Chapter 13 introduces the important concept of "4-vectors." The material in this chapter could alternatively be placed in the previous two, but for various reasons I thought it best to make one. For the most part, however, the remarks address questions that arise naturally in the course of the discussion.
Strategies for solving problems
General strategies
It's very easy to misspell 8 for 9 on a calculator, but you probably won't misspell aq on a piece of paper. The main point here is to learn what to do with the output of the problem once you get it.
Units, dimensional analysis
We claimed above that the only combination of our given dimensional quantities that has unit 1/T is√. The energy has units of ML2/T2, and the only given combination of dimensional constants of this form is Bkx02, where Bis is a dimensionless number.
Approximations, limiting cases
When making approximations, how do you know how many terms in the Taylor series to keep. So in this case we'll know we'll have to go back and include the x2/2 term in the series.
Solving differential equations numerically
If you have enough computing time, you can achieve any desired accuracy (assuming the system is not chaotic, but we won't have to worry about that for the systems we'll be dealing with). In Chapter 4, we will find that the solution can be written as, among other things.
Problems
Exercises
Jupiter, and that the radius of Jupiter is 11 times that of Earth, which is vJ/vE. It can be shown that the tangent F = equation (where θ is measured with respect to the vertical).
Solutions
A mass is subject to a damping force proportional to its velocity, which means that the equation of motion takes the form x¨ = −A˙x, where A is a constant. The answer involves m/(Vfb), which diverges as Vf→0. b) You can show that the only combination of the quantities that have units ofLis =g(n) m.
Statics
Balancing forces
The total force acting on the surface is usually a combination of the normal force and the frictional force (see below). For the most part, the only difference between "tension" and "normal force" is the direction of the force.
Balancing torques
In this problem, there are two forces acting at the bottom of the ladder, so this is the point we will choose for the pivot point (but you should verify that other choices for the pivot point, for example the middle or top of the ladder, give same result). This is the direction that causes the lines of the three forces on the ladder to be parallel (that is, pass through a common point), as shown in Fig.2.8.
Problems
A rope with lengthLand mass density per length unitρ lies on a plane inclined at an angle θ (see Fig.2.11). Each of the following planar objects is placed, as shown in Fig.2.13, between two frictionless circles of radius R. The mass density per unit. the area of each object is σ and the radii of the contact points make an angle θ with the horizontal.
Exercises
A rope rests on two platforms, both inclined at an angle θ (which you are free to choose), as shown in Fig.2.30. A massless string connects the base of the left stick with the right stick, perpendicularly as shown in Fig.2.37.
Solutions
Therefore, you can think that the tension at the top of the rope is ρLgsinθ−µρLgcosθ. The normal force at the pivot point of the stick (which is equal to the vertical component of F, because the stick is massless) is equal to MgLsinθcosθ/2.
Newton’s laws
For example, it fails in the turntable frame. 2 Intuitively, an inertial frame is one that moves with constant velocity. You can define it for every particle in the world and for every acceleration, but the point is that the definitions have nothing to do with each other. In this case, the momentum is carried away in the electromagnetic field (so the total momentum of the particles and the field is conserved).
Free-body diagrams
The masses of the platform, the person and the pulley6 are M,m and µ respectively.7The rope is massless. This string has to go somewhere, so it ends up in the part of the string that touches m1 (see Figure 3.6). Note that these random movements undoubtedly do not correspond to the actual movements of the mass.
Solving differential equations
Assuming we can do the integral on the left, this equation gives a function of x. Assuming we can perform this integrally, it yields as a function of v, and thus (in principle) as a function oft, that is, v(t). This makes sense, because the drag force at the start is negligible, leaving the ball essentially in free fall.
Projectile motion
Assume the ground is horizontal and the ball is released from ground level. If the ground slopes at an angle β, then the equation for the line of the ground is y=(tanβ)x. We need to solve that makey=(tanβ)x, because this gives the time at which the path of the ball intersects the line of the ground.
Motion in a plane, polar coordinates
For circular motion, it says that F = ma along the tangential direction because rθ¨ is the second derivative of the distance rθ along the circumference. For circular motion, it says that the radial force is −m(rθ)˙ 2/r= −mv2/r, which is the known force that causes centripetal acceleration, v2/r. The forces on the mass are the tension in the string, T, and gravity, mg (see Figure 3.10).
