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Circular Motion of Point-Mass Objects

 

 

 

 

 

 


Previously Asked Questions

Q:     Is the angular velocity, , a vector or a scalar?

A:     One can travel on the circumference of the circle either clockwise" or "anticlockwise" which gives the arc traveled, s, a direction and makes it a vector.  In fact, the element of arc, ds, is in the direction of the tangent to the circle in any given point, and has a direction. Conclusion: ds is a vector.  Since, by definition, angular velocity is = ds/dt = (1/dt) ds, it means that, ds being a vector and dt being a scalar, the product (1/dt) ds is a vector.  Conclusion, is a vector.

Q:     What is the difference between the centripetal and centrifugal force?

A:     The difference between centripetal and centrifugal forces is quite simple - centrifugal forces do not exist while centripetal forces do.  As with most simple statements, there is a great deal more to understanding this issue than simply memorizing which of the forces does or does not exist.  To understand the centripetal force and the fictitious centrifugal force, let's first examine the words centripetal and centrifugal.

centri is derived from the Latin centr meaning "center."

petal is derived from the Latin petere meaning "seek."

fugal is derived from the Latin fugere meaning "to flee" as in fugitive.

So, literally, the centripetal force is a "center-seeking" force.  The fictitious centrifugal force is, literally, a non-existing "center-fleeing" force.  Now that we know the definition of the terms, we pose the question: "Which force, the 'center-seeking' force or the 'center-fleeing' force, keeps an object in a circular path?"

Figure 1: Diagram of a point-mass object moving along a circular path.

Let us impose on a point-mass object the condition that it be on a circular path at any time. The vector velocity of this object is always tangent to the circle, therefore it changes direction in time, as the object moves along the circle. Consequently, the circular motion is a accelerated motion, simply because the direction of vector velocity changes, even if its magnitude (speed) remains constant. The problem now becomes to find out what force generates the acceleration that keeps the object moving on a circle.

Consider an initial object position A and a position B where the object reaches after a time interval t (Fig 1). The velocity vectors at A and B are shown in the figure as v0 and vf , respectively. The arc of circle traveled by the object in the time interval t is called s.  Force simplicity, in this example, the magnitude of velocity (speed) to be constant.  Such a motion of an object on a circle, with constant speed (v), is called "uniform circular motion".  Again, such a motion is an "accelerated motion" just because the direction of the vector velocity changes.

Note that in the equation F = ma, there are two vectors F and a, m is merely a scalar and thus has no effect on the direction of F.  So, finding the direction of a will also give us the direction of F.

Acceleration, a, is the change in velocity over time:

a = v / t

We notice that acceleration is a vector, v, multiplied by a scalar, 1/t.  So, the direction of the acceleration will be the same as the direction of the change in velocity, v.  This means that the direction of the force, F, will also be in the direction of v.

F || a || v.

We are finally ready to find the direction of the force F, which is also the direction of the acceleration, a, and the direction of the change of velocity, v.  So what is the vector "change of velocity" v?

The "change" in any physical quantity is defined as the final quantity minus the initial quantity.  So, the change in velocity is the final vector velocity minus the initial vector velocity:

v = vf - v0

Figure 2: Evaluation of v = vf. - v0.

To find the vector v we graphically subtract vector v0 from vector vf (Fig. 2).  Note: for clarity, we have moved the points of origin of both vectors vf and v0 to a common point, out of Figure 1.  (Remember, since a vector is defined only by magnitude and direction, its point of origin is irrelevant.)  Figure 2 shows the vector v in red color.  We can now transport it back into the initial figure (Fig.1), preserving its magnitude and direction as given in Figure 2.  After executing this operation, we call this figure Figure 3.

We note that F || a || v.  So, we can draw in Figure 3 all three vectors (a - blue, F- green, and v - red) parallel to each other.

Notice the direction of all these three vectors; they are all directed toward the center of the circle and are, therefore, "center-seeking" or centripetal.  Both the centripetal acceleration a and the centripetal force F are labeled with a sub-index "r" (ar and Fr) to indicate that they are oriented in the direction of the radius of the circle, r.

Figure 3:  Same as Figure 1, showing the vectors F, a, and v.

Conclusion: This Fr, obtained from the only initial condition that the object be on the circle at all times, and is found to be oriented along the radius r and pointing towards the center.  This Fr is the only force needed to generate the acceleration ar which represents the change of velocity from v0 to vf , keeping the object on the circle at all times.  No other force is needed to keep the object on the circle.  No force was obtained in a direction away from the center.  Therefore, no "centrifugal" force exists.

Derivation of an Analytic Expression for Fr

Now, in order to derive an analytic expression for Fr, we use the following property of two "similar" isosceles triangles, as show below:

                   
Figure 4: The relation between the sides of "similar"
isosceles triangles.

For the triangles in Figure 1 and Figure 2, this relationship is:

       (1)           from which:

    (2)

Divide this (2) by the time interval t:

   (3)

    (4)

   (5)     and therefore

   (6)

These expressions above (equations 5 and 6) represent the magnitudes of ar (centripetal acceleration) and Fr (centripetal force).  If we want to express them as vectors, these formulas become:

  and  .

Where (r-hat) represents the "unit vector" along r.
 

Q:    Do satellites remain in orbit forever?  Why or why not?

A:    Satellites are placed in orbits inside the earth's atmosphere (not outside the atmosphere, where they would be in vacuum.)  Therefore, there is friction which results in generation of heat.  This amount of heat energy is being dissipated and lost into the environment.   Its only source is the kinetic energy of the satellite which gradually decreases.   This means a decrease in the linear velocity of the satellite in orbit, which results in a decrease of the radius of the orbit.  The final result is a spiraling of the satellite toward the surface of the earth.  If this would really happen, the satellite would probably burn even before falling to ground.  In order to keep satellites in their designed orbits, booster rockets have to be activated from time to time to provide the extra energy needed to compensate for heat losses.  Launching and maintaining satellites in orbits involve excellent high-technology profession for the present and the future.

[Top] [Previously Asked Questions] [References]


References

Equations

Uniform Circular Motion Acceleration a = v2 / r
The Period of Revolution. T = {2 pi2.gif (831 bytes) r) / v
Centripetal Force F = ma = mv2 / r

 

[Top] [Previously Asked Questions] [References]


Mechanics List of Topics

Measurements Newton's Laws Potential Energy and Conservation of Energy Rotation of
Rigid Bodies
Elasticity
Vectors Forces and Fields Linear Momentum Angular Momentum Mechanical
Oscillations
Motion of Point-Mass Objects in One Dimension The Gravitational Field Collisions Torque Mechanical Waves
Motion of Point-Mass Objects in Two and Three Dimensions Kinetic Energy
and Work
Circular Motion of Point-Mass Objects Equilibrium Sound

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