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Chapter 3: Problem 2

A particle of mass \(m\) is subject to a force \(\boldsymbol{F}\). Obtain the equations of motion in cylindrical polar coordinates.

Short Answer

Expert verified
Answer: The equations of motion in cylindrical polar coordinates for a particle of mass m subject to a force F are: 1. ρ-component: m[(d^2ρ/dt^2) - ρ(dφ/dt)^2] = F_ρ 2. φ-component: m[2(dρ/dt)(dφ/dt) + ρ(d^2φ/dt^2)] = F_φ 3. z-component: m(d^2z/dt^2) = F_z

Step by step solution

01

Write the position vector in cylindrical polar coordinates

The position vector in cylindrical polar coordinates can be written as: r(ρ, φ, z) = ρ(cos(φ)î + sin(φ)ĵ) + zk^
02

Compute the velocity vector in cylindrical polar coordinates

To compute the velocity vector in cylindrical polar coordinates, differentiate the position vector with respect to time: v(ρ, φ, z) = dr/dt = (dρ/dt)(cos(φ)î + sin(φ)ĵ) + ρ(-sin(φ)(dφ/dt)î + cos(φ)(dφ/dt)ĵ) + (dz/dt)k^
03

Compute the acceleration vector in cylindrical polar coordinates

Now we differentiate the velocity vector with respect to time to obtain the acceleration vector in cylindrical polar coordinates: a(ρ, φ, z) = dv/dt = [(d^2ρ/dt^2) - ρ(dφ/dt)^2] (cos(φ)î + sin(φ)ĵ) + [2(dρ/dt)(dφ/dt) + ρ(d^2φ/dt^2)](-sin(φ)î + cos(φ)ĵ) + (d^2z/dt^2)k^
04

Express the force in cylindrical polar coordinates

Now we need to express the force F in terms of its components in the cylindrical polar coordinates. The force can be written as: F(ρ, φ, z) = F_ρ(cos(φ)î + sin(φ)ĵ) + F_φ(-sin(φ)î + cos(φ)ĵ) + F_zk^
05

Write the equations of motion in cylindrical polar coordinates

Now we can write the equations of motion by setting the force F equal to the mass m times the acceleration a in cylindrical polar coordinates. This will give us three separate equations, one for each coordinate: Equation 1 (ρ-component): m[(d^2ρ/dt^2) - ρ(dφ/dt)^2] = F_ρ Equation 2 (φ-component): m[2(dρ/dt)(dφ/dt) + ρ(d^2φ/dt^2)] = F_φ Equation 3 (z-component): m(d^2z/dt^2) = F_z These are the equations of motion for the particle of mass m subject to the force F in cylindrical polar coordinates.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equations of Motion
When dealing with different coordinate systems, such as cylindrical polar coordinates, the equations of motion adapt to describe a particle's movement efficiently. These coordinates are particularly useful for problems with rotational symmetry. In cylindrical polar coordinates, the movement of a particle is described by three main components: radial distance \( \rho \), angular position \( \phi \), and height \( z \). Each of these components has its corresponding differential equation of motion.
  • Radial Component: It describes the motion towards or away from the origin, along the line that radiates outward.
  • Angular Component: This describes the rotational movement around the origin, a critical factor in analyzing circular paths.
  • Vertical Component: It handles movements along the axis perpendicular to the rotation plane.
In practice, by setting the force \( \boldsymbol{F} \) equal to the mass \( m \) times the acceleration \( \boldsymbol{a} \) in each direction, we obtain three distinct equations. These equations help to express the dynamics of any particle moving under a force in this coordinate system. Understanding these equations allows for accurate modeling of systems like spiraling galaxies or tornadoes.
Force Components
In cylindrical polar coordinates, any force acting on a particle can be broken down into three components: radial, angular, and vertical force components. This breakdown simplifies the complex forces into manageable parts, allowing for precise analysis in each direction.
  • Radial Force \( F_{\rho} \): This is the force component that acts along the radial direction toward or away from the center. It influences the radial acceleration.
  • Angular Force \( F_{\phi} \): Acting perpendicular to the radial force, this component affects rotational acceleration around the axis.
  • Vertical Force \( F_{z} \): This is responsible for linear movement upwards or downwards along the vertical axis.
By treating forces as such components, the analysis becomes simpler and more intuitive. Each force component is aligned with one of the coordinate system's axes, allowing direct application of Newton's second law: \( F = ma \). This strategy is key to solving physical problems where forces have different impacts along different directions.
Acceleration Vector
The acceleration vector in cylindrical polar coordinates takes into account the differentiation of both radial and angular positions along with vertical height. This vector is essential as it illustrates how the velocity of a particle changes over time considering all possible pathways.
  • Radial Acceleration: Comprising terms like \( \frac{d^2 \rho}{dt^2} \) and \( \rho \left(\frac{d \phi}{dt}\right)^2 \), it accounts for both the linear acceleration towards the origin and the centrifugal effects due to rotation.
  • Angular Acceleration: This includes terms such as \( 2 \frac{d \rho}{dt} \frac{d \phi}{dt} \) and \( \rho \frac{d^2 \phi}{dt^2} \), capturing the dynamics of rotational motion. Particularly, it considers the changes in angular speed and the interaction with radial motion.
  • Vertical Acceleration: The term \( \frac{d^2 z}{dt^2} \) sticks to Newton's motion laws for linear dynamics, addressing vertical movements without interference from rotation.
Together, these components form a complete picture of how forces and accelerations act in scenarios needing cylindrical coordinates. Grappling with each term's meaning helps clarify how motion is holistically represented, crucial for any physics or engineering student dealing with three-dimensional dynamic systems.

