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.

A particle \(P\) of mass \(m\) is attached to points \(A\) and \(B\), where \(A\) is a distance \(9a\) vertically above \(B\), by elastic strings, each of which has modulus of elasticity \(6mg\). The string \(AP\) has natural length \(6a\) and the string \(BP\) has natural length \(2a\). Let \(x\) be the distance \(AP\). The system is released from rest with \(P\) on the vertical line \(AB\) and \(x = 6a\). Show that the acceleration \(\ddot{x}\) of \(P\) is \(\ds{4g \over a}(7a - x)\) for \(6a < x < 7a\) and \(\ds{g \over a}(7a - x)\) for \(7a < x < 9a\,\). Find the time taken for the particle to reach \(B\).