Problems

Filters
Clear Filters

2 problems found

2005 Paper 3 Q11
D: 1700.0 B: 1500.0
circular motion rigid body energy equilibrium stability trigonometric identity angular velocity potential energy spindle vertical plane rotation

A horizontal spindle rotates freely in a fixed bearing. Three light rods are each attached by one end to the spindle so that they rotate in a vertical plane. A particle of mass \(m\) is fixed to the other end of each of the three rods. The rods have lengths \(a\), \(b\) and \(c\), with \(a > b > c\,\) and the angle between any pair of rods is \(\frac23 \pi\). The angle between the rod of length \(a\) and the vertical is \(\theta\), as shown in the diagram. \vspace*{-0.1in}

\psset{xunit=0.45cm,yunit=0.45cm,algebraic=true,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-6.49,-3.44)(6.06,9.38) \psline[linestyle=dashed,dash=1pt 1pt](0,9.15)(0,-3.05) \psline(0,2.51)(2.41,8.3) \psline(0,2.51)(3.89,-0.76) \psline(0,2.51)(-4.39,0.87) \parametricplot{-2.7855695569416454}{1.176155335856138}{1*1.77*cos(t)+0*1.77*sin(t)+0|0*1.77*cos(t)+1*1.77*sin(t)+2.51} \parametricplot{1.1701030633139027}{1.5707963267948966}{1*2.47*cos(t)+0*2.47*sin(t)+0|0*2.47*cos(t)+1*2.47*sin(t)+2.54} \rput[tl](0.08,4.53){\(\theta\)} \rput[tl](0.59,3.24){\(\frac{2}{3}\pi\)} \rput[tl](-0.46,2.08){\(\frac{2}{3}\pi\)} \rput[tl](1.56,6.08){\(a\)} \rput[tl](2.57,1.3){\(b\)} \rput[tl](-2.8,2.31){\(c\)} \begin{scriptsize} \psdots[dotsize=6pt 0,dotstyle=*](2.41,8.3) \psdots[dotsize=6pt 0,dotstyle=*](3.89,-0.76) \psdots[dotsize=6pt 0,dotstyle=*](-4.39,0.87) \end{scriptsize} \end{pspicture*}
Find an expression for the energy of the system and show that, if the system is in equilibrium, then \[ \tan \theta = -\frac{(b-c) \sqrt{3}}{2a-b-c}\;. \] Deduce that there are exactly two equilibrium positions and determine which of the two equilibrium positions is stable. Show that, for the system to make complete revolutions, it must pass through its position of stable equilibrium with an angular velocity of at least \[ \sqrt{\frac{4gR}{a^2+b^2+c^2}} \, , \] where \(2R^2 = (a-b)^2+(b-c)^2 +(c-a)^2 \;\).

View
1996 Paper 2 Q9
D: 1600.0 B: 1500.0
centre of mass solid cone hemisphere equilibrium stability toppling composite body optimisation

A child's toy consists of a solid cone of height \(\lambda a\) and a solid hemisphere of radius \(a\), made out of the same uniform material and fastened together so that their plane faces coincide. (Thus the diameter of the hemisphere is equal to that of the base of the cone.) Show that if \(\lambda < \sqrt{3}\) the toy will always move to an upright position if placed with the surface of the hemisphere on a horizontal table, but that if \(\lambda > \sqrt{3}\) the toy may overbalance. Show, however, that if the toy is placed with the surface of the cone touching the table it will remain there whatever the value of \(\lambda\). [The centre of gravity of a uniform solid cone of height \(h\) is a height \(h/4\) above its base. The centre of gravity of a uniform solid hemisphere of radius \(a\) is at distance \(3a/8\) from the centre of its base.]

Solution:

TikZ diagram
By symmetry the centre of mass will lie on the main axis. Taking the plane faces as \(x = 0\) we have the following centers of mass: \begin{align*} && \text{COM} && \text{Mass} \\ \text{Hemisphere} && -\frac{3a}{8} && \frac{2\pi a^3}{3} \\ \text{Cone} && \frac{\lambda a}{4} && \frac{\lambda \pi a^3}{3} \\ \text{Toy} && \bar{x} && \frac{(\lambda + 2)\pi a^3}{3} \\ \end{align*} Therefore, \begin{align*} && \frac{(\lambda + 2)\pi a^3}{3} \cdot \bar{x} &= -\frac{3a}{8} \cdot \frac{2\pi a^3}{3} + \frac{\lambda a}{4} \cdot \frac{\lambda \pi a^3}{3} \\ \Rightarrow && (\lambda + 2) \bar{x} &= \frac{(\lambda^2 -3)a}{4} \end{align*} Therefore the centre of mass will be inside the hemisphere (and it will always move to an upright position) iff \(\bar{x} < 0 \Leftrightarrow \lambda < \sqrt{3}\).
TikZ diagram
For the toy to topple from this position, \(\bar{x}\) must be longer than it would need to be to form a right-angled triangle with the vertical at the plane face. The angle at this point will be \(\theta\), so we need: \(\bar{x} > a\tan \theta = a \frac{a}{\lambda a} = \frac{a}{\lambda}\) \begin{align*} && \bar{x} &> \frac{a}{\lambda} \\ \Leftrightarrow && \frac{(3-\lambda^2)a}{4(\lambda + 2)} &> \frac{a}{\lambda} \\ \Leftrightarrow && {(3-\lambda^2)\lambda} &> {4(\lambda + 2)} \\ \Leftrightarrow && -\lambda^3-\lambda -8 &> 0 \\ \end{align*} Contradiction! Therefore it can never topple when laid on its side.

View