PHYS 101 Lecture 13: Chapter 3-3

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22 Mar 2017

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A feather and a diecast car are dropped at the same time. Which one, taking into account air resistance, hits the ground first: the feather, the diecast car, they both hit at the same time, the feather rises instead, like a helium balloon. A feather and a diecast car are dropped at the same time in a vacuum. Which hits the ground first: the feather, the diecast car, they both hit at the same time, the feather rises instead, it depends on what model of car the diecast car is. To a place where earth is just one planet in a system of planets, rotating around a star, following the same laws of gravitation as all other stars and planets. Where everything is made out of atoms, and where most of the space occupied by each atom is just empty space. And where motion is controlled by the principle of inertia.

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How fast can a racecar travel through a circular tum without skidding and hitting the wall? The answer could depend on several factors:

The weight of the car;
The friction between the tires and the road;
The radius of the circle;
The “steepness” of the turn.

In this project, we investigate this question for NASCAR racecars at the Bristol Motor Speedway in Tennessee. Before considering this track in particular, we use vector functions to develop the mathematics and physics necessary for answering questions such as this.

A car of mass m moves with constant angular speed to around a circular curve of radius R (Figure 3.20). The curve is banked at an angle . If the height of the car off the ground is h, then the position of the car at time t is given by the function

As the car moves around the curve, three forces act on it: gravity, the force exerted by the road (this force is perpendicular to the ground), and the friction force (Figure 3.21). Because describing the frictional force generated by the tires and the road is complex, we use a standard approximation for the frictional force. Assume that for some positive constant . The constant is called the coefficient of friction.

Let denote the maximum speed the car can attain through the curve without skidding. In other words, is the fastest speed at which the car can navigate the turn. When the car is navigate at this speed, the magnitude of the centripetal force is

The next three questions deal with developing a formula that relates the speed to the banking angle .

Now that we have a formula relating the maximum speed of the car and the banking angle, we are in a position to answer the questions like the one posed at the beginning of the project.

The Bristol Motor Speedway is a NASCAR short track in Bristol, Tennessee. The track has the approximate shape shown in Figure 3.22. Each end of the track is approximately semicircular, so when cars make turns they are traveling along an approximately circular curve. If a car takes the inside track and speeds along the bottom of turn 1, the car travels along a semicircle of radius approximately with a banking angle of . If the car decides to take the outside track and speeds along the top of turn 1, then the car travels along a semicircle with a banking angle of . (The track has variable angle banking.)

The coefficient of friction for a normal fire in Elly conditions is approximately . Therefore, we assume the coefficient for a NASCAR tire in dry conditions is approximately .

Before answering the following questions, note that it is easier to do computations in terms of feet and seconds, and then convert the answers to miles per hour as a final step.

11. Suppose the measured speed of a car going along the outside edge of the turn is . Estimate the coefficient of friction for the car’s tires.

goldpug777

Huygens' principle (first described by the Dutch scientist Christiaan Huygens in 1678) uses geometry to determine the shape of a wavefront at a time t, given some initial wavefront at an earlier time. This can be constructed by imagining that the initial wavefront is a source of wavelets that propagate from each point at the speed of light. From this principle, one can determine the angles of reflection and refraction by considering the speed of light at each point on the wavefront. In this problem, we will explore some of these concepts.

Part A

First, let's look at a plane wave that is incident on a flat piece of material with an index of refraction n. Part of the light is transmitted through the material and part of it is reflected. For the moment, let's just look at the part that is transmitted. At time t=0, the wavefront is a distance d away from the surface of the material. At time t=t₁, the wavefront is at the position of the material interface. Use the Huygens' principle to determine how far into the material (d') the wavefront has propagated by time t=2t₁.

Express your answer in terms of the variables n, t₁, and the speed of light in a vacuum c.

Note that there would be no change in the direction of the wavefront at any point because the wavefront encountered the material interface at the same time at all points. Thus, all of the individual wavefronts all propagated at the same speed, thereby maintaining the flat wavefront.

Now, instead of having a flat wavefront propagating normal to the material interface, we have a flat wavefront propagating toward the material at an angle of 55° to the axis perpendicular to the material interface. In this part, we will look at the relative positions of a few points—A, B, and C—on the wavefront to illustrate Huygens' principle. (Intro 1 figure) Point C touches the vacuum/material interface at time t=0 whereas point B is a distance d and point A is a distance 2d away from the interface.

Part B

What is the time t_B it will take for point B of the wavefront to encounter the vacuum/material interface?

Express your answer numerically to two decimal places in units of t₁ (the time it takes light to travel a distance d in a vacuum).

It should be noted that the accuracy of the method increases the smaller the distance is before making a new wavefront and redoing the wavelets. If you tried to just draw a large wavelet from point B, then it would hit the surface after only traveling a distance d. This large wavelet would make it difficult to visualize how to add up all of the other wavelets to make a coherent wavefront. As a result, one should just draw a line perpendicular to the wavefront at point B until it hits the surface, at which point the wavelets will change owing to the new index of refraction.

Part C

How far did point C go into the material interface in the time t=t_B that it took for point B to get to the interface? For this part we are looking for the distance traversed by the point C in the material (d_C).

Give your answer in terms of the time t_B, c, and n.

Part D

What is the new angle θ at which point C of the wavefront is propagating (relative to a line perpendicular to the vacuum/material interface)? Try to use the fact that you have a spherical wavefront propagating from t=0 at the point where C met the vacuum/material interface until time t_B when the wavefront at point B reached the interface.

Express your answer in terms of inverse trigonometric functions and n.

greenduck607

PHYS 101 Lecture 13: Chapter 3-3

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