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The hallmark of an inertial reference frame is that any object which is subject to zero net force will travel in a straight line at constant speed. To illustrate this, consider the following: I am standing on a level floor at the origin of an inertial frame \(\mathcal{S}\) and kick a frictionless puck due north across the floor. (a) Write down the \(x\) and \(y\) coordinates of the puck as functions of time as seen from my inertial frame. (Use \(x\) and \(y\) axes pointing east and north respectively.) Now consider two more observers, the first at rest in a frame \(\mathcal{S}^{\prime}\) that travels with constant velocity \(v\) due east relative to \(\mathcal{S},\) the second at rest in a frame \(\mathcal{S}^{\prime \prime}\) that travels with constant acceleration due east relative to \(\mathcal{S}\). (All three frames coincide at the moment when I kick the puck, and \(\mathcal{S}^{\prime \prime}\) is at rest relative to \(\mathcal{S}\) at that same moment.) (b) Find the coordinates \(x^{\prime}, y^{\prime}\) of the puck and describe the puck's path as seen from \(\mathcal{S}^{\prime} .\) (c) Do the same for \(\mathcal{S}^{\prime \prime}\) Which of the frames is inertial?

Short Answer

Expert verified
Frames \( \mathcal{S} \) and \( \mathcal{S}^{\prime} \) are inertial, \( \mathcal{S}^{\prime\prime} \) is not.

Step by step solution

01

Identify Inertial Frame Coordinates

In the inertial frame \( \mathcal{S} \), where no forces act on the object, the puck moves in a straight line at constant speed. Assuming the puck was given an initial velocity \( v_0 \) towards the north (the positive \( y \)-direction), the position in coordinates as a function of time \( t \) are: \[ x(t) = 0 \, \text{(since there is no motion in the x-direction)} \] \[ y(t) = v_0 t \, \text{(constant speed motion)} \]
02

Transition to Observer's Frame with Constant Velocity

For the frame \( \mathcal{S}^{\prime} \), which is moving with velocity \( v \) east (positive x-direction) relative to \( \mathcal{S} \), the transformation will adjust only the x-coordinate, as: \[ x^{\prime}(t) = x(t) - vt = -vt \] \[ y^{\prime}(t) = y(t) = v_0 t \] The puck in frame \( \mathcal{S}^{\prime} \) moves in a straight line to the north, appearing to be moving east due to the frame’s motion.
03

Transition to Observer's Frame with Constant Acceleration

For the frame \( \mathcal{S}^{\prime\prime} \), which is accelerating east with some acceleration \( a \), we adjust the coordinates by factoring in acceleration: \[ x^{\prime\prime}(t) = x(t) - \frac{1}{2}at^2 = -\frac{1}{2}at^2 \] \[ y^{\prime\prime}(t) = y(t) = v_0 t \] The puck's path appears to curve due to the frame's acceleration, even though it moves straight in \( \mathcal{S} \).
04

