Uniform circular motion and conservation of energy

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A point mass in an uniform circular motion is continuously changing the velocity direction. To do it, it continuously need force (energy).

If we don't give any energy to the system it will anyhow continues its uniform circular motion. How it's possible, who gives the energy ?

(It's seems a simple question, but i'm trying to understand it very profoundly )
 

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  • #2
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If we don't give any energy to the system it will anyhow continues its uniform circular motion.
Who told you that?
 
  • #3
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Who told you that?
Mhmm.. probability i'm wrong, but i thought that a variation of momentum ( even in the case only the direction of momentum changes) needs a force and a force needs energy to be generated
 
  • #4
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Mhmm.. probability i'm wrong, but i thought that a variation of momentum needs a force and a force needs energy to be generated
Well it mostly depends on how or if the circular motion is caused. If the particle is "on a stick", then it can continue to move in a circle without added energy (ignoring friction). If not, it will move inertially according to Newton's laws (i.e. not in a circle).
Are you asking about one of these situations?
 
  • #5
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force needs energy to be generated
There lies your problem. This is not true. Force and energy are different concepts, you can't say that force needs energy to be generated.
 
  • #6
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I'm i bit confuse now. I tried to derived uniform circular motion starting only from :

1) Conservation of energy
2) Knowing that a change in direction or intensity of momentum need a force ( so energy ) to be produced.

But i this point i think my two premises are incorrect :(
 
  • #7
Dale
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To do it, it continuously need force (energy).
Force, yes. Energy, no. Force and energy are not the same thing. Continuous energy is not required.
 
  • #8
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need a force ( so energy )
Again, your "so" is wrong. Energy and force are different concepts. There are situations when energy is not transfered even though force acts on a point mass. It happens for example in uniform circular motion.
 
  • #10
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Probably i was wrong because i analized momentum from the origin of axis
Again, your "so" is wrong. Energy and force are different concepts. There are situations when energy is not transfered even though force acts on a point mass. It happens for example in uniform circular motion.
So, it's like : force needs energy only when it changes the intensity of the momentum ?
 
  • #11
nasu
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Force does not need anything. :)
If the energy of the system increases, the energy of another system will decrease. The energy transfer between the two systems is due to the interaction between them. One of the quantities describing the interaction is force. So the interaction between systems is measured by interaction forces. This interaction may (or may not) result in a transfer of energy.
 
  • #12
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Your question boils down to whether work is being done on an object in uniform circular motion.

##W=\int{F \cdot ds}##

The dot product in the integral is the key to answering your question.

Recall ##|a \cdot b| = ab cos(\theta)## So what is ##\theta##, the angle between the force and the infinitesimal displacement, in this case?
 
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  • #13
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So, it's like : force needs energy only when it changes the intensity of the momentum ?
No. A force does work when it's applied parallel to the velocity. Whether the momentum of the mass changes doesn't matter, as there might be other forces.
 
  • #14
jbriggs444
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No. A force does work when it's applied parallel to the velocity. Whether the momentum of the mass changes doesn't matter, as there might be other forces.
A force applied parallel to the velocity will increase the magnitude of the momentum (aka the "intensity" of the momentum as I read @mfig's wording). A force applied at right angles to a velocity will neither increase nor decrease its magnitude.

So I would say that the magnitude of the momentum is indeed a key indicator. (Assuming the absence of other forces).
 
  • #15
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So I would say that the magnitude of the momentum is indeed a key indicator. (Assuming the absence of other forces).
That assumption in brackets means we are talking about the most trivial case only. Hence the statement is not general and leads to confusion in analyzing work done by forces, as soon as there is more than one force. It's similar to the common confusion when people don't understand that the F in F=ma is the net force.
 
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  • #16
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I have difficulties in understand intuitively what a perpendicular force is. To me, since it doesn't imply an energy transfer , it seem something that doesn't exist.
I'm sorry if it's a trivial doubt, but to me it's a very strange concept...
 
  • #17
jbriggs444
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I have difficulties in understand intuitively what a perpendicular force is. To me, since it doesn't imply an energy transfer , it seem something that doesn't exist.
I'm sorry if it's a trivial doubt, but to me it's a very strange concept...
A rock on a string changes direction without changing energy. The string has tension. A force most certainly does exist. It is perpendicular to the rock's direction of travel.
 
  • #18
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I have difficulties in understand intuitively what a perpendicular force is. To me, since it doesn't imply an energy transfer , it seem something that doesn't exist.
I'm sorry if it's a trivial doubt, but to me it's a very strange concept...
A book sitting motionless on a table experiences both a downward force due to gravity and an upward force from the table, yet no energy is transferred to or from the book, the table, or from the Earth. As has been said already, forces do not necessarily require that energy be transferred. You are actually intimately familiar with this concept. It takes no energy to keep yourself asleep in bed, right? Otherwise sleeping wouldn't be nearly as popular as it is. :wink:

In both this example and in the example of uniform circular motion, the force is not acting in the direction of motion. Either because there is no motion or because the force is perpendicular to the direction of motion. Because of this, no work is done in either example and so no energy is transferred anywhere.
 
