Mechanics problem, mass sliding over a sphere

In summary, the mass will move in a circle if the centripetal force is equal to the gravitational force. If the centripetal force is greater than the gravitational force, the mass will rise and if the centripetal force is less than the gravitational force, the mass will move closer to the center of the sphere.
  • #1
fluidistic
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Homework Statement


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.


Homework Equations



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

The Attempt at a Solution


I found out the period of the mass to be [itex]T=\frac{2\pi R}{v}[/itex] 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 [itex]F_g=\frac{GMm}{R^2}[/itex]. 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.
 
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  • #2
Rotational motion: What's the centripetal force?
 
  • #3
Thanks for the big insight!
gneill said:
Rotational motion: What's the centripetal force?

OH I totally forgot about this. From memory, [itex]a_c=v^2/R[/itex] so that [itex]F_c=mv^2/R[/itex]. This force is equal to the gravitational one in this case.
So I have that [itex]v=\sqrt {\frac{GM}{R}}[/itex] thus the period is [itex]T=\frac{2 \pi R^{3/2}}{\sqrt {GM}}[/itex].
So... there's only 1 possible value for the speed of the mass?! Hard to swallow for me.
 
  • #4
What happens if the velocity is less than the one you calculated? How about if the velocity exceeds it?
 
  • #5
gneill said:
What happens if the velocity is less than the one you calculated? How about if the velocity exceeds it?
"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
Your quoted text looks fine.

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
gneill said:
Your quoted text looks fine.

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...).
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
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
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
fluidistic said:
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).

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. :smile:
 
  • #11
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.
 

Related to Mechanics problem, mass sliding over a sphere

1. What is the equation for calculating the acceleration of a mass sliding over a sphere?

The equation for calculating the acceleration of a mass sliding over a sphere is a = g * sin(θ), where a is the acceleration, g is the gravitational constant, and θ is the angle of the slope of the sphere.

2. How does the mass of the sliding object affect its acceleration?

The mass of the sliding object does not affect its acceleration. The acceleration is solely dependent on the angle of the slope of the sphere and the gravitational constant.

3. What factors can affect the friction between the mass and the sphere?

The factors that can affect the friction between the mass and the sphere include the surface texture of the sphere, the weight of the mass, and any external forces acting on the mass.

4. How can the coefficient of friction be calculated for a mass sliding over a sphere?

The coefficient of friction can be calculated by dividing the force of friction by the normal force. The normal force can be calculated by multiplying the mass of the sliding object by the gravitational constant.

5. How does the angle of the slope of the sphere affect the motion of the sliding mass?

The steeper the slope of the sphere, the greater the acceleration of the sliding mass will be. This is because the sine of the angle increases as the angle becomes steeper, resulting in a larger value for the acceleration equation.

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