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Most popular questions from this chapter

In Example \(3.8\), find \(\tilde{Q}_{1}\) and \(\tilde{Q}_{2}\) in terms of the components of \(\boldsymbol{F}\) by considering the work done under suitable small displacements. Check that the same expressions for the \(\tilde{Q}_{a}\) s follow from the chain rule in the case that \(\boldsymbol{F}=-\boldsymbol{\nabla} U\).

\({ }^{\dagger}\) Two particles \(P_{1}\) and \(P_{2}\) have respective masses \(m_{1}\) and \(m_{2}\) and are attracted to each other by a force with time- independent potential \(U(\boldsymbol{r})\), where \(\boldsymbol{r}\) is the vector from \(P_{1}\) to \(P_{2}\). (1) Show that the motion of the centre of mass is the same as in the case \(U=0\). (2) Show that the motion of \(P_{2}\) relative to \(P_{1}\) is the same as in the case that \(P_{1}\) is fixed and \(P_{2}\) is attracted to \(P_{1}\) by a force with potential $$ V=\frac{m_{1}+m_{2}}{m_{1}} U . $$

A particle of unit mass is moving in the \(x, y\) plane under the influence of the potential $$ U=-\frac{1}{r_{1}}-\frac{1}{r_{2}}, $$ where \(r_{1}\) and \(r_{2}\) are the respective distances from the points \((1,0)\) and \((-1,0)\). Show that if new coordinates are introduced by putting $$ x=\cosh \varphi \cos \theta, \quad y=\sinh \varphi \sin \theta, $$ then the Lagrangian governing the motion becomes $$ L=\frac{1}{2} \Omega\left(\dot{\theta}^{2}+\dot{\varphi}^{2}\right)+2 \Omega^{-1} \cosh \varphi, $$ where \(\Omega=\cosh ^{2} \varphi-\cos ^{2} \theta\). Write down the equations of motion and show that if the particle is set in motion with \(T+U=0\), then $$ \frac{1}{2}\left(\theta^{2}+\dot{\varphi}^{2}\right)-\frac{2 \cosh \varphi}{\Omega^{2}}=0 $$ Deduce that $$ \frac{\mathrm{d}^{2} \theta}{\mathrm{d} \tau^{2}}=0, \quad \frac{\mathrm{d}^{2} \varphi}{\mathrm{d} \tau^{2}}=2 \sinh \varphi $$ where \(\mathrm{d} t / \mathrm{d} \tau=\Omega\). Hence find the path of the particle.

A particle of unit mass is constrained to move on the surface of a unit sphere, but is otherwise free. Show that the dynamical trajectories are great circles traversed at uniform speeds. Show that if \(\gamma\) is a complete circuit of a great circle in time \(t\), then $$ J_{L}(v)=2 \pi^{2} / t $$ Does \(\gamma\) minimize \(J_{L}\) over all curves on the sphere that start and end at a point \(P\) on the equator and take time \(t\) for the round trip from \(P\) back to \(P ?\)

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 \(\varphi\).
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