An auto-rickshaw takes a sharp left turn at a crossing in Chennai. The three passengers on the back seat slide to the right, and the one on the right is pressed against the side bar. A bag on the seat slides to the right too. To the passengers, it feels as if a force suddenly pushed everything outwards, away from the turn. A traffic policeman at the crossing sees nothing of the kind.
One block, two observers
Put a small block on the floor of a large horizontal platform, a distance \(r\) from its axis. Turn the platform at a steady angular speed \(\omega\). The block does not slide; it turns with the floor. Describe it first from the ground, which is an inertial frame. The block moves on a circle, so it accelerates at \(\omega^2 r\) towards the axis. The only horizontal force on it is static friction from the floor, and friction points towards the axis. Newton's second law holds: \(f = m\omega^2 r\).
Now describe the same block from the platform, a rotating frame. An observer standing on the platform turns with it, and to that observer the block is at rest. But friction has not gone away. Friction is a real force between two real surfaces, and it still pulls the block towards the axis. So this observer sees an unbalanced force on a body at rest. Newton's laws fail. A rotating frame is a non-inertial frame, like the accelerating bus of the chapter on accelerated objects.
The outward term
The observer on the platform wants to keep using Newton's laws. The chapter on accelerated objects showed how. In a frame whose acceleration is \(\vec a_0\), add a pseudo force \(-m\vec a_0\) to every body. Each point of the platform at a distance \(r\) from the axis accelerates at \(\omega^2 r\) towards the axis. So the pseudo force on a body there is \(m\omega^2 r\), pointing away from the axis. It is called the centrifugal force, "fleeing the centre".
In a frame rotating at a steady angular speed \(\omega\) about an axis, the centrifugal force on a body of mass \(m\) at a distance \(r\) from the axis is \(m\omega^2 r\), directed away from the axis. It is a pseudo force. It exists only in the rotating frame, and it has no agent and no reaction.
With the centrifugal force added, the block on the platform is in balance: friction \(m\omega^2 r\) inwards, the centrifugal force \(m\omega^2 r\) outwards. The observer on the platform gets the right answer, \(f = m\omega^2 r\), from a balance of forces. The observer on the ground got the same answer from the second law. The two descriptions are one equation, with the term \(m\omega^2 r\) written on different sides.
The centrifugal force is not the reaction to the centripetal force. A third-law pair acts on two different bodies. The centripetal force is the net real force on the block; its reaction is the block's friction on the platform. The centrifugal force acts on the block itself, and it has no agent at all. Also, never use both in one frame. From the ground there is no centrifugal force. From the platform the block has no acceleration, so there is no \(mv^2/r\) on the right-hand side.
A student says, "The centrifugal force is the reaction to the centripetal force, by Newton's third law." What is wrong?
Back in the auto-rickshaw
Seen from the road, the auto turns left and the passenger's body tends to go straight on, by the first law. The seat's friction on the passenger, and the side bar's push if the passenger reaches it, must supply the inward force \(mv^2/R\). When the seat's friction is not enough, the auto turns left under the passenger, who slides towards the right side. Nothing pushed the passenger outwards.
Seen from inside the auto, the passenger feels the centrifugal force \(mv^2/R\) pushing outwards. Friction from the seat balances it, up to its limit. When the centrifugal force passes that limit, the passenger slides outwards relative to the auto. For a \(60\,\text{kg}\) passenger in a turn of radius \(10\,\text{m}\) at \(5\,\text{m/s}\), the centrifugal force is \(60 \times 25/10 = 150\,\text{N}\). A seat with \(\mu_s = 0.3\) can hold \(180\,\text{N}\). At \(6\,\text{m/s}\) the centrifugal force is \(216\,\text{N}\), and the passenger slides.
The auto turns left. A passenger describes the slide from inside the auto, a rotating frame. Which forces does the passenger list along the horizontal?
When to use it
The centrifugal force is a tool, and it has one rule. Use it only when you describe the motion from the rotating frame. Then a body that turns with the frame is at rest, and every problem of this chapter becomes a balance. The rider in the rotor: the centrifugal force \(m\omega^2 R\) presses the rider into the wall, the wall's \(N\) balances it, and friction balances the weight. The conical pendulum: \(T\), \(mg\) and \(m\omega^2 r\) balance, which again gives \(\tan\theta = \omega^2 r/g\).
