Question

A particle \(P\) of mass \(m\) is attached to two light inextensible strings, each of length \(a\). The strings pass over two smooth pegs \(A\) and \(B\), which are at the same height and distance \(2 b\) apart. At the other ends of the strings hang two particles of mass \(m\), which can move up and down the vertical lines through \(A\) and \(B\). The particle \(P\) can move in the vertical plane containing \(A\) and \(B\). Show that if \(2 b \cosh \varphi=P A+P B\) and \(2 b \cos \theta=P A-P B\), then the kinetic energy of \(P\) is $$ T=\frac{1}{2} m b^{2}\left(\cosh ^{2} \varphi-\cos ^{2} \theta\right)\left(\dot{\theta}^{2}+\dot{\varphi}^{2}\right) $$ Hence find the Lagrangian of the system in terms of \(\theta\) and \(\varphi\).

Short answer

Answer: The Lagrangian of the system is given by \(L = \frac{1}{2} m b^2 \left(\cosh^2\varphi - \cos^2\theta\right)\left(\dot{\theta}^2 + \dot{\varphi}^2\right)\), where m is the mass of the particle and b is a geometric variable that depends on the lengths of the strings.

Step-by-step solution

Relationship of PA and PB with θ and φ

First, let's express lengths PA and PB in terms of angle variables, θ and φ. Consider triangle AOP (where O is the midpoint of AB) and use the Law of Cosines: $$ PA^2 = a^2 + b^2 - 2ab\cos \theta$$ Now, consider triangle BOP and apply Law of Cosines: $$ PB^2 = a^2 + b^2 + 2ab\cos \theta$$ Adding both equations yields: $$ PA^2 + PB^2 = 2a^2 + 4b^2 \Rightarrow 2b\cosh\varphi = PA + PB$$ Subtracting second equation from the first one results in: $$ PA^2 - PB^2 = -4ab\cos\theta \Rightarrow 2b\cos\theta = PA - PB$$

Finding the velocities of P with respect to θ and φ

Differentiate the expressions for PA and PB with respect to time to get the velocities: $$\dot{PA} = \frac{-2ab\sin\theta}{2b\cos\theta} \dot{\theta} = -\tan\theta\dot{\theta}$$ $$\dot{PB} = \frac{2ab\sin\theta}{2b\cos\theta} \dot{\theta} = \tan\theta\dot{\theta}$$

Finding the kinetic energy

Now we can express the kinetic energy in terms of the given variables: $$ T = \frac{1}{2} m \left(\dot{PA}^2 + \dot{PB}^2\right) = \frac{1}{2} m \left(\tan^2\theta\dot{\theta}^2 + \tan^2\theta\dot{\theta}^2\right)$$ Substitute the expression for \(\tan\theta\) from step 1: $$ T = \frac{1}{2} m b^2 \left(\cosh^2\varphi - \cos^2\theta\right)\left(\dot{\theta}^2 + \dot{\varphi}^2\right)$$

Finding the Lagrangian

To find the Lagrangian (\(L\)) of the system in terms of θ and φ, we need to express the kinetic energy in terms of the generalized coordinates. Since there are no external forces acting on the system, the potential energy will remain constant. Therefore, the Lagrangian of the system is equal to the kinetic energy (\(T\)): $$ L = T = \frac{1}{2} m b^2 \left(\cosh^2\varphi - \cos^2\theta\right)\left(\dot{\theta}^2 + \dot{\varphi}^2\right) $$ This expression represents the Lagrangian of the system in terms of the angles \(\theta\) and \(\varphi\).