Acceleration on the surface of a sphere

[freddy]

Hello everyone!

I need some help with problem 2.2 out of Classical Dynamics of Particles and Systems by Thornton & Marion. I can't believe I got stuck on the second problem

The text of the problem is:

A particle of mass m is constrained to move on the surface of a sphere of radius R by an applied force F(theta, phi). Write the equation of motion.

Seemed simple to me at first, F_theta = m * d2/dt2 theta and I'm done. I figure since the surface of a sphere is a two dimensional surface and thetahat and phihat are perpindicular to each other I should be able to just apply forces in either thetahat or phihat and expect the particle to stay on the surface of the sphere. Looking at the answer in the back of the book I can see that I'm very wrong:

F_theta = m*R*(d2/dt2 theta - d/dt(phi^2)*sin(theta)*cos(theta))
F_phi = m*R*(2*d/dt(theta) * d/dt(phi) * cos(theta) + d2/dt2(phi) * sin(theta)

After having given it some thought I think I understand why the two accelerations must be dependent upon each other, but I still don't know how I'm supposed to work it mathematically.

The accelerations must depend on one another because of how phi and theta are set up, almost anywhere on the sphere if I want to travel solely through either phi or theta I'm not actually traveling in a straight line; a straight line on a sphere is a line that bisects the sphere. So, in general, if I have a particle on the surface of a sphere and want it to travel in only one of theta or phi I must exert a force in both in order to keep it on track.

Is my reasoning correct? If not, what's wrong with it, and if so, how the heck am I supposed to get there mathematically?

Thanks in advance for any help :)

 

[arildno]

You need a vector representation of your surface in order to manage this.
In addition, you'll need to know the unit vectors in spherical coordinates.

Any point on the sphere with radius R may be given the vector representation:
R(\sin\gamma\cos\psi\vec{i}+\sin\gamma\sin\psi\vec{j}+\cos\gamma\vec{k})
0\leq\psi\leq2\pi,0\leq\gamma\leq\pi
(You may find out how \gamma,\psi are related to the used angle variables \theta,\phi)

The unit vectors are:
\vec{i}_{r}= (\sin\gamma\cos\psi\vec{i}+\sin\gamma\sin\psi\vec{j}+\cos\gamma\vec{k})
\vec{i}_{\gamma}=\frac{\partial\vec{i}_{r}}{\partial\gamma}=(\cos\gamma\cos\psi\vec{i}+\sin\gamma\sin\psi\vec{j}-\sin\gamma\vec{k})
\vec{i}_{\psi}=\frac{1}{\sin\gamma}\frac{\partial\vec{i}_{r}}{\partial\psi}=-\sin\psi\vec{i}+\cos\psi\vec{j}

Clearly, the position of the particle with respect to time can be written as:
\vec{r}(t)=R\vec{i}_{r}(t)
with angular functions of time \gamma(t),\psi(t)
Use the chain rule to determine the acceleration components.

 

[wais]

Hi there, it looks like you have a good understanding of the problem and the reasoning behind the answer. The key concept here is that the particle is constrained to move on the surface of the sphere, meaning it can only move in the two dimensions of theta and phi. This means that any force applied to the particle must take into account the motion in both directions.

To understand the mathematical derivation of the answer, it might be helpful to think about the forces acting on the particle. In this case, we have the applied force F(theta, phi) and the constraint force that keeps the particle on the surface of the sphere. This constraint force can be broken down into two components, one in the theta direction and one in the phi direction.

Using Newton's second law, we can write the equations of motion as:

F(theta, phi) + F_constraint(theta) = m*a(theta)
F(theta, phi) + F_constraint(phi) = m*a(phi)

Where a(theta) and a(phi) are the accelerations in the theta and phi directions, respectively. From here, we can use some trigonometry and the fact that the constraint forces must be perpendicular to the surface of the sphere to derive the equations given in the book.

So, in summary, your reasoning is correct and the key is understanding the constraint forces and how they affect the motion of the particle on the surface of the sphere. I hope this helps!