Note: This section uses vectors, and some way of denoting them on the board and in the notebook must be agreed on by the class. In this lesson plan, all vector quantities will be in boldface.

(23) Accelerated Frames of Reference: Inertial Forces

Index

22.Reference Frames

22a.Starlight Aberration

22b. Relativity

22c. Flight (1)

22d. Flight (2)

23. Inertial Forces

23a. The Centrifugal Force

23b. Loop-the-Loop

24a.The Rotating Earth

24b. Rotating Frames

The Sun

S-1. Sunlight & Earth

S-1A. Weather

S-1B. Global Climate

S-2.Solar Layers

The other case of a moving frame of reference studied here is an accelerated frame of reference.

Roller coaster loop

An earlier section described motion in a circle and introduced the centripetal force which makes such motion possible. That is a force towards the center of rotation, of magnitude

_centripetal

where R is the radius of the circle of rotation and V is the velocity of motion around it.

To anyone who has not studied motions and accelerations, "centripetal" is probably a new word. However, most people are aware of the outwards-pushing centrifugal force, the force which flings you away from the center of rotation in a car speeding around a corner. In a popular carnival ride, that is the force which pushes you against a rotating drum, holding you in place even when you are upside down. What is the connection?

Inertial Forces

As will be seen, the centrifugal force is not a "real" force, in the sense that in any motion calculated "in the frame of the universe" (or in one moving uniformly with respect to the universe) it does not appear at all. In such a frame, if an object moves around a circle, a centripetal force is needed to maintain that motion--otherwise it flies off at a tangent, with constant velocity along a straight line.

Unfortunately, if you sit (for instance) in a roller coaster car going around a vertical loop, as in the picture above, from an amusement park in Japan, it is a bit difficult to visualize your motion with respect to the fixed Earth. It is much simpler to orient yourself with respect to the car in which you sit.

The roller-coaster car, however, is undergoing various accelerations, and as a rule, when we try to apply the laws of motion inside an accelerated frame of reference, extra forces enter the picture, known as inertial forces. You could call them "fictitious" forces, if you wish, because when the same motion is calculated in the frame of the outside world, they do not appear at all. Inside the accelerated frame of reference, however, they can't be told apart from real forces, and they need real forces to balance them.

The centrifugal force is one such force, but before studying it, we consider a simpler case. You are sitting in a bus, moving in a straight line. Suddenly the driver applies brakes and all passengers feel pushed forward. Why?

Unit Vectors

Forces are vectors, and F=ma is a vector equation. Rather than separate it into components, we introduce here a simple notation which allows vectors to be handled as single entities.

Suppose we work in an (x,y) system of coordinates, with x pointing down and y pointing in the direction of the motion of the bus (if you are used to an (x,y) system with the y-axis pointing up, rotate it clockwise by 90°). Then unit vectors x_u and y_u would then be vectors of unit length, in the x and y directions.

[Note: marking unit vectors with a subscript "u" is a non-standard notation, forced by the limitation of the HTML code used on the world-wide web. The standard notation is a caret on top, as in â, ê or û, but unfortunately, HTML does not allow carets to be placed on most letters. If presenting this material in a classroom, it may be better to use carets.]

So, in (x,y) coordinates, a vector V with components (V_x,V_y) could be written

V = V_xx_u + V_y y_u

The Decelerating Bus

Let us examine the forces on some passenger inside the bus; the x direction is down, y is in the direction in which the bus is moving. Two forces are involved: the weight F₁=mg x_u, pulling the passengers down, and the reaction F₂ of the seat, which does not allow any motion in that direction. As long as the bus moves in a straight line and with a constant speed, these two are the only ones that matter and we get as condition of equilibrium

F₁+ F₂ = 0

What happens when the brakes are applied? Let us look first from the frame of reference of the outside world. If the bus accelerates forward, the acceleration is

a = a y_u When the brake is applied, therefore

a = − a y_u and the forces on a mass inside must obey Newton's law

F₁+ F₂ = − ma y_u

If equilibrium is to be maintained, the forces must change. The weight F₁ is fixed, but to balance horizontal forces, the reaction force F₂ must change. For instance, the passenger may grab the seat in front, pushing her or his body back, with a force equal to what the acceleration added on the right side of the equation.

To see how the situation looks in the frame of the bus, we add +ma y_u to both sides. On the right now, what is added equals what is subtracted, leaving zero, so the equation becomes

F₁+ F₂ + ma y_u = 0

This may be interpreted in the frame of the decelerating bus as follows. For forces to stay in equilibrium, all forces (as before) must add up to zero, but now they must include an inertial force may_u pushing forward, in the direction of y_u. Such an inertial force is only felt in the moving frame. You may call it a "fictitious force" if you wish. But when you need a seatbelt or an airbag to stop it from throwing you through a windshield, it seems real enough!

Taking Off

If you ever take a trip on a jetliner, notice how during take-off you are pushed back in your seat. That is the inertial force acting in the frame of the accelerating airplane. In a decelerating bus, you are pushed forward, but in an accelerating airplane you are pushed back. For a simple experiment, take with you a weight on a string (e.g. a fishing sinker or a stack of screw-nuts tied to the end of the string) and let it hang before take-off, defining the "down" direction. During take-off, the string will slant backwards, by perhaps 5-10 degrees.

All this is rather mild compared to the inertial forces felt by astronauts on a rocket or on the space shuttle during launch. The rocket accelerates at about 2-3g, so the added force felt by the astronauts, in their frame of reference, is 2-3 times their weight.

Next Stop: #23a The Centrifugal Force

Timeline Glossary Back to the Master List

Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .

Updated: 9-22-2004 ; reformatted 25 March 2006 ; later update 14 October 2016