Equation of trajetory+conserved quantities of a motion over a sphere

In summary: Thanks!In summary, the author attempted to solve a homework problem involving the motion of a particle constrained to move freely over the surface of a sphere. They used spherical coordinates (r, \phi , \theta) and found that a rotation over any diameter of the sphere shouldn't change the dynamics of the particle. They then wrote the Lagrangian and found that E= \sum _i \frac{\partial L}{\partial \dot q_i} \dot q_i -L. They explained that the unit vectors i, j and k can be derived by taking the time derivative of the unit vectors in cylindrical and spherical coordinates and that it is easier to remember this method if x= rho cos (phi),
  • #1
fluidistic
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Homework Statement


Determine the equation of the trajectory and the conserved quantities in the motion of a particle constrained to move freely over the surface of a sphere.

Homework Equations


Not sure.

The Attempt at a Solution


I think it is convenient to use spherical coordinates [tex](r, \phi , \theta)[/tex].
I notice that a rotation over any diameter of the sphere shouldn't change the dynamics of the particle, hence the angular momentum is conserved?
Anyway, I want to write the Lagrangian. My problem resides in writing the position vector of the particle in spherical coordinates. I know that [tex]r=r \hat r[/tex]. But I'm not sure how to write [tex]\phi[/tex] and [tex]\theta[/tex] in terms of [tex]\hat r[/tex], [tex]\hat \theta[/tex] and [tex]\hat \phi[/tex].
I realize that the Lagrangian must depend explicitly on [tex]\phi[/tex] and [tex]\theta[/tex] and not on [tex]r[/tex] since they are variables depending on time. I also believe the energy is conserved, but I have to show it I believe using the Lagrangian of the particle.
Any correction of my thoughts and help about how to write the position vector is welcome.Edit: OK I just saw in wikipedia that [tex]\vec r=r\hat r[/tex] and "thus" [tex]\dot \vec r = \dot r \hat r + r \dot \theta \hat \theta + r \dot \theta \sin (\theta) \hat \phi[/tex]. I'm actually trying to understand this implication.Edit 2: Ok, assuming the last formula for [tex]\dot \vec r[/tex], since r is constant I have that [tex]\dot \vec r =r \dot \theta \hat \theta + r \dot \theta \sin (\phi) \hat \phi[/tex]. I can get the Lagrangian. I know that [tex]E= \sum _i \frac{\partial L}{\partial \dot q_i} \dot q_i -L[/tex].
Now if someone can explain me how to get the expression given in wikipedia, you'll save me hours. Thanks in advance.
 
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  • #2
Taking the time derivative of the unit vectors in spherical coords can be tricky. I find it easiest to convert the unit vector in spherical coords to cartesian coords. Take the time derivative there, and convert back to spherical coords.
 
  • #3
nickjer said:
Taking the time derivative of the unit vectors in spherical coords can be tricky. I find it easiest to convert the unit vector in spherical coords to cartesian coords. Take the time derivative there, and convert back to spherical coords.

It might be nice and easy method, yet still very complicated to me. Unfortunately the conversion of spherical unit vectors into cartesian's one cannot be found in this table: http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates.
And I've tried the derivation, I get lost. I've found no website with such a derivation.
 
  • #5
A handy reference for converting between Cartesian, Cylindrical and Spherical coordinates can be found in the back cover of Griffiths Introduction to Electrodynamics, if you have that text.
 
  • #6
otherwise for the general formulation, my math method text (riley hobson bence) has a good section under - general curvilinear coordinates, which i found pretty informatiev to understand how all the derivatives are calaulated - though I'm sure you could probably google it
 
  • #7
Thanks a lot. I've checked out both Griffith and Riley's book. It is much of help. Especially Riley's since it is explained how can one derive the unit vectors i, j and k expressed in cylindrical and spherical coordinates. So instead of memorizing the whole formulas given in wikipedia and Griffith, if I keep in mind that x= rho cos (phi), y=rho sin (phi) and z=z for cylindrical coordinates, I can derive the corresponding unit vectors with simple partial derivatives (Jacobian) and a normalization. It also allow me to derive them without making a single sketch of the different coordinate systems, which is pretty nice since I get confused a lot on 2 dimensional paper sheets.
So your suggestions really helped me.
 

FAQ: Equation of trajetory+conserved quantities of a motion over a sphere

What is the equation of trajectory for motion over a sphere?

The equation of trajectory for motion over a sphere is given by r(t) = Acos(ωt + φ)i + Asin(ωt + φ)j + Bt + Ck, where A, B, and C are constants, ω is the angular velocity, and φ is the initial phase angle.

What are conserved quantities of motion over a sphere?

The conserved quantities of motion over a sphere are angular momentum, linear momentum, and energy. These quantities remain constant throughout the motion and are conserved due to the symmetry of the system.

How is angular momentum conserved in motion over a sphere?

Angular momentum is conserved in motion over a sphere because the direction of the angular momentum vector remains constant in the absence of external torques. This is due to the rotational symmetry of the system.

What is the significance of the constant term in the equation of trajectory for motion over a sphere?

The constant term in the equation of trajectory represents the height of the motion above the sphere's surface. It is a conserved quantity and can be used to calculate the maximum height reached by the object during its motion.

How does energy conservation apply to motion over a sphere?

Energy conservation applies to motion over a sphere because the total energy of the system, which includes kinetic energy and potential energy, remains constant. As the object moves along its trajectory, the kinetic energy may change, but the potential energy, which is related to the height above the sphere, will also change to keep the total energy constant.

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