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Centripetal force
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The centripetal force is the external force required to make a body follow a curved path. Hence centripetal force is a kinematic force requirement, not a particular kind of force like gravity or electromagnetism. Isaac Newton's description is found in the Principia. Many different examples are examined, one being the following:
Any force (gravitational, electromagnetic, etc.) or combination of forces can act to provide a centripetal force.
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The centripetal force is the external force required to make a body follow a curved path. Hence centripetal force is a kinematic force requirement, not a particular kind of force like gravity or electromagnetism. Isaac Newton's description is found in the Principia. Many different examples are examined, one being the following:
Any force (gravitational, electromagnetic, etc.) or combination of forces can act to provide a centripetal force. An example for the case of uniform circular motion is shown in Figure 1.
Centripetal force is directed inward, toward the center of curvature of the path. The term centripetal force comes from the Latin words centrum ("center") and petere ("tend towards", "aim at"), and can also be derived from Newton's original definitions described in the Principia.
Centripetal force should not be confused with centrifugal force (see the Common misunderstandings section below).
Simple example: uniform circular motion
The velocity vector is defined by the speed and the direction of motion. Objects experiencing no net force do not accelerate and hence move in a straight line with constant speed; they have a constant velocity. However, an object moving in a circle, even at constant speed, has a changing direction of motion. The rate of change of the object's velocity vector in this case is the centripetal acceleration (see Figure 1).
The centripetal acceleration varies with the radius of curvature of the path (R) and speed (v) of the object, becoming larger for greater speeds and smaller radii. If an object is traveling in a circle with a varying speed, its acceleration can be divided into two components: a radial acceleration (the centripetal acceleration that changes the direction of the velocity) and a tangential acceleration that changes the magnitude of the velocity.
Source of centripetal force
Centripetal force is inferred from the trajectory of the object, without regard for how the path was arrived at (regardless of the origin of the forces involved). The studies of trajectories and the forces they imply is kinematics, while the study of which motions result from given physical forces is kinetics, the other branch of dynamics.
Supposing the analysis of a trajectory (kinematics) has concluded that for an object to follow the observed path a centripetal force is required, one might reasonably ask the (kinetic) question: "Where is the centripetal force coming from?"
For a satellite in orbit around a planet, the centripetal force is supplied by the gravitational attraction between the satellite and the planet, and acts toward the center of mass of the two objects. For an object at the end of a rope rotating about a vertical axis, the centripetal force is the horizontal component of the tension of the rope, which acts towards the center of mass between the axis of rotation and the rotating object. For a spinning object, internal tensile stress is the centripetal force that holds the object together in one piece.
Common misunderstandings
Centripetal force should not be confused with centrifugal force. Centripetal force is a kinematic force requirement deduced as necessarily present if an object is to track an observed trajectory. It is not a kinetic force like gravity or electrical forces. Centripetal force requirements may be deduced from a trajectory in any frame of reference (although the trajectory of an object and the deduced centripetal force will vary from one frame to another). Because centripetal force is a kinematic force requirement inferred from an established trajectory, it is not used to deduce a trajectory from a physical situation. Centripetal force is therefore not included in the inventory of forces that are used in applying Newton's second law to calculate a trajectory.
Centrifugal force, on the other hand, is treated in a rotating frame as a kinetic force – it is part of the inventory of forces used in Newton's second law to predict motion. Centrifugal force is a fictitious force, which arises only when motion is described or experienced in a rotating reference frame, and it does not exist in an inertial frame of reference.
Centripetal force should not be confused with central force, either. To re-iterate, centripetal force is a force requirement necessary for a curved trajectory to be possible; it is not a type of force, such as a nuclear or gravitational force. In contrast, the term "central forces" does refer to a type of force, more exactly to a class of "physical forces" between two objects that meet two conditions: (1) their magnitude depends only on the distance between the two objects and (2) their direction points along the line connecting the centers of these two objects. Examples of central forces include the gravitational force between two masses and the electrostatic force between two charges.
As an example relating these terms, the centripetal force implied by the circular motion of an object often is provided by a central force.
Analysis of several cases
Below are three examples of increasing complexity, with derivations of the formulas governing velocity and acceleration.
Uniform circular motion
Uniform circular motion refers to the case of constant rate of rotation. Here are two approaches to describing this case.
Geometric derivation
The circle in the left of Figure 2 shows an object moving on a circle at constant speed at two different times in its orbit. Its position is given by the vector R and its velocity by the vector v.
