IIT JEE , AIEEE Forum - Physics , Mechanics

ayush

what is a psuedo force?what are different types of it & how it is dealt with newton's laws in mechanics

edison

Fictitious force/pseudo force

A fictitious force is an apparent force that acts on all masses in a non-inertial frame of reference, e.g., a rotating reference frame. The force

F

does not arise from any physical interaction, but rather from the acceleration

a

of the non-inertial reference frame itself. Due to Newton's second law

F = m a

, fictitious forces are always proportional to the mass

m

being acted upon.

Role as calculational tool

It is sometimes convenient to solve physical problems in a non-inertial reference frame. In such cases, it is necessary to introduce fictitious forces to account for the acceleration of the reference frame. For example, the surface of the Earth is a rotating reference frame. To solve classical mechanics problems exactly in an Earth-bound reference frame, two fictitious forces must be introduced, the Coriolis force and the centrifugal force (described below), of which the Coriolis force is dominant on Earth. Both of these fictitious forces are weak compared to most typical forces in everyday life, but they can be detected under careful conditions. For example, Léon Foucault was able to show the Coriolis force that results from the Earth's rotation using the Foucault pendulum. If the Earth were to rotate a thousand-fold faster (making each day only ~86 seconds long).

Detection of non-inertial reference frame

If a person living inside a closed box that is rotating (or otherwise accelerating) can detect their own rotation/acceleration. Careful observers within the box can detect that they are in a non-inertial reference frame from the fictitious forces that arise from the acceleration of the box. They can even map out the magnitude and direction of the acceleration at every point within the box. For example, a Foucault pendulum in a science museum will precess in exactly the same manner, regardless of whether the museum has walls or not.

For comparison, observers living inside a closed box that is moving uniformly (i.e., without acceleration) cannot detect their own motion. That is the essential physics of Newton's first two laws of motion.

Newtonian examples of fictitious forces

Acceleration in a straight line

When a car accelerates hard, the common human response is to feel "pushed back into the seat." In an inertial frame of reference attached to the road, there is no physical force moving the rider backward. However, in the rider's non-inertial reference frame attached to the accelerating car, there is a backward fictitious force. We mention two possible ways of analyzing the problem:

From the viewpoint of an inertial reference frame with constant velocity matching the initial motion of the car, the car is accelerating. In order for the passenger to stay inside the car, a force must be exerted on him. This force is exerted by the seat, which has started to move forward with the car and compressed against the passenger until it transmits the full force to keep the passenger inside. Thus the passenger is accelerating in this frame, due to the unbalanced force of the seat.
From the point of view of the interior of the car, an accelerating reference frame, there is a fictitious force pushing the passenger backwards, with magnitude equal to the mass of the passenger times the acceleration of the car. This force pushes the passenger back into the seat, until the seat compresses and provides an equal and opposite force. Thereafter, the passenger is stationary in this frame, because the fictitious force and the (real) force of the seat are balanced.

This serves as an illustration of the manner in which fictitious forces arise from switching to a non-inertial reference frame. Calculations of physical quantities made in any frame give the same answers, but in some cases calculations are easier to make in a non-inertial frame. (In this simple example, the calculations are equally easy in either of the two frames described.)

Circular motion

A similar effect occurs in circular motion, circular for the standpoint of an inertial frame of reference attached to the road, with the fictitious force called the centrifugal force, fictitious when seen from a non-inertial frame of reference. If a car is moving at constant speed around a circular section of road, the occupants will feel pushed outside, away from the center of the turn. Again the situation can be viewed from inertial or non-inertial frames:

From the viewpoint of an inertial reference frame stationary with respect to the road, the car is accelerating toward the center of the circle. This is called centripetal acceleration and requires a centripetal force to maintain the motion. This force is maintained by the friction of the wheels on the road. The car is accelerating, due to the unbalanced force, which causes it to move in a circle.
From the viewpoint of a rotating frame, moving with the car, there is a fictitious centrifugal force that tends to push the car toward the outside of the road (and the occupants toward the outside of the car). The centrifugal force is balanced by the acceleration of the tires inward, making the car stationary in this non-inertial frame.

To consider another example, taking as our reference frame the surface of the rotating earth, centrifugal force reduces the apparent force of gravity by about one part in a thousand, depending on latitude. This is zero at the poles, maximum at the equator.

Another fictitious force that arises in the case of circular motion is the Coriolis force, which is ordinarily visible only in very large-scale motion like the projectile motion of long-range guns or the circulation of the earth's atmosphere. Neglecting air resistance, an object dropped from a 50 m high tower at the equator will fall 7.7 mm eastward of the spot below where it was dropped because of the Coriolis force.

Both the centrifugal and the Coriolis force are needed to explain the motion of distant objects relative to rotating reference frames. Consider a distant star observed from a rotating spacecraft. In the reference frame co-rotating with the spacecraft the distant star appears to rotate around the spacecraft. The apparent motion of the star requires a fictitious centripetal force acting on the star. Just like in the example of the car in circular motion above, the centrifugal force acting on the star has the same magnitude as the centripetal force, but is directed in the opposite direction. In this case the Coriolis force has twice the magnitude of the centrifugal force and is directed oppositely to the centrifugal force.

