Circular Motion By Prof. Massimiliano Galeazzi, University of Miami If you spin a bucket full of water in a vertical circle fast enough, the water will stay in the bucket. How fast do you need to spin it to make sure you don’t get showered? What is the force that makes your car turn when you turn the steering wheel? How big is that force? These are some of the questions we will answer in this chapter. The general topic is that of circular motion, that is, objects moving on a circular trajectory. Along the way we will learn about ferris wheels, cars turning without going off the road, etc. But first, we need to improve our math understanding by reviewing the product between vectors. --------------------------------------------------------------------------------------------------------------------------------------------------- MATH INSERT In chapter 1 we have introduced the concept of vectors and found how vectors can be added and or subtracted. We also mentioned the fact that vectors can be multiplied with each other in two different ways, what are called the scalar (or dot) product and the vector (or cross) product. In this chapter we will make use, for the first time, of the dot-product, so it is appropriate to take a look at it before we proceed. 1. The scalar (dot) product between vectors FIG. 1 Let’s consider two vectors, = + + and = + + . The scalar product between and , denoted (“A dot B”, hence the퐴⃗ term퐴푥 횤“̂ dot퐴 product푦횥̂ 퐴푧”),푘� is a 퐵�scalar⃗ 퐵 푥qu횤̂ antity퐵푦횥̂ equal퐵푧푘� to the product between the magnitudes퐴⃗ 퐵�⃗ of the two vectors, times the cosine of the angle between the two: 퐴⃗ ∙ 퐵�⃗ = cos , 휙 [1] where 퐴the⃗ ∙ 퐵� ⃗angle�퐴⃗ �� 퐵�is⃗� measured휙 by drawing the two vectors with the tails in the same point (see Fig. 1a-b). We notice that, from Eq. 1, the dot product, like the “regular” product between scalars is commutative, i.e., 휙 = [2] Also,퐴⃗ ∙ 퐵�⃗ we퐵� ⃗can∙ 퐴⃗ write Eq. [1] as: = cos , [3] ⃗ �⃗ ⃗ �⃗ That is,퐴 the∙ 퐵 scalar�퐴�� product�퐵� 휙between� and is equal to the magnitude of times the projection of in the direction of , i.e., cos (see Fig. 1c). Similarly, we can write: 퐴⃗ 퐵�⃗ 퐴⃗ 퐵�⃗ 퐴⃗ �퐵�⃗� 휙 1 Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden. = cos , [4] ⃗ �⃗ �⃗ ⃗ That is,퐴 the∙ 퐵 scalar�퐵�� product�퐴� 휙between� and is equal to the magnitude of times the projection of in the direction of (see Fig. 1d). 퐴⃗ 퐵�⃗ 퐵�⃗ 퐴⃗ 퐵�⃗ From the definition of the scalar product it is possible to derive a few simple properties: • The scalar product between two vectors can be positive, negative, or zero. In particular, if 0 < 90 then the product is positive, while if 90 < 180 then the product is negative. 표 ≤ 휙 • The scalar product of two parallel vectors표 is equal표 to the product of the magnitudes of the two ( = 0). 휙 ≤ • The scalar product of two vectors that are anti-parallel is equal to the negative of the product of the two 휙 magnitudes ( = 180 ). • The scalar product of two표 perpendicular vectors is equal to zero ( = 90 ). 휙 • From the previous result it follows that, if the scalar product between two표 vectors is zero and their magnitudes 휙 are different from zero, then the two vectors must be perpendicular. This provides a simple, yet very powerful way to determine whether two vectors are perpendicular. While the definition of scalar product in Eq. [1] is very useful, sometimes it is easier to work with vector components, rather than magnitude and direction. It is also possible to calculate the scalar product of two vectors in terms of their components: = + + + + . [5] 푥 푦 푧 푥 푦 푧 To퐴⃗ ∙ 퐵find�⃗ the�퐴 value횤̂ 퐴 of횥̂ the퐴 scalar푘�� ∙ � product퐵 횤̂ 퐵 in횥 ̂ terms퐵 푘� of� components, we simply expand the product in Eq. [5]: = + + + + + + + [6] 퐴⃗ ∙ 퐵�⃗ 퐴푥횤̂ ∙ 퐵푥횤̂ 퐴푥횤̂ ∙ 퐵푦횥̂ 퐴푥횤̂ ∙ 퐵푧푘� 퐴푦횥̂ ∙ 퐵푥횤̂ + + + 퐴푦횥̂ ∙ 퐵푦횥̂ 퐴푦횥̂ ∙ 퐵푧푘� 푧 � 푥 푧 � 푦 푧 � 푧 � And taking퐴 푘into∙ 퐵 account횤̂ 퐴 푘 that,∙ 퐵 횥usinĝ 퐴 the푘 ∙ 퐵properties푘 of the scalar product and of unit vectors = = = 1, and = = = 0, we get: 횤̂ ∙ 횤̂ 횥̂ ∙ 횥̂ 푘� ∙ 푘� � � 횤̂ ∙ 횥̂ 횤̂ ∙=푘 횥̂ ∙ 푘+ + . [7] 푥 푥 푦 푦 푧 푧 END OF퐴⃗ ∙ MATH퐵�⃗ 퐴 퐵INSERT퐴 퐵 퐴 퐵 --------------------------------------------------------------------------------------------------------------------------------------------------- 2. Motion along a circular path Consider a particle with mass m that is constrained to move on a circular path, as shown in Fig. 2. To describe the motion of the particle we can set up a coordinate system with the origin coincident with the center of the circle and then use the x and y coordinate of the particle. We could, however, also use the angle that the particle position makes with the positive x-axis, or the distance of the particle from the positive x-axis along the circular path. Note that in our choice 휃 of coordinates the particle is moving around the z-axis which is perpendicular to the page and passes through the center of 푙 the circle. A general convention is to consider the angle and the distance l as positive if measured counterclockwise. The Cartesian coordinates x and y of the particle and the vector position can then be easily calculated as: 휃 푟⃗ 2 Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden. FIG. 2 = cos [8] 푥 = 푅 sin 휃 [9] 푦 = 푅 + 휃 = ( cos ) + ( sin ) , [10] where 푟⃗R is푥 the횤̂ 푦radius횥̂ 푅of the휃 circle.횤̂ 푅 Moreover,휃 횥̂ since the distance l traveled