The velocity vector is always perpendicular to the position vector (since the velocity vector is always tangent to the circle of motion). Since R moves in a circle, so does v. The circular motion of the velocity is shown in the circle on the right of Figure 2, along with its acceleration a. Just as velocity is the rate of change of position, acceleration is the rate of change of velocity.
Since the position and velocity vectors move in tandem, they go around their circles in the same time T. That time equals the distance traveled divided by the velocity
and, by analogy,
Setting these two equations equal and solving for |a|, we get
The angular rate of rotation in radians per second is:
Comparing the two circles in Figure 2 also shows that the acceleration points toward the center of the R circle. For example, in the left circle in Figure 2, the position vector R pointing at 12 o'clock has a velocity vector v pointing at 9 o'clock, which (switching to the circle on the right) has an acceleration vector a pointing at 6 o'clock. So the acceleration vector is opposite to R and toward the center of the R circle.
Derivation using vectors
Figure 3 shows the vector relationships for uniform circular motion. The rotation itself is represented by the vector O, which is normal to the plane of the orbit (using the right-hand rule) and has magnitude given by:
with ? the angular position at time t. In this subsection, d?/dt is assumed constant, independent of time. The displacement l of the particle in time dt along the circular path is
which, by properties of the vector cross product, has magnitude rd? and is in the direction tangent to the circular path.
Consequently,
In other words,
Differentiating with respect to time,
Lagrange's formula states:
Applying Lagrange's formula with the observation that O • r(t) = 0 at all times,
-
In words, the acceleration is pointing directly opposite to the radial displacement r at all times, and has a magnitude:
-
where vertical bars |...| denote the vector magnitude, which in the case of r(t) is simply the radius R of the path. This result agrees with the previous section if the substitution is made for rate of rotation in terms of the period of rotation T:
Not surprisingly, this result also agrees with that of the nonuniform circular motion when the rate of rotation is made constant.
A merit of the vector approach is that it is manifestly independent of any coordinate system.
Example: The banked turn
Figure 4 shows a ball in circular motion on a banked curve. The curve is banked at an angle ? from the horizontal, and the surface of the road is considered to be slippery. The object is to find what angle the bank must have so the ball does not slide off the road (the so-called "angle of bank"). Intuition tells us that on a flat curve with no banking at all, the ball will simply slide off the road; while with a very steep banking, the ball will slide to the center unless it travels the curve rapidly.
The right side of Figure 4 indicates the forces on the ball. There are two forces; one is the force of gravity vertically downward through the center of mass of the ball mg where m is the mass of the ball and g is the gravitational acceleration; the second is the upward normal force exerted by the road perpendicular to the road surface man. The centripetal force shown in Figure 4 is the net force obtained by vector addition of the normal force and the force of gravity, and is not a third force applied to the ball.
The horizontal net force on the ball is the horizontal component of the force from the road, which has magnitude |Fh| = m|an|sin?. The vertical component of the force from the road must counteract the gravitational force, that is |Fv| = m|an|cos? = m|g|. Accordingly one finds the net horizontal force to be:
On the other hand, at velocity |v| on a circular path of radius R, kinematics says that the force needed to turn the ball continuously into the turn is the radially inward centripetal force Fc of magnitude:
Consequently the ball is in a stable path when the angle of the road is set to satisfy the condition:
or,
As the angle of bank ? approaches 90°, the tangent function approaches infinity, allowing larger values for |v|2/R. In words, this equation states that for faster speeds (bigger |v|) the road must be banked more steeply (a larger value for ?), and for sharper turns (smaller R) the road also must be banked more steeply, which accords with intuition. When the angle ? does not satisfy the above condition, the horizontal component of force exerted by the road does not provide the correct centripetal force, and an additional frictional force tangential to the road surface is called upon to provide the difference. If friction cannot do this (that is, the coefficient of friction is exceeded), the ball slides to a different radius where the balance can be realized.
These ideas apply to air flight as well. See the FAA pilot's manual.
Example: The yo-yo
Another interesting example involving centripetal force occurs in the yo-yo.
As the string unwinds, the yo-yo rolls down one side of the string. When the string is totally unwound, the yo-yo continues rolling while performing a complete one hundred and eighty degree U-turn in its translation motion. It then rolls back up the other side of the string, while simultaneously causing the string to wind up again. In the figure this behavior is shown on the left at three different points in the motion; arrows indicate the direction of motion and the rotation. The potential energy of the yo-yo at the top of its motion is converted to rotational kinetic energy as it falls, and then back to gravitational potential energy again as it rises.
