigid body angular velocity reaction force vertical axis rotation equilibrium condition

Two particles of mass \(M\) and \(m\) \((M>m)\) are attached to the ends of a light rod of length \(2l.\) The rod is fixed at its midpoint to a point \(O\) on a horizontal axle so that the rod can swing freely about \(O\) in a vertical plane normal to the axle. The axle rotates about a vertical axis through \(O\) at a constant angular speed \(\omega\) such that the rod makes a constant angle \(\alpha\) \((0<\alpha<\frac{1}{2}\pi)\) with the vertical. Show that \[ \omega^{2}=\left(\frac{M-m}{M+m}\right)\frac{g}{l\cos\alpha}. \] Show also that the force of reaction of the rod on the axle is inclined at an angle \[ \tan^{-1}\left[\left(\frac{M-m}{M+m}\right)^{2}\tan\alpha\right] \] with the downward vertical.

Solution:

TikZ diagram
The accelerations of \(M\) and \(m\) are \(l \sin \alpha \omega^2\) and \(-l \sin \alpha \omega^2\) so the forces \(R_M\) and \(R_m\) are \(M\binom{l \sin \alpha \omega^2}{g}, \,m \binom{-l \sin \alpha \omega^2}{g}\). Since the axle is rotating freely, the moment about \(O\) for the rod must be \(O\). The moment for \(M\) will be \(M\binom{l \sin \alpha\omega^2}{g} \cdot \binom{-l\cos \alpha}{l \sin \alpha} = lM\sin\alpha (g - l \cos \alpha\omega^2)\). The moment for \(m\) will be \(m \binom{-l \sin \alpha\omega^2}{g} \cdot \binom{-l\cos \alpha\omega^2}{l \sin \alpha} = lm \sin \alpha(g+l \cos \alpha\omega^2)\) Therefore \begin{align*} && lM\sin\alpha (g - l \cos \alpha\omega^2) &= lm \sin \alpha(g+l \cos \alpha\omega^2) \\ && M(g - l \cos \alpha \omega^2) &= m(g + l \cos \alpha \omega^2 ) \\ \Rightarrow && g(M-m) &= l \cos \alpha (M+m) \omega^2 \\ \Rightarrow && \omega^2 &= \left (\frac{M-m}{M+m} \right) \frac{g}{l \cos \alpha} \end{align*} as required. The total force on the rod is \(\mathbf{0}\) so the reaction force must be \(M\binom{l \sin \alpha \omega^2}{g}+ \,m \binom{-l \sin \alpha \omega^2}{g} = \binom{l \sin \alpha \omega^2 (M-m)}{(M+m)g}\) Therefore the angle this makes with downward vertical is: \begin{align*} \theta &= \tan^{-1} \left ( \frac{l \sin \alpha \omega^2 (M-m)}{(M+m)g} \right) \\ &= \tan^{-1} \left ( \frac{l \sin \alpha (M-m)}{(M+m)g} \omega^2\right) \\ &= \tan^{-1} \left ( \frac{l \sin \alpha (M-m)}{(M+m)g} \left (\frac{M-m}{M+m} \right) \frac{g}{l \cos \alpha}\right) \\ &= \tan^{-1}\left[\left(\frac{M-m}{M+m}\right)^{2}\tan\alpha\right] \end{align*} as required.

View