Problems

A particle \(P\) of mass \(m\) is connected by two light inextensible strings to two fixed points \(A\) and \(B\), with \(A\) vertically above \(B\). The string \(AP\) has length \(x\). The particle is rotating about the vertical through \(A\) and \(B\) with angular velocity \(\omega\), and both strings are taut. Angles \(PAB\) and \(PBA\) are \(\alpha\) and \(\beta\), respectively. Find the tensions \(T_A\) and \(T_B\) in the strings \(AP\) and \(BP\) (respectively), and hence show that \(\omega^2 x\cos\alpha \ge g\). Consider now the case that \(\omega^2 x\cos\alpha = g\). Given that \(AB=h\) and \(BP=d\), where \(h>d\), show that \(h\cos\alpha \ge \sqrt{h^2-d^2}\). Show further that \[ mg < T_A \le \frac{mgh}{\sqrt{h^2-d^2}\,}\,. \] Describe the geometry of the strings when \(T_A\) attains its upper bound.

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Solution:

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\begin{align*} \text{N2}(\uparrow): && T_A \cos \alpha - T_B \cos\alpha - mg &= 0 \\ \Rightarrow && T_A \cos \alpha - T_B \cos\beta &= mg \\ \text{N2}(\leftarrow, \text{radially}): && T_A \sin \alpha + T_B \sin \beta &= m x \sin \alpha \omega^2 \\ \Rightarrow && T_A(\cos \alpha \sin \beta+\sin \alpha \cos \beta) &= mg \sin \beta + mx \sin \alpha \omega^2 \cos \beta \\ \Rightarrow && T_A &=\frac{mg\sin \beta + m x \sin \alpha \omega^2 \cos \beta }{\sin(\alpha + \beta)} \\ \Rightarrow && T_B(\sin \beta \cos \alpha- \cos \beta \sin \alpha)&= mx \sin \alpha \omega^2 \cos \alpha -mg \sin \alpha \\ \Rightarrow && T_B &= \frac{m x \sin \alpha \omega^2 \cos \alpha - mg \sin \alpha}{\sin(\beta - \alpha)} \\ &&&= \frac{m \sin \alpha(\omega^2 \cos\alpha - g)}{\sin (\beta - \alpha)} \end{align*} Since \(T_B \geq 0 \Rightarrow \omega^2 \cos\alpha - g \geq 0\) as required.

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\(\sqrt{h^2-d^2}\) is the length of the final side on the dashed right angle triangle with hypotenuse \(AB\). \(h \cos \alpha\) will be clearly longer as the angle \(\alpha\) will be smaller and so \(\cos \alpha\) will be larger. When \(\omega^2 x \cos \alpha = g\) we must have \(T_B = 0\). \(T_A\cos \alpha = mg \Rightarrow T_A > mg\) since \(\alpha \neq 0\). \(T_A = \frac{mg}{\cos \alpha} \leq \frac{mgh}{\sqrt{h^2-d^2}}\) \(T_A\) will attain it's upper bound when \(\angle APB\) is a right angle.

\(\,\) \setlength{\unitlength}{1cm}

1. An attempt is made to move a rod of length \(L\) from a corridor of width \(a\) into a corridor of width~\(b\), where \(a \ne b.\) The corridors meet at right angles, as shown in Figure 1 and the rod remains horizontal. Show that if the attempt is to be successful then $$ L \le a \cosec {\alpha} + b \sec {\alpha} \;, $$ where \({\alpha}\) satisfies $$ \tan^3\alpha =\frac a b \;. $$
2. An attempt is made to move a rectangular table-top, of width \(w\) and length \(l\), from one corridor to the other, as shown in the Figure 2. The table-top remains horizontal. Show that if the attempt is to be successful then $$ l\le a \cosec {\beta} + b \sec {\beta} -2w \cosec 2{\beta}, $$ where \({\beta}\) satisfies $$ w= \left(\frac {a -b \tan^3 \beta} {1 - \tan^2 \beta} \right) \cos \beta \;. $$

A bead \(B\) of mass \(m\) can slide along a rough horizontal wire. A light inextensible string of length \(2\ell\) has one end attached to a fixed point \(A\) of the wire and the other to \(B\,\). A particle \(P\) of mass \(3m\) is attached to the mid-point of the string and \(B\) is held at a distance \(\ell\) from~\(A\,\). The bead is released from rest. Let \(a_1\) and \(a_2\) be the magnitudes of the horizontal and vertical components of the initial acceleration of \(P\,\). Show by considering the motion of \(P\) relative to \(A\,\), or otherwise, that \(a_1= \sqrt 3 a_2\,\). Show also that the magnitude of the initial acceleration of \(B\) is \(2a_1\,\). Given that the frictional force opposing the motion of \(B\) is equal to \(({\sqrt{3}}/6)R\), where \(R\) is the normal reaction between \(B\) and the wire, show that the magnitude of the initial acceleration of \(P\) is~\(g/18\,\).