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By learning Physics Topics, we can gain a deeper appreciation for the natural world and our place in it.
Centrifugal Force – A Pseudo Force
Let us consider a merry-go-round, capable of revolving in a horizontal plane, to be at rest. A person is sitting in that merry-go-round and has a stone in his hand. When the merry-go-round begins to revolve with uniform angular velocity, the stone also begins revolving in a circular path along with the person [Fig.] The following two observers will describe the motion of the stone in two different ways.
Observer standing on the surface of the earth at rest: Since the stone is revolving in a circular path, according to this observer, a centripetal force is acting on the stone.
Due to application of centripetal force, a body can move in a circular path. If the person sitting in the merry-go-round releases the stone from his hand at the point A, then the observer will see the stone flying off along the tangent (AB) of the circular path. Due to the inertia of the stone, it seems to move along the path AB according to the observer.
Observer sitting in the merry-go-round: This observer is moving with the same angular velocity as that of the stone, and hence, he observes the stone to be always at the same place. This means that the stone seems to be at rest to this observer. If the observer releases the stone from his hand at the point A, then after sometime when the stone reaches the point B, the observer revolving along with the merry-go-round will reach the point C. So, to him the motion of the stone will be along the straight line CB, away from the centre. The stone seems to be moving radially outwards under the influence of some force. This is called the centrifugal force.
If the person sitting in the whirling merry-go-round holds the stone tightly in his hands, then the stone cannot fly off. Here, the person exerts a force on the stone and it seems to him that this real force (which is the real centripetal force to the observer standing on the surface of the earth at rest) and the centrifugal force keep the stone in equilibrium. Hence, these two forces are equal but opposite in direction.
We know that, Newton’s laws of motions are not valid in the non-inertial frames. If the frame translates with respect to an inertial frame with an acceleration \(\vec{a}^{\prime}\), an apparent force -m\(\vec{a}^{\prime}\) acts on a particle of mass m ,whose motion is described using a non-inertial frame of reference or rotating frame of reference. This force is called pseudo force. Once this pseudo force is included, one can use Newton’s laws in their usual form. This is not a real force. This force only exists in the non-inertial frame of reference or rotating frame of reference.
If a frame of reference rotates at a constant angular velocity ω with respect to an inertial frame and we analyse the dynamics of particle of mass m kept at a distance r from the axis of rotation, we have to assume that a force mω2r acts radially outward on the particle. Only then we can apply Newton’s laws of motion in the rotating frame. This radially outward force is the centrifugal force which is an example of pseudo force. It should be mentioned that centrifugal force acts on a particle because we describe the particle from a rotating frame which is non-inertial and still we use Newton’s laws. So,
centrifugal force = ma’
[where a’ is the centripetal acceleration]
= \(\frac{m v^2}{r}\) = mω2r
This force can be defined in the following way:
Definition: When a body rotates with an angular velocity along a circular path and an observer rotates with the same angular velocity with the body then to that observer, a force equal but opposite to the centripetal force appears to be acting on the body. This force is called the centrifugal force.
When a bus, full of passengers, turns at a bend on the road, the passengers feel a push opposite in direction to that of the bend and they lean towards that side. A person standing on the road, i.e., in an inertial frame of reference, explains this occurrence as an effect of inertia. While taking a turn at the bend, the bus behaves as a rotating frame of reference and the passengers belonging to that frame of reference it appears to that a force pushes them away from the centre of the circular path. This force is nothing but centrifugal force.
A Few Examples of Centrifugal Force
The existence of centrifugal force in a rotational motion is assumed in various aspects of our daily lives.
i) Centrifuge: It is usually a container capable of rotating about an axle with a high angular velocity Particles suspended in a liquid can be separated with this instrument. Generally the density of a liquid and that of the particles suspended in that liquid are different. From the expression mω2r of centrifugal force, it can be said that greater the mass (m) of a body, greater the centrifugal force acting on it if ω and r are kept constant. As a result, the heavier particles move farther from the axle than the lighter particles. This instrument is used for separating cream from milk blood corpuscles from blood, etc. When separating cream from milk, cream particles, being lighter than milk, gather round the axle rod and skimmed milk gets separated.