Problems
What should θ be so that the block travels a given horizontal distance in minimum time. Let θ0 be the angle at which the ball must be thrown so that the length of the path is maximum. Your task is to pull the string so that the mass keeps moving along.
Exercises
How fast must the ball be thrown so that it lands at the foot of the cliff. What must θ be so that the ball travels the greatest horizontal distance, d, when it returns to the height of the top of the plane. A driver encounters a large tilted parking lot where the angle of the ground to the horizontal is θ.
The coefficient of friction between the tires and the ground is µ. The motorcycle starts from rest. What is the minimum distance it must travel to reach its maximum allowable speed, that is, the speed above which it will slip out of the circular path?27Solve this in two ways:. Write the radial and tangential F =ma equations (you'll want to write a as dv/dt), then take the derivative of the radial equation.
Solutions
Then the combination of the mass f(x) and the following masses can be considered as the effective mass f(f(x)). This tells us that the speed of the chain when it leaves the table (that is, wheny=) is v=α. If the acceleration of the ball is written as aa=dv/dt, then F =magives−mg−mαv =m dv/dt.
Oscillations
Linear differential equations
So if we find one solution, then we know we've found the whole thing. Example 2 (x¨=ax): If it is negative, we will see that this equation describes the oscillatory motion of, for example, a spring. For example, we will see that for a spring,ω=. k/m, where is the spring constant.ω is independent of the initial conditions.
Simple harmonic motion
Let be the length of the string and θ(t) be the angle that the string makes with the vertical (see Figure 4.2). Exercise 4.23 deals with higher-order motion corrections in the case where the amplitude is not small. No matter how complicated the system looks at first glance, if you end up with an equation that looks like z¨+ω2z =0, then you know that the system is undergoing simple harmonic motion with a frequency equal to the square root of the coefficient ofz , regardless of what this coefficient is.
Damped harmonic motion
Notes: If γ is very small (more precisely, if γ is ω), then ω˜ ≈ω, which makes sense because we have a nearly damped oscillator. Of course, for very small ω˜ it is difficult to even say that the oscillations exist, because they will be damped on a time scale of the order of 1/γ ≈1/ω, which is short compared to the long time scale of the oscillations, 1 /ω˜.
- Driven (and damped) harmonic motion
- Coupled oscillators
- Problems
- Exercises
- Solutions
Find the ratio between the maximum speed in the former case and that in the latter case. Show that the maximum speed of the mass in the first case is equal to the maximum speed in the second case.7. In terms of the angle θ in Figure 4.28, the velocity components relative to the ground.
Conservation of energy and momentum
Conservation of energy in one dimension
Since this equation is true at all points of particle motion, the sum of kinetic energy and potential energy is constant. Therefore, whenever one talks about the gravitational potential energy in a structure on the surface of the earth, it is generally understood that the earth is the reference point. The general work-energy theorem states that the work done on a system by external forces is equal to the change in energy of the system.
Small oscillations
Equation (5.20) is an important result because any function V(x) essentially looks like a parabola (see Figure 5.4) in a sufficiently small region around the minimum (except in the special case where V(x0)=0).
Conservation of energy in three dimensions
With dr ≡ (dx,dy,dz), this work can be written as F·dr (see Appendix B for the definition of "dot product"). B.2), we see that work can also be written as F|dr|cosθ, where θ is the angle between F and dr. Integrating this as (Fcosθ)|dr|shows that work is equal to the distance moved times the force component along the displacement. Alternatively, we can group it as F(|dr|cosθ) and show that work is also equal to the magnitude of the force multiplied by the displacement component in the direction of the force.
Given a force F(r), a necessary and sufficient condition for the potential,
- Gravity
- Momentum
- The center of mass frame .1 Definition
- Collisions
- Inherently inelastic processes
- Problems
- Constant y ˙ **
- Exercises
The only degree of freedom in the CM frame is the angle of the line containing the final (oppositely directed) velocities. Then quickly pull on one of the ends, in the direction of the bend. What is the speed of the chain when it loses contact with the table?