On the right, the figure shows an idealized version of the orbit of the yo-yo's center of mass as a elongated ellipse. The motion around this orbit requires a centripetal force, which reaches its maximum strength at the bottom of its orbit where the curvature is very sharp. Blue arrows indicate roughly the inward, centripetal force, which is vertical at the bottom of the orbit. There the reversal in the translational motion is provided by the tension of the fully unwound string. This tension continues in a line from the point of connection of the string through the center of mass of the yo-yo, giving rise to a centripetal force that causes the yo-yo to do a U-turn around the end of the string.
Nonuniform circular motion
As a generalization of the uniform circular motion case, suppose the angular rate of rotation is not constant. The acceleration now has a tangential component, as shown in Figure 5. This case is used to demonstrate a derivation strategy based upon a polar coordinate system.
Let r(t) be a vector that describes the position of a point mass as a function of time. Since we are assuming circular motion, let r(t) = R·ur, where R is a constant (the radius of the circle) and ur is the unit vector pointing from the origin to the point mass. The direction of ur is described by ?, the angle between the x-axis and the unit vector, measured counterclockwise from the x-axis. The other unit vector for polar coordinates, u? is perpendicular to ur and points in the direction of increasing ?. These polar unit vectors can be expressed in terms of Cartesian unit vectors in the x and y directions, denoted i and j respectively:
- ur = cos? i + sin? j
and
- u? = sin? i + cos? j.
We differentiate to find velocity:
-
-
-
where ? is the angular velocity d?/dt.
This result for the velocity matches expectations that the velocity should be directed tangential to the circle, and that the magnitude of the velocity should be ?R. Differentiating again,
and noting that
we find that the acceleration, a is:
Thus, the radial and tangential components of the acceleration are:
and
where |v| = R? is the magnitude of the velocity (the speed).
These equations express mathematically that, in the case of an object that moves along a circular path with a changing speed, the acceleration of the body may be decomposed into a perpendicular component that changes the direction of motion (the centripetal acceleration), and a parallel, or tangential component, that changes the speed.
General planar motion
Polar coordinates
The above results can be derived perhaps more simply in polar coordinates, and at the same time extended to general motion within a plane, as shown next. Polar coordinates in the plane employ a radial unit vector u? and an angular unit vector u?, as shown in Figure 6. A particle at position r is described by:
where the notation ? is used to describe the distance of the path from the origin instead of R to emphasize that this distance is not fixed, but varies with time. The unit vector u? travels with the particle and always points in the same direction as r(t). Unit vector u? also travels with the particle and stays orthogonal to u?. Thus, u? and u? form a local Cartesian coordinate system attached to the particle, and tied to the path traveled by the particle. By moving the unit vectors so their tails coincide, as seen in the circle at the left of Figure 6, it is seen that u? and u? form a right-angled pair with tips on the unit circle that trace back and forth on the perimeter of this circle with the same angle ?(t) as r(t).
When the particle moves, its velocity is
To evaluate the velocity, the derivative of the unit vector u? is needed. Because u? is a unit vector, its magnitude is fixed, and it can change only in direction, that is, its change du? has a component only perpendicular to u?. When the trajectory r(t) rotates an amount d?, u?, which points in the same direction as r(t), also rotates by d?. See Figure 6. Therefore the change in u? is
or
In a similar fashion, the rate of change of u? is found. As with u?, u? is a unit vector and can only rotate without changing size. To remain orthogonal to u? while the trajectory r(t) rotates an amount d?, u?, which is orthogonal to r(t), also rotates by d?. See Figure 6. Therefore, the change du? is orthogonal to u? and proportional to d? (see Figure 6):
Figure 6 shows the sign to be negative: to maintain orthogonality, if du? is positive with d?, then du? must decrease.
Substituting the derivative of u? into the expression for velocity:
To obtain the acceleration, another time differentiation is done:
Substituting the derivatives of u? and u?, the acceleration of the particle is:
-
As a particular example, if the particle moves in a circle of constant radius R, then d?/dt = 0, v = v?, and:
-
These results agree with those above for nonuniform circular motion. See also the article on non-uniform circular motion. If this acceleration is multiplied by the particle mass, the leading term is the centripetal force and the negative of the second term related to angular acceleration is sometimes called the Euler force.
For trajectories other than circular motion, for example, the more general trajectory envisioned in Figure 6, the instantaneous center of rotation and radius of curvature of the trajectory are related only indirectly to the coordinate system defined by u? and u? and to the length |r(t)| = ?. Consequently, in the general case, it is not straightforward to disentangle the centripetal and Euler terms from the above general acceleration equation.
To deal directly with this issue, local coordinates are preferable, as discussed next.