Fictitious forces and work

Fictitious forces can be considered to do work, provided that they move an object on a trajectory that changes its energy from potential to kinetic. For example, consider a person in a rotating chair holding a weight in his outstretched arm. If he pulls his arm inward, from the perspective of his rotating reference frame he has done work against centrifugal force. If he now lets go of the weight, from his perspective it spontaneously flies outward, because centrifugal force has done work on the object, converting its potential energy into kinetic. From an inertial viewpoint, of course, the object flies away from him because it is suddenly allowed to move in a straight line. This illustrates that the work done, like the total potential and kinetic energy of an object, can be different in a non-inertial frame than an inertial one.

Gravity as a fictitious force

All fictitious forces are proportional to the mass of the object upon which they act, which is also true for gravity. This led Albert Einstein to wonder whether gravity was a fictitious force as well. He noted that a freefalling observer in a closed box would not be able to detect the force of gravity; hence, free falling reference frames are equivalent to an inertial reference frame (the equivalence principle).

Mathematical derivation of fictitious forces

General derivation

Consider a particle with mass m and position vector x_a(t) in a particular inertial frame A. Consider a non-inertial frame B whose position relative to the inertial one is given by X(t). Since B is non-inertial, we must have that d²X/dt² (the acceleration of frame B with respect to frame A) is non-zero. Let the position of the particle in frame B be x_b(t). Then we have

$old{x}_a(t) = old{x}_b(t) + old{X}(t)$

Taking two time derivatives, this gives

$rac{d^2old{x}_{a}}{dt^2} = rac{d^2old{x}_{b}}{dt^2} + rac{d^2old{X}}{dt^2}$

Now consider the forces in the problem. By Newton's Second Law, F = ma. The true force is of course the one in frame A (the inertial one), so

$old{F}_{mbox{true}} = m rac{d^2old{x}_{a}}{dt^2}$

However, suppose we are working to solve a problem in frame B. It may be useful to consider the apparent force in this frame, which is given by

$old{F}_{mbox{apparent}} = m rac{d^2old{x}_{b}}{dt^2} = m rac{d^2old{x}_{a}}{dt^2} - m rac{d^2old{X}}{dt^2} = old{F}_{mbox{true}} - m rac{d^2old{X}}{dt^2}$

Now we define

$old{F}_{mbox{fictitious}} = - m rac{d^2old{X}}{dt^2}$

giving finally:

$old{F}_{mbox{apparent}} = old{F}_{mbox{true}} + old{F}_{mbox{fictitious}}$

Thus we can solve problems in frame B by assuming that Newton's Second Law holds (with respect to quantities in that frame) and treating F_fictitious as an additional force.

Rotating coordinate systems

A common situation in which noninertial reference frames are useful is when the reference frame is rotating. Since such rotational motion is non-inertial, due to the acceleration present in any rotational motion, a fictitious force can always be invoked by using a rotational frame of reference. Despite this complication, the use of fictitious forces often simplifies the calculations involved.

The relationship between acceleration in an inertial frame, and that in a coordinate frame rotating with angular velocity $oldsymbolomega$ can be expressed as

$mathbf{a}_{mbox{in}}= left( rac{dmathbf{v}_{mbox{in}}}{dt} ight)_{mbox{in}} =left( rac{dmathbf{v}_{mbox{in}}}{dt} ight)_{mbox{rot}} + oldsymbolomega imes mathbf{v}_{mbox{in}}$

where we have used the relationship for the time derivative of a vector in rotating coordinates

$left( rac{dmathbf{B}}{dt} ight)_{mbox{in}} = left( rac{dmathbf{B}}{dt} ight)_{mbox{rot}} + oldsymbolomega imes mathbf{B}$ , for any vector $mathbf{B}$

Since $mathbf{v}_{mbox{in}} = mathbf{v}_{mbox{rot}}+ oldsymbolomega imes mathbf{r}\$ , the acceleration becomes

$mathbf{a}_{mbox{in}} = left( rac{d ( mathbf{v}_{mbox{rot}} + oldsymbolomega imes mathbf{r})}{dt} ight)_{mbox{rot}} + oldsymbolomega imes mathbf{v}_{mbox{rot}} + oldsymbolomega imes (oldsymbolomega imes mathbf{r} )$

or, equivalently,

$mathbf{a}_{mbox{in}} = mathbf{a}_{mbox{rot}} + rac{d oldsymbolomega}{dt} imes mathbf{r} + 2 oldsymbolomega imes mathbf{v}_{mbox{rot}} + oldsymbolomega imes (oldsymbolomega imes mathbf{r} )$

The acceleration in the rotating frame equals

$mathbf{a}_{mbox{rot}} = mathbf{a}_{mbox{in}} - 2 oldsymbolomega imes mathbf{v}_{mbox{rot}} - oldsymbolomega imes (oldsymbolomega imes mathbf{r} ) - rac{d oldsymbolomega}{dt} imes mathbf{r}$

Since the force in the rotating frame is $mathbf{F}_{mbox{rot}} = m mathbf{a}_{mbox{rot}}\$ and, by definition, $mathbf{F}_{mbox{rot}} = mathbf{F}_{mbox{in}} + mathbf{F}_{mbox{fict}}\$ , the fictitious force equals

$mathbf{F}_{mbox{fict}} = - 2 m oldsymbolomega imes mathbf{v}_{mbox{rot}} - m oldsymbolomega imes (oldsymbolomega imes mathbf{r} ) - m rac{d oldsymbolomega }{dt} imes mathbf{r}$

Here, the first term is the Coriolis force, the second term is the centrifugal force, and the third term is the Euler force. When the rate of rotation doesn't change, as is typically the case for a planet, the Euler force is zero.

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