Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Physics 3309 April 4, 2017 (21-1) LAST TIME: Introduced noninertial reference frames and examples. 2 . Setting up the reference frame on the surface of the earth is shown on the figure to the right. y points north, x points east, and z points outward or up. In the rotating reference frame, is given by Ω sin Ω cos . Note that the x-component is zero. here is the colatitude (measured down from the north pole) or 90o – latitude. Our next job is to use our general equation to write the component equations for x, y, and z in the rotating frame of the earth. How big is Using the period of the earth’s rotation as 24 hours, we get about 7.29×10-5 rad/s. You can see why we get by without these corrections for many applications. Here is a figure that more clearly shows the basic geometry. Let be the distance from point O to the line connecting the center of the earth to the north pole. Comment on selecting an inertial reference frame here since the earth rotates around the sun and the sun with our solar system rotates around the center of the Milky Way galaxy. Consider the omegas for these rotations. Then sin .This is important because it is the radius of the circle that the origin of our coordinate system has with respect to an inertial frame at the center of the earth. Now let’s apply our equation to the reference frame on the surface of the earth for a static situation; i.e., a mass attached to a string near the surface of the earth. Let 0; 0; 0; 0; sin .Now we assume that a mass m is supported on a string so that the only real forces acting on mass m are T .Our equation now takes the form and 0 sin 1 . We usually identify T with mg, which is what we measure when we hang something on the end of .This means that the figure a string near the surface of the earth. We will also identify that expresses each of these forces is as shown. If we let a be the angle between mg and mgo, then we may set up the following relationship using the law of sines. sin sin sin 90 since90 istheanglebetween and –mA. At the equator, 2R = 0.034 m/s2. max is about 0.1o and this occurs at a latitude and colatitide of 45 degrees. For dynamic cases near the surface of the earth, the Coriolis force is by far the most important as we will see now. r is very small compared to R, so any centrifugal force will be small compared to the acceleration of the origin of our frame attached to the surface of the earth. Let’s do the problem in its entirety to first order in . Now our equation for motion near the surface of the earth is given by 2 .F and –mA have already been combined into mg, where g varies slightly depending on latitude. Therefore our equation becomes 2 . to get the x, y, and z components of the motion. We need to calculate Therefore det cos 0 sin sin cos sin cos Now we write everything in component form and try to solve the equations. Here are the three equations, where m cancels. 2 cos sin ; 2 cos ; and 2 sin Let’s check the zeroth order where = 0. It is easy to see that 0 .These are our usual equations that we got when we neglected the earth’s rotation. It is worth noting that depend on ,so if we solve for , we can directly substitute that value in and get the remaining equations solved. 2 cos sin ; We substitute the values for 2 tcos cos 2 sin sin 2 2 ; 2 sin backinto . We neglect terms that are of order 2 to get 2 cos Integrate to get 2 2 sin sin . . This may be integrated one more time to get 1 3 sin cos sin . We use this value of x(t) to get the values for y and z. Again 2 terms are neglected. Here are the equations. 2 2 cos ; sin cos 1 2 ; sin Now let’s go back to check on our intuition concerning dropping an object form rest at the equator. At the equator, = 90o. 0; ; 0.We want to find the drift that occurs as a result of the earth’s rotation. First, find the time required for the object to hit the earth. This set z(t) = 0. ; Our equations become Therefore ; . .Using h = 100 m gives a value of x = 2.2 cm. Suppose that we give the object an initial velocity in the easterly direction so that 0.You will then notice that the time of flight to the ground is not quite the same as before since the second term in the equation for z(t) no longer vanishes. Let’s look at the Coriolis force in different locations and with different velocities as another qualitative example. (a) Move south near the north pole. (b) Move east on the equator. (c) Move south on the equator. (a) (b) (c) so only the z-component of contributes and so = 90 degrees and 2 2m so = 90 degrees and 2 3 2 is west (along –y). Ω 2m Ω ,so up here. Ω 0. One of the most interesting applications that clearly shows the effects of the earth’s rotation is the Foucault pendulum. Typically, L is quite long, and the connection to point P so as to allow torque-free operation. Small ocsillations are considered as usual. In our notation, the equation of motion is 2 . g is the local value with the correction for the rotation made. When is small, the following approximations are valid. ≅ ; ; ; ≅ 0. This is equivalent to saying that mass m is moving in a horizontal plane under the action of Tx and Ty. Since ≅ 0,the equations of motion are given by 2 cos , 2 sin . and Reminder – we have solved a similar set of equations before. Do you remember what they were? . Suppose Bx = 0 and look at the equations. det 0 so you can see the similarities between and . Our equations for the Foucault pendulum are 2Ω 0 and 0. 2Ω Back in lecture 6, I showed you how to solve the equations for the motion of a charged particle ini a magnetic field, but these equations have an extra wrinkle. They do not involve only the first and second derivatives, but they also have the function itself in the equation. Remind about earlier solution. These equations seem ideally suited to solve using a complex variable given by so that and .Multiply the equation for y by i and add the two equations to get 4 0. 2Ω Were it not for the factor of i, this would be a damped harmonic oscillator, but the factor of i changes things. As usual, we search for a solution of the form and see that . and Therefore, 2 0 Now we may get the equation for 2 0 So 2 4 ⁄ 4 ⁄ 2 Usually, ≫ soweuse 0.57 . . .To see this consider L = 30 m. This gives This gives a value for of . To understand what happens with this equation, let’s give some initial conditions. We need to remember that is a complex variable, so we need 4 initial conditions, one set for x and one set for y. Choose the initial conditions as 0 ; 0 0; 0.Remember that C1 and C2 are complex as well to get the four unknowns. Now . We also need the general expression for . . 0 0 . and Therefore, Using 0 0 0so . 0 Now we equate the real and imaginary parts of the left- and right-hand sides to get 0 ⇒ 0 5 0. . Now go back to the equation a + c = A to get 1 and 2 Now a = A – c so 2 . 2 Now we may write the final result for (t). 2 2 2 cos sin . Recall that x(t) = Re (t) so cos cos sin sin ≅ cos cos . ≅ sin cos Now we get y(t) in the same way. Im cos sin sin cos The behavior is shown on the next page in graphical form. 6 . The period of the oscillations in z is given by 2/z and at 42 degrees latitude is given by z = cos 48o, which implies a period of about 36 hours. At the north pole, we would get a period of 24 hours. NEXT TIME: Begin Chapter 10 on Rotation of Rigid Bodies 7