  • #19
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I believe your confusion may stem from your problem statement. You mention a point mass in uniform circular motion and deriving the equation.
First, are you trying to derive the equation (i.e. expression) for the centripetal force, or the centripetal acceleration. If you are attempting to derive the expression for the centripetal acceleration, you do not need (and cannot use) force, or energy as useful concepts. You only need the definitions of position, velocity, and acceleration as vectors and motion in the plane. The simplest route (if you know calculus) is to describe the circular motion (position) as
x = R cos (omega * time) I + Y sin (omega * time) J, where I and J are unit vectors in the x and y directions. Then take two derivatives to get
a = d2x/ dt2 = (-) R * omega * omega (cos (omega * time) I + Y sin (omega * time) J) = - R omega * omega times a unit vector in the radial direction cos (omega * time) I + Y sin (omega * time) J.
So acceleration is radially inward (note the minus sign) and has magnitude R * omega squared. Hence the name centripetal which means "center seeking"
because it is directed towards the center.

Without calculus you can still derive the centripetal acceleration kinematically, but you need to remember similar triangles from high school geometry, and be clever.

The centripetal force is merely the point mass (mass) times the centripetal acceleration. Energy plays no role at all.

I have found (and I use to tell my students), that the best idea is to recall Newton's laws.

1. A body at rest tends to remain at rest and a body in motion tends to remain in motion in a straight line unless acted upon by a force.

Your proposed point mass is in circular motion so it has a force acting on it.
What provided the force?
You did not say. In could be the Earth's gravity if you are treating a satellite. It could be electrical if you are treating an electron in orbit around a proton in a hydrogen atom and using the old quantum theory, it could be your muscles if you are spinning the rock on a string. Etc.

2. It turns out you only have half a problem here without defining the force holding the point mass in circular motion.
By the way force is a vector and momentum is a vector.

Newton's second law (often misquoted as F = ma) is actually F = dp/dt. This law holds for all three components x, y, and z separately so
Fx = d px / dt; Fy = d py / dt. and Fz = d pz / dt.

Your construct "intensity of momentum" , which I take you mean the "magnitude" of the momentum vector is a concept that Newton did not address, and I cannot see where it is useful. Surely, Newton's second law holding for each component separately is a tighter requirement and will be more useful.

3. Insofar as energy is concerned, energy is a scalar, "centripetal" meaning center seeking implies a direction. Centripetal force and centripetal acceleration makes sense, but centripetal energy does not.

4. I sort of blame the misconception that energy is needed to "preserve" motion on Star Trek.

The starship Enterprise was always in danger of plummeting to the planet when their dilithium crystals gave out. In reality, a spacecraft in orbit without a source of energy could take years to decay from even a modest 100 mile altitude. (The decay is eventually from frictional losses of the spacecraft energy from the few atmosphere molecules) The Earth has been in orbit around the Sun for billions of years, (and without dilithium crystals) too.

5. Actually though, your thoughts on energy transfer did motivate me to revisit some concepts on energy transfer, that I remember reading more than 30 years ago in an unexpected source. I encourage any reader to examine what is written in "The Classical Electromagnetic Field" by Leonard Eyges. Dover edition 1972. The interested reader need not have the prerequisites in EM theory to explore the last paragraph on page 201,and first two paragraphs on page 202, referring to energy transfer. I felt it is very revealing.

I hope this lengthy reply helps and especially provides some humor.
 
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  • #21
bobob
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A point mass in an uniform circular motion is continuously changing the velocity direction. To do it, it continuously need force (energy).

If we don't give any energy to the system it will anyhow continues its uniform circular motion. How it's possible, who gives the energy ?

(It's seems a simple question, but i'm trying to undestand it very profoundly )
OK, here is one way to see it. A mass in a circular orbit has a velocity tangential to that circle. However, it will also fall radially due to gravity. The result is that it cannot continue moving in the same straight line without applying a force to make it do so, nor can it just stop and fall. It simply changes the direction that it is moving to maintain that circular orbit. Such an object is in a state of constant free fall. In free fall, there are no forces acting.
 
  • #22
jbriggs444
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OK, here is one way to see it. A mass in a circular orbit has a velocity tangential to that circle. However, it will also fall radially due to gravity. The result is that it cannot continue moving in the same straight line without applying a force to make it do so, nor can it just stop and fall. It simply changes the direction that it is moving to maintain that circular orbit. Such an object is in a state of constant free fall. In free fall, there are no forces acting.
This is the classical physics sub-forum. Here, gravity is a force. A force that does no work because, in the case of a circular orbit, it always acts at right angles to the direction of motion.
 
  • #23
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In free fall, there are no forces acting.
1) The OPs question is about circular motion in general, not just orbits due to gravity. Bringing up alternative gravity models doesn't adress circular motion in general.

2) Even in the model where gravity is not an interaction force, it is still an inertial force which could do work, but doesn't because it is perpendicular to velocity in the frame where the orbit is a circle.
 

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