A potter's wheel turns about a vertical axis. A small lump of clay sits on it, \(20\,\text{cm}\) from the axis, with \(\mu_s = 0.5\). Take \(g = 10\,\text{m/s}^2\). Working in the frame of the wheel, find the greatest angular speed at which the clay stays put.
In the frame of the wheel the clay is at rest. Horizontally, the centrifugal force \(m\omega^2 r\) points outwards, and static friction balances it, up to \(\mu_s mg\).
The clay is about to slide when \(m\omega^2 r = \mu_s mg\), so \(\omega_{\max} = \sqrt{\mu_s g/r} = \sqrt{0.5 \times 10/0.2} = \sqrt{25} = 5\,\text{rad/s}\), a little under one turn per second. From the ground the same equation reads "friction is the centripetal force".
A coin lies on a turntable \(10\,\text{cm}\) from its axis, and \(\mu_s = 0.4\) between the coin and the turntable. Take \(g = 10\,\text{m/s}^2\). Above what angular speed, in rad/s, does the coin slide?
In which of these descriptions is it correct to include a centrifugal force?
The spin dryer, from both frames
The drum of a washing machine spins at more than ten turns a second, and its wall is full of small holes. Seen from the ground, the wall pushes each wet cloth inwards and keeps it on its circle. A drop of water lying over a hole has nothing to push it inwards. So, like the stone of Lesson 1 whose string was cut, the drop leaves along the tangent, through the hole.
Seen from the drum, the same drop is at rest until it leaves. The centrifugal force \(m\omega^2 r\) pulls it outwards. The cloth's grip holds it back only up to a limit, and at high spin the centrifugal force wins. The drop moves outwards through the hole. Both stories are right; the second is simply told from the frame of the drum.
Seen from the ground, why does a water drop leave the spinning drum of a dryer through a hole?
The simple rule \(m\omega^2 r\) is complete only for bodies at rest in the rotating frame. A body that moves in the rotating frame needs a second pseudo force, the Coriolis force, \(2m\omega v_{\text{rel}}\), perpendicular to its velocity relative to the frame. On the turning Earth it bends winds into the great spirals of cyclones over the Bay of Bengal. Class 11 problems use only bodies at rest in the rotating frame, so the centrifugal force alone is enough.
The Earth turns once a day, so a scale on the ground turns with it. At the equator a person is \(6400\,\text{km}\) from the axis, and \(\omega = 2\pi/86\,164\,\text{s} \approx 7.3 \times 10^{-5}\,\text{rad/s}\). The centrifugal force per kilogram, \(\omega^2 R\), is about \(0.034\,\text{m/s}^2\), a third of one per cent of \(g = 9.8\,\text{m/s}^2\). A \(60\,\text{kg}\) person weighed at the equator reads about \(0.2\,\text{kg}\) less than the same person at a pole, from the Earth's spin alone.
The Dutch physicist Christiaan Huygens, who built the first accurate pendulum clock, studied a ball whirled on a string. In a manuscript of 1659, De vi centrifuga, he found that the pull of the string grows as the square of the speed. We now write this rule as \(mv^2/r\). He described it from the point of view of a person turning with the ball, who feels the ball tug outwards. He named that tug the vis centrifuga. Newton later turned the idea round, calling the inward pull the centripetal force. Huygens's treatise was published only in 1703, after his death.
Explain the slide of the passengers in the turning auto-rickshaw twice, once from the road and once from inside the auto. Why are both explanations correct?
From the road, the passengers tend to keep moving straight, by the first law. The auto turns left, so the seat must push them inwards with friction \(mv^2/R\). If friction cannot supply that much, the auto turns under them and they slide to the right; no outward force acts. From inside the auto, the passengers are at rest in a rotating frame, so a centrifugal force \(mv^2/R\) outwards is added. When it exceeds the limit of seat friction, they slide outwards. Both are correct. The centrifugal force is only the term \(mv^2/R\), moved from the acceleration side of the ground-frame equation to the force side.
- Drawing a centrifugal force on a free-body diagram in the ground frame.The ground frame uses real forces only. Add \(m\omega^2 r\) outwards only in the rotating frame.
- Using both a centrifugal force and \(mv^2/r\) in one equation.In the rotating frame the body is at rest, so the forces, including the centrifugal force, balance to zero.
- Calling the centrifugal force the reaction to the centripetal force.A third-law pair acts on two different bodies. The centrifugal force has no agent and no reaction.