Local coordinates
By local coordinates is meant a set of coordinates that travel with the particle,
and have orientation determined by the path of the particle. Unit vectors are formed as shown in Figure 7, both tangential and normal to the path. This coordinate system sometimes is referred to as intrinsic or path coordinates or nt-coordinates, for normal-tangential, referring to these unit vectors. These coordinates are a very special example of a more general concept of local coordinates from the theory of differential forms.
Distance along the path of the particle is the arc length s, considered to be a known function of time.
A center of curvature is defined at each position s located a distance ? (the radius of curvature) from the curve on a line along the normal un (s). The required distance ?(s) at arc length s is defined in terms of the rate of rotation of the tangent to the curve, which in turn is determined by the path itself. If the orientation of the tangent relative to some starting position is ?(s), then ?(s) is defined by the derivative d?/ds:
The radius of curvature usually is taken as positive (that is, as an absolute value), while the curvature ? is a signed quantity.
A geometric approach to finding the center of curvature and the radius of curvature uses a limiting process leading to the osculating circle. See Figure 7.
Using these coordinates, the motion along the path is viewed as a succession of circular paths of ever-changing center, and at each position s constitutes non-uniform circular motion at that position with radius ?. The local value of the angular rate of rotation then is given by:
with the local speed v given by:
As for the other examples above, because unit vectors cannot change magnitude, their rate of change is always perpendicular to their direction (see the left-hand insert in Figure 7):
Consequently, the velocity and acceleration are:
and using the chain-rule of differentiation:
with the tangential acceleration
In this local coordinate system the acceleration resembles the expression for nonuniform circular motion with the local radius ?(s), and the centripetal acceleration is identified as the second term.
Extension of this approach to three dimensional space curves leads to the Frenet-Serret formulas.
Alternative approach
Looking at Figure 7, one might wonder whether adequate account has been taken of the difference in curvature between ?(s) and ?(s + ds) in computing the arc length as ds = ?(s)d?. Reassurance on this point can be found using a more formal approach outlined below. This approach also makes connection with the article on curvature.
To introduce the unit vectors of the local coordinate system, one approach is to begin in Cartesian coordinates and describe the local coordinates in terms of these Cartesian coordinates. In terms of arc length s let the path be described as:
Then an incremental displacement along the path ds is described by:
where primes are introduced to denote derivatives with respect to s. The magnitude of this displacement is ds, showing that:
(Eq. 1)
This displacement is necessarily tangent to the curve at s, showing that the unit vector tangent to the curve is:
while the outward unit vector normal to the curve is
Orthogonality can be verified by showing the vector dot product is zero. The unit magnitude of these vectors is a consequence of Eq. 1. Using the tangent vector, the angle of the tangent to the curve, say ?, is given by:
and
The radius of curvature is introduced completely formally (without need for geometric interpretation) as:
The derivative of ? can be found from that for sin?:
Now:
in which the denominator is unity. With this formula for the derivative of the sine, the radius of curvature becomes:
where the equivalence of the forms stems from differentiation of Eq. 1:
With these results, the acceleration can be found:
-
as can be verified by taking the dot product with the unit vectors ut(s) and un(s). This result for acceleration is the same as that for circular motion based on the radius ?. Using this coordinate system in the inertial frame, it is easy to identify the force normal to the trajectory as the centripetal force and that parallel to the trajectory as the tangential force. From a qualitative standpoint, the path can be approximated by an arc of a circle for a limited time, and for the limited time a particular radius of curvature applies, the centrifugal and Euler forces can be analyzed on the basis of circular motion with that radius.
This result for acceleration agrees with that found earlier. However, in this approach the question of the change in radius of curvature with s is handled completely formally, consistent with a geometric interpretation, but not relying upon it, thereby avoiding any questions Figure 7 might suggest about neglecting the variation in ?.
Example: circular motion
To illustrate the above formulas, let x, y be given as:
Then:
which can be recognized as a circular path around the origin with radius a. The position s = 0 corresponds to [a, 0], or 3 o'clock. To use the above formalism the derivatives are needed:
With these results one can verify that:
The unit vectors also can be found:
which serve to show that s = 0 is located at position [?, 0] and s = ?p/2 at [0, ?], which agrees with the original expressions for x and y. In other words, s is measured counterclockwise around the circle from 3 o'clock. Also, the derivatives of these vectors can be found:
To obtain velocity and acceleration, a time-dependence for s is necessary. For counterclockwise motion at variable speed v(t):
where v(t) is the speed and t is time, and s(t = 0) = 0. Then:
where it already is established that a = ?. This acceleration is the standard result for non-uniform circular motion.
Further reading
- vs. , from an online Regents Exam physics tutorial by the Oswego City School District
External links
See also
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