Antonin Prandti invented the first dairy centrifuge.
ii) A drying machine used for drying wet clothes is another kind of centrifuge. The wet clothes are placed in a container consisting of a large number of perforations. This container is rotated with a high angular velocity. Due to centrifugal force, the water droplets present in the wet clothes are driven off, and the clothes are gradually dried up.
iii) Loss of weight of a body due to the earth’s diurnal motion: The earth rotates about its own axis in 24 hours. This is known as the diurnal motion or daily rotation of the earth. Due to this rotational motion of the earth about its own axis, every object on the earth’s surface rotates with the same angular velocity
Let the latitude of a point A on the earth’s surface be θ and a body of mass m be situated at that point [Fig.]. The weight of the body mg acts along the line AO towards the centre of the earth. Again, due to diurnal motion of the earth, the body revolves around the axis of the earth in a circular path of radius r with angular velocity ω and feels an outward centrifugal force mω2r.
The component of this centrifugal force in the direction AC is mω2rcosθ. As this component acts in the opposite direction of the weight mg, there is an effective loss of weight. It should be noted that if the earth was at rest, then no centrifugal force would act, and hence, there would have been no apparent loss of weight.
So, the weight of the body situated at the point A,
W= mg – mω2rcosθ = m(g – ω2Rcos2θ)
[R is the radius of the earth and r = Rcosθ]
At the equatorial region, θ = 0
∴ W = m(g – w2R)
So, the decrease in weight of a body is maximum at the equatorial region due to the diurnal motion of the earth.
iv) Reason for flattening of the earth at the poles: The shape of the earth is not a perfect sphere, but an oblate spheroid, i.e., flattened slightly at the poles and bulged at the equator. This is caused by the daily rotation of the earth about its axis and the generation of corresponding centrifugal force.
At the time of formation of the earth, its temperature was very high and it was mainly made up of fused and gaseous matter. Since the magnitude of the centrifugal force acting outwards is maximum at the equatorial region, the fused and gaseous particles in this region had the tendency to move away from the axis. On the other hand, as the magnitude of the centrifugal force at the poles is zero, the particles at the poles did not tend to move away from the axis.
From the beginning, the mutual force of cohesion between the material particles on the earth was large. As a result, the material particles at the equatorial region bulged out due to centrifugal force, whereas the material particles at the polar regions were pulled inwards due to the force of cohesion. Later, when the earth’s crust hardened, this specific irregularity in shape became a permanent feature. For this reason, the earth is slightly flattened at the poles and bulged at the equator.
Numerical Examples
Example 1.
A tube of length L is filled completely with an incompressible liquid of mass M and its two open ends are closed. The tube is then rotated with an angular velocity ω with one end of it as the centre. What will be the force exerted by the liquid on the other end?
Solution:
Mass of the liquid = M. The centre of mass for the whole liquid is situated at the mid-point of the tube. Hence, the radius of the circular path along which the centre of mass rotates is, r = \(\frac{L}{2}\)
So, the centrifugal force acting on the centre of mass
= Mω2r = Mω2\(\frac{L}{2}\)
This is the centrifugal force acting on the liquid in the tube. This is the force that acts on the other end of the tube.
Example 2.
A hemispherical bowl of radius 0.1 m is rotated about a vertical axis passing through the centre of the bowl with an angular velocity ω. A particle of mass m = 10-2 kg placed inside the bowl also rotates with the bowl. If the height of the position of the particle from the bottom of the bowl is h, find the relation between h and ω.
Solution:
The particle is at a height h from the bottom of the bowl. With the rotation of the bowl, the particle also rotates about the vertical axis along a circular path of radius r [Fig.].
In this situation, the weight mg of the particle of mass m, the centrifugal force mω2r and the normal force (R) on the particle by the surface of the bowl keep the particle in equilibrium.
So, Rsinθ = mω2r and Rcosθ = mg
∴ tanθ = \(\frac{\omega^2 r}{g}\)
If the radius of the bowl is a, then from Fig., we get
tanθ = \(\frac{r}{a-h}\) ∴ \(\frac{r}{a-h}\) = \(\frac{\omega^2 r}{g}\) or, a – h = \(\frac{g}{\omega^2}\)
or, h = a – \(\frac{g}{\omega^2}\) = 0.1 – \(\frac{9.8}{\omega^2}\) = 0.1(1 – \(\frac{9.8}{\omega^2}\))m
Example 3.
The radius of the earth is 6400 km. What will be the value of the centrifugal acceleration at the equatorial region due to the earth’s diurnal motion?
Solution:
If the radius of the earth is R and its angular velocity is ω, then at the equatorial region,
centrifugal acceleration
= ω2R = \(\left(\frac{2 \pi}{T}\right)^2\) × R = \(\left(\frac{2 \pi}{24 \times 60 \times 60}\right)^2\) × 6400 × 103
= 0.0338 m ᐧ s-2.