EN PH131 Chapter 12: EN PH 131 12.5 Curvilinear Motion - Rectangular Components

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amethystjellyfish251

21 Apr 2016

School

University of Alberta

Department

Engineering Physics

Course

EN PH131

Professor

Douglas Gingrich

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EN PH131 Full Course Notes

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If the particle is at a point (x, y, z) on the curved path s shown in the below figure, then its location is defined by the position vector. The first time derivative of r yields the velocity of the particle. When taking this derivative, you need to account for changes in both the magnitude and direction of each of the vector"s components. E. g. the derivative of the i component of r is: The second term on the right side is zero, provided the x, y, z reference frame is fixed, and therefore the direction (and the magnitude) of i does not change with time. Differentiation of the j and k components may be carried our similarly: The dot notation represents the first time derivatives of x = x(t), etc. The velocity has a magnitude that is found from: And direction is specified by the unit vector uv = v/v. This direction is always tangent to the path.

Related Questions

which of the vectors below best represents the direction of the impulse vector j⃗ ?

To learn about the impulse-momentum theorem and its applications in some common cases.

Using the concept of momentum, Newton's second law can be rewritten as

?F? =dp? dt, (1)

where ?F? is the net force F? net acting on the object, and dp? dt is the rate at which the object's momentum is changing.

If the object is observed during an interval of time between times t1 and t2, then integration of both sides of equation (1) gives

?t2t1?F? dt=?t2t1dp? dtdt. (2)

The right side of equation (2) is simply the change in the object's momentum p2??p1?. The left side is called the impulse of the net force and is denoted by J? . Then equation (2) can be rewritten as

J? =p2??p1?.

This equation is known as the impulse-momentum theorem. It states that the change in an object's momentum is equal to the impulse of the net force acting on the object. In the case of a constant net force F? net acting along the direction of motion, the impulse-momentum theorem can be written as

F(t2?t1)=mv2?mv1. (3)

Here F, v1, and v2 are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in question is two-dimensional, it is often useful to apply equation (3) to the x and y components of motion separately.

( The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass m moving along the x-axis. The net force F is acting on the particle along the x-axis. F is a constant force.)

A) The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0.

B) The particle starts from rest at t=0. What is the magnitude v of the velocity of the particle at time t? Assume that t>0.

C) The particle has the momentum of magnitude p1 at a certain instant. What is p2, the magnitude of its momentum ?t seconds later?

D) The particle has the momentum of magnitude p1 at a certain instant. What is v2, the magnitude of its velocity ?t seconds later?

E) Find the magnitude of the impulse J delivered to the particle.

F) Which of the vectors below best represents the direction of the impulse vector J??
(Figure 1)

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puceworm982

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.