Identify Inertial and Non-Inertial Frames

Frame \( \mathcal{S} \) is an inertial frame because the puck continues its motion in a straight line without deviation when no external force affects it. Frame \( \mathcal{S}^{\prime} \), traveling at constant velocity relative to \( \mathcal{S} \), also qualifies as inertial. However, frame \( \mathcal{S}^{\prime\prime} \), being under constant acceleration, is not inertial.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coordinate Transformation
Coordinate transformation is the process of converting the description of motion or positions from one reference frame to another.
This is essential in physics to understand how different observers perceive the same event or movement.In the exercise, we look at two additional reference frames, aside from the inertial frame \( \mathcal{S} \): a frame moving at a constant velocity east (\( \mathcal{S}' \)), and a frame accelerating east (\( \mathcal{S}'' \)).
The key steps in coordinate transformation include:
  • Identifying initial positions and velocities: Recognize whether the object is initially at rest or in motion in the original frame.
  • Applying transformations: For velocity transformations, use the relative velocity to adjust coordinates. For acceleration, factor in dragging effects due to the frame's acceleration.
  • Resultant coordinates: These are expressed in terms of time (\( t \)) and will differ based on the observer's motion.
For a moving frame like \( \mathcal{S}' \), the x-coordinate adjustments account for perceived motion due to relative velocity. Similarly, for an accelerating frame like \( \mathcal{S}'' \), the change in coordinates involves terms that depend on acceleration \( a \).
Velocity
Velocity is the rate of change of position of an object, usually described in terms of speed and direction. In an inertial frame, the puck travels north at a constant speed. Therefore, the velocity can be defined simply by its northern route and constant speed \( v_0 \).
When viewed from temporal frames, the velocity observed changes due to the movement of the frame itself. In frame \( \mathcal{S}' \), moving east with velocity \( v \), the puck's northward progress remains unchanged, but an additional perceived velocity component appears due east. This reflects the relative motion of the frame.
  • In \( \mathcal{S}' \), while the true velocity north is still \( v_0 \), the puck appears to drift eastwards at velocity \( v \).
  • Velocity adjustments between frames are linear in nature for frames like \( \mathcal{S}' \) when compared with original reference frame \( \mathcal{S} \).
Acceleration
Acceleration is the change in velocity over time, and it is a vector quantity involving changes in both speed and direction. In the context of this exercise:
  • In the inertial frame \( \mathcal{S} \), the puck has no net force acting on it and hence no acceleration beyond the initial force applied. It continues in a straight line at constant speed.
  • In the frame \( \mathcal{S}'' \), which accelerates east, the puck appears to curve due to the acceleration of the frame. This is because the frame itself is gaining velocity in the east direction over time.
The transformation in this non-inertial, accelerating frame makes it seem as if there is an apparent force affecting the puck's motion. In non-inertial frames, such as \( \mathcal{S}'' \), pseudo-forces appear due to the acceleration of the frame itself, giving the illusion of curved paths even in the absence of external forces acting on the object.
Non-Inertial Frames
Non-inertial frames are those which are accelerating or rotating with respect to an inertial frame. They cannot be considered as at rest or moving at a constant velocity.
The exercise shows how the puck is perceived differently in such frames, particularly \( \mathcal{S}'' \). In this frame, which accelerates east, the puck's seemed path is curved. Despite the puck not having actual forces acting on it other than the initial impulse:
  • Non-inertial frames introduce "pseudo" or "fictitious" forces. This makes straight motion observed as curved.
  • Understanding non-inertial frames is crucial in contexts where acceleration or rotation cannot be ignored, such as inside a moving vehicle.
  • Analyzing physics problems in such frames requires careful consideration of additional apparent forces resulting from the frame's acceleration or rotation.
Thus, while inertial frames tend to simplify motion analysis, non-inertial frames require adjustments for these effects.

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Most popular questions from this chapter

Conservation laws, such as conservation of momentum, often give a surprising amount of information about the possible outcome of an experiment. Here is perhaps the simplest example: Two objects of masses \(m_{1}\) and \(m_{2}\) are subject to no external forces. Object 1 is traveling with velocity \(\mathbf{v}\) when it collides with the stationary object \(2 .\) The two objects stick together and move off with common velocity \(\mathbf{v}^{\prime}\). Use conservation of momentum to find \(\mathbf{v}^{\prime}\) in terms of \(\mathbf{v}, m_{1},\) and \(m_{2}\)

By evaluating their dot product, find the values of the scalar \(s\) for which the two vectors \(\mathbf{b}=\hat{\mathbf{x}}+s \hat{\mathbf{y}}\) and \(\mathbf{c}=\hat{\mathbf{x}}-s \hat{\mathbf{y}}\) are orthogonal. (Remember that two vectors are orthogonal if and only if their dot product is zero.) Explain your answers with a sketch.

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Prove that if \(\mathbf{v}(t)\) is any vector that depends on time (for example the velocity of a moving particle) but which has constant magnitude, then \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t) .\) Prove the converse that if \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t),\) then \(|\mathbf{v}(t)|\) is constant. [Hint: Consider the derivative of \(\mathbf{v}^{2}\).] This is a very handy result. It explains why, in two- dimensional polars, \(d \hat{\mathbf{r}} / d t\) has to be in the direction of \(\hat{\boldsymbol{\phi}}\) and vice versa. It also shows that the speed of a charged particle in a magnetic field is constant, since the acceleration is perpendicular to the velocity.

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