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 \(\varphi\).

Key concepts

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. For a particle, this energy depends on its mass and velocity. In Lagrangian mechanics, the kinetic energy is often one of the primary components used to analyze a system's dynamics.

In the original exercise, the particle \(P\) was analyzed to determine its kinetic energy when suspended between two pegs. The kinetic energy \(T\) of a particle of mass \(m\) is generally given by:

• \(T = \frac{1}{2} m v^2\) where \(v\) is the velocity of the particle.

However, in the context of the particle \(P\) suspended by strings, its velocity and therefore its kinetic energy depend on the angles \(\theta\) and \(\varphi\). By expressing the particle's motion in terms of these angles, the kinetic energy of \(P\) was found to be:

• \(T = \frac{1}{2} m b^{2}(\cosh^{2} \varphi - \cos^{2} \theta)(\dot{\theta}^{2} + \dot{\varphi}^{2})\)

This formula shows that the kinetic energy takes into account both the position of the particle in the vertical plane and its change over time (time derivatives of the angles).

Potential Energy

Potential energy is the stored energy of an object due to its position in a force field, such as gravity. It is an essential concept in Lagrangian mechanics because it helps determine the total energy of a system.

For this exercise, although the potential energy was mentioned briefly, it plays a crucial role. The potential energy of particle \(P\) mainly comes from its height in relation to its surroundings. In general, potential energy can be calculated as:

• \(U = mgh\) for gravitational situations, where \(h\) is height.

In the system described, potential energy remains constant since the strings allow certain movements that always return the particle \(P\) to a specific alignment or height. In problems using Lagrangian mechanics like this one, calculating potential energy might not always be straightforward. That's why it's sometimes abstracted or not explicitly calculated, focusing on the more dynamic parts such as kinetic energy. Understanding how gravitational forces work on the involved particles will shed light on potential energy's constant state in this context, allowing a direct analysis of other dynamic energies.

Generalized Coordinates

Generalized coordinates are variables used in Lagrangian mechanics to describe the configuration of a system relative to some reference frame. These coordinates simplify the analysis, especially for systems with constraints like the particle \(P\) held by strings.

In many physical scenarios, ordinary coordinates (like Cartesian coordinates) are not practical for analysis due to system constraints. Hence, generalized coordinates like angles can provide a more straightforward description.

• Angles \(\theta\) and \(\varphi\) are used in this scenario as generalized coordinates.
• They describe the position of particle \(P\) relative to the vertical lines through pegs \(A\) and \(B\).

These angles serve to express the particle motion without requiring more complex positional variables. The choice of using \(\theta\) and \(\varphi\) simplifies the equations of motion for particle \(P\) and enables an easier computation of energies. In Lagrangian mechanics, this transformation into generalized coordinates is often essential for solving problems where the motion is constrained or the system's configuration is intricate.