Homework Help: Mechanics problem, mass sliding over a sphere

1. Jul 6, 2011

fluidistic

1. The problem statement, all variables and given/known data
I'm stuck on a relatively simple problem. I can't use Lagrangian mechanics, only Newtonian one.
A mass m slides without friction over a sphere of radius r and mass M. Find its period and the allowed values of its speed.

2. Relevant equations

The sphere isn't in an external gravitational field. So there's the gravitational force between the mass m and the sphere.

3. The attempt at a solution
I found out the period of the mass to be $T=\frac{2\pi R}{v}$ where v is the speed of the mass.
And now this is where I'm stuck. I know that the modulus of the gravitational force between the mass and the sphere is $F_g=\frac{GMm}{R^2}$. I also know that for a critical value of v, the mass will start to be in orbit over the sphere. This happens when the normal force is worth 0N. So I think I must express the modulus of the normal force acting on the mass in function of the speed of the mass. But I don't know how to "include v" in the expression for the normal force. Hmm.
Is that a reasonable way to approach the problem? Could you give me any tip? Thanks in advance.

2. Jul 6, 2011

Staff: Mentor

Rotational motion: What's the centripetal force?

3. Jul 6, 2011

fluidistic

Thanks for the big insight!
OH I totally forgot about this. From memory, $a_c=v^2/R$ so that $F_c=mv^2/R$. This force is equal to the gravitational one in this case.
So I have that $v=\sqrt {\frac{GM}{R}}$ thus the period is $T=\frac{2 \pi R^{3/2}}{\sqrt {GM}}$.
So... there's only 1 possible value for the speed of the mass?! Hard to swallow for me.

4. Jul 6, 2011

Staff: Mentor

What happens if the velocity is less than the one you calculated? How about if the velocity exceeds it?

5. Jul 6, 2011

fluidistic

"If the velocity is lesser than the one I just calculated, the centripetal force is lesser than the gravitational force so that the mass is still over the sphere.
If the velocity of the mass is greater than the one I just calculated, then the centripetal is greater than the gravitational force so that the mass leaves the sphere."

I know this is wrong. Hmm... considering the vector forces I'm confused. If I draw the forces, the centripetal force and the gravitational force acting on the mass m have the same direction.

6. Jul 6, 2011

Staff: Mentor

The gravitational force supplies the centripetal force to hold the mass in circular motion. If the required centripetal force exceeds that supplied by gravity, the mass can no longer maintain circular motion; it must depart from the sphere's surface (in fact it's motion will become elliptical rather than circular if no other forces are applied, but that's a story for another day...).

7. Jul 6, 2011

fluidistic

I understand what you mean, but I don't get it from the equations. In the critical case, when the centripetal force is equal in magnitude to the gravitational force, the normal force acting on the mass is 0N? It seems like the gravitational force and the centripetal force have opposite direction. Is this true? If so, then I'd understand what you mean also by equations rather than only words.

8. Jul 6, 2011

Staff: Mentor

In the critical case, the gravitational force *is* the centripetal force. And yes, the normal force exerted by the Earth on the mass will be zero at the critical velocity. So at the critical velocity, the normal force is zero and the body is in circular motion because the gravitational force on the mass is exactly equal to the required centripetal force for circular motion.

The centripetal force of circular motion is the center-directed force that maintains the circular motion. Without other constraints (like running into the Earth's surface in this case), if the inward directed force exceeds this value then the mass will move closer to the center. If the inward directed force is less than this value then the mass will want to rise. In either event the motion will no longer be circular.

The reason we use centripetal force in our discussions is that there is a certain prejudice, if you will, against using what are called pseudo-forces in calculations in an inertial reference frame. Thus you only hear "centrifugal force" mentioned in whispers and with apologies, even though it may in some circumstances make the analysis more obvious. A centrifugal force would be directed outward, opposing the gravitational force.

I know it sounds strange, but there you are. The equations that result from the use of the centripetal force paradigm turn out to be the same as those that would use the centrifugal force.

9. Jul 6, 2011

fluidistic

Ok thanks a lot for all your help, I really think I understand now.
For the fun of it, I calculated the critical speed for the Earth to be around 790596302 m/s. This seems too huge to be right so I might have made a mistake (I used the Ubuntu calculator, not sure it worked out as I wanted).

10. Jul 6, 2011

Staff: Mentor

You're quite welcome.

The speed you've found for the Earth has the right digits, but the order of magnitude is a tad high. Must have been a unit conversion slip-up along the way. 790596302 m/s is about 2.6 times the speed of light! The result should be closer to 7905 m/s.

11. Jul 6, 2011

fluidistic

Oh that's hilarious. I looked at the number and didn't count the digits but I thought it was less than the speed of light (actually I did think about comparing it to the speed of light). So yes, it was WAY too huge.
I think I took R in kilometers instead of meters.