Rotational Motion. Description of the motion. is the relation between ω and the speed at which the body travels along the circular path.

Transcription

1 Rotational Motion We are now going to study one of the most common types of motion, that of a body constrained to move on a circular path. Description of the motion Let s first consider only the description of rotational motion, without worrying about the cause of the motion. This means that what we are now doing is mathematics and not physics. We will consider the physics later, when we talk about the nature of forces which give rise to circular motion, and discuss some examples. Suppose the radius of the circle is r, and that the angle θ is changing at a rate (so many radians per second) symbolized by ω (the Greek letter omega ): d θ = ω. dt This is called the angular velocity. It may or may not be constant. If the body is speeding up or slowing down as it goes around, then ω will change in time. One of the first important things to know is what is the relation between ω and the speed at which the body travels along the circular path. θ r direction of motion The distance covered along the circumference is rθ. The rate of change of this distance is v = d dt ( r θ ) = r d θ dt, where we used the fact that r is constant to perform the last step. Hence, the speed is v = r ω It turns out that there s another way to obtain this result, using an expression for the velocity in polar coordinates in two dimensions derived in an earlier section on vectors. Because the circular path lies in a single plane, these are the rθ.

2 appropriate coordinates to use. In order to get some practice with polar coordinates, let s see how it works. In the general case where both coordinates r and θ are changing in an arbitrary way, we showed that the velocity vector is v P ( t ) = dr dt r ˆ + r d θ dt θ ˆ. In the present case, the radial coordinate r is constant, so dr/dt=0. Also, dθ/dt=ω. Therefore, the velocity is v P = r ω θ ˆ where the unit vector θ ˆ points around the circle in the direction of increasing θ. Taking the magnitude of both sides, we find that the speed is v = rω, which agrees with the above. How large is the acceleration? To answer this, we will use another general expression derived in the section on vectors: a P ( t ) = r ˆ ä å d 2 r ã dt r ω 2 ë ì + θ ˆ 1 2 í r, d dt á r 2 ω é In our particular case, the radial coordinate r is constant, so we get a P = r ω 2 r ˆ + r α θ ˆ centripetal tangential, where we have introduced a new symbol α = d ω dt = d 2 θ dt 2 denoting the angular acceleration (in the same sense that ω=dθ/dt is the angular velocity). The first term points towards the center of the circle (in the r ˆ direction). For this reason, it is called centripetal acceleration. (In Latin, the word petere means to go toward, to seek.) The second term, the tangential acceleration, points along the circumference of the circle, and is proportional to the angular acceleration. In order to better understand centripetal acceleration, let s consider for a moment the special case where the body is moving with constant speed along the circumference of the circle. Then its angular acceleration is zero, and the body s entire acceleration is centripetal. Let s try to see why the acceleration points towards the center of the circle. The next figure shows the velocity vector at a

3 particular time t, and then at a later time t+ t. The difference v P ( t + t ) v P ( t ), shown in red, points inward although not directly through the center of the circle. P v ( t + t ) P v ( t ) P v ( t + t ) P v ( t ) As t is made smaller and smaller, the difference points more and more towards the center: P v ( t + t ) P v ( t ) P v ( t + t ) P v ( t ) In the limit as t goes to zero, the difference points directly towards the center. The acceleration is defined as the limit of this difference, divided by t: a P ( t ) = d v P ( t ) dt v P ( t + t ) v P ( t ) = lim. t 60 t Hence, the acceleration points directly towards the center of the circle, for a body moving with constant speed along the circle. In the general case, the body is also changing speed as it travels along the circle. In that case, the acceleration has a tangential component as well. The angular velocity vector Until now, we haven t been concerned about the orientation of the circular path in space. We have only been looking at the plane containing the motion. But sometimes we would like to keep track of both the angular velocity and the orientation of the circle. To do this in a concise way, it is useful to define a vector called the angular velocity vector ω P = ω n ˆ where n ˆ is a unit vector perpendicular to the plane of the circle. Its direction is given by the,

4 right-hand rule: you curl the fingers of your right hand around in the direction of the body s motion and stick out your thumb. Your thumb points in the direction of n ˆ : n ˆ The magnitude of the angular velocity vector is the angular velocity ω. There is a nice expression for the velocity in terms of the angular velocity vector: v P = ω P H P r, as shown in the following figure: ω P Pr ω P HPr The angular velocity vector is particularly useful when we are talking about the rotation of a rigid body. For example, consider this blob-shaped plate: ω P Each piece of the plate has a different velocity, but all have the same angular velocity. The angular velocity vector is a property of the plate as a whole. Once we know the angular velocity vector of the plate, we can get the velocity of any piece of the plate just by knowing its position, thanks to the last formula. The same goes for the rotation of a threedimensional rigid body, of course. The above formula for the velocity in terms of the angular velocity is still valid, except now there is one small subtlety involving the position vector of a given piece of the body. Notice that we don t really have to measure P r from the center of the circle. We can measure it from any point on the axis of the circle:

5 ω P Pr ω P HPr We can do this because any part of P r which is parallel to ωp does not contribute to the cross product ω P H P. r This is helpful because we usually want to refer all the position vectors of the pieces of the rigid body to the same point within the body. We can choose any point on the axis of rotation of the body: ω P Pv Pr 2 2 P r 1 Pr 3 Pv 1 Pv 3 Then the velocity of that piece of the body located at position i is v P i = ω P H P r i. We will make use of this formula when we talk about the inertia tensor in a later section. Cause of the motion Suppose we are now concerned with how to constrain a body to move on a circular path. We would like to know what force is necessary to achieve this. Why is a force necessary, in the first place? Well, we have just learned that a body moving on a circular path is always accelerating, even if its speed is constant. But, according to Newton s second law, if a body accelerates with respect to an inertial frame, a force must be acting. What force is causing this acceleration? Whatever is constraining the body to move on its circular path must be exerting a force on the body. This force of constraint is just the body s mass times its acceleration. Using what we learned above about the acceleration, we deduce that the force splits into two pieces. The piece which is always present even if the speed is constant is called the centripetal force:

6 F P centripetal = m ω 2 r r ˆ and the extra piece which is present if the speed along the circle is changing is the tangential force: F P tangential = m r α θ ˆ (Recall that α is the angular acceleration.) The following diagram shows the centripetal force P v F P centripetal 1 0 For example, if we attach a rock to a string and twirl it around, this is the force we must exert in order to keep the rock on a circular path with constant speed. We exert this force by pulling on the string. (Here s a short note about terminology: there s., another force called the centrifugal force, which we will discuss later. All you have to know right now is that it is not the same as the centripetal force. We will explain the distinction in more detail later, but for now just remember that the two forces point in opposite directions. The centrifugal force points outwards from the center of the circle. In Latin, the word fugere means to run away from, to flee - as in the English word fugitive.) Example: a rock on a string Suppose you have tied a rock to a piece of string, and are twirling it around at constant speed in a circle. Then the above analysis applies, and a centripetal force must be acting on the rock. In this case, the centripetal force is the tension in the string. Its magnitude is just m ω 2 r, where m is the rock s mass, ω is the angular frequency of rotation, and r is the radius of the rock s orbit. Example: circular orbit around earth The gravitational force on a body of mass m orbiting earth at a distance r from earth s center is F P = GMm r ˆ, where G is Newton s constant and M is earth s mass. Because this force is in the negative radial r 2

7 direction, it can give rise to circular motion around earth s center (although the motion need not be circular, as we will see later.) We ll suppose that we know the frequency with which the body orbits earth, and calculate the radius of its orbit. To do this, we use the fact that the centripetal force on the body is the gravitational force: m ω 2 r r ˆ = GMm r ˆ. r 2 Solving for the radius, we find r = ä å GM ë ì 1 / 3 ã ω 2 í. Sub-example: geosynchronous orbit Suppose a satellite orbits earth in such a way that it remains stationary above a particular spot on the equator. What is the radius of its orbit? The angular frequency is the same as that of the earth: ω = 2 π 24 hours = 2 π 86, 400 s = 7.27 H s The value of the gravitational constant is G = H m 3 kg 1 s 2 and earth s mass is M = H kg. Inserting these values, we find r H 10 7 m = 42, 000 km. For comparison, earth s radius is about 6,400 km. Example: charged particle in magnetic field Suppose an electrically charged particle moves in a region containing a magnetic field BP. If the particle s velocity is v P and its electric charge is q, then the force on the particle is F P = q P v H BP. Suppose now that the particle moves in a region where the magnetic field is constant in time and uniform in space. Then it turns out to be possible for the particle to be executing uniform circular motion in a plane perpendicular to B P, as we will now see. Let the magnetic field point in the positive z direction: B P = Bz ˆ, B > 0. A particle undergoing uniform circular motion with radius r in the plane perpendicular to the z direction has velocity

8 v P = r ω θ ˆ, as we showed before. Hence, the force on such a particle is q á r ω θ ˆ é H á B z ˆ é = qrω B r ˆ. If the product qω is negative, this force is directed radially inward. It is then a centripetal force. To find the frequency of rotation, we equate the magnetic force to our previous expression for the centripetal force m ω 2 r r ˆ = qbrω r ˆ. Solving for the frequency, we find ω = qb m. The radius increases with increasing particle speed v, and decreases with increasing field strength. If the field goes to zero, then the radius is infinite (a straight line). As in the gravitational case, circular motion turns out not to be the most general kind of motion in a uniform magnetic field. It turns out that the most general path in this situation is a spiral, or helix. The projection of the helix onto a plane perpendicular to the magnetic field is the circular path just described. The component of the velocity parallel to the magnetic field is constant (since the component of the magnetic force qv P H BP in that direction is zero). Here is a plot of such a path: If q is positive, then the particle goes around the magnetic field in the negative (i.e. left-handed) sense. The magnitude of ω is called the cyclotron frequency. Notice that it increases with increasing magnetic field strength, and that it does not depend on the particle s speed. The radius of the circular orbit is r=v/ ω, or r = mv q B. B v o

9 Exercise: what is the sign of the electric charge of the above particle? We will treat the more general case in which both electric and magnetic fields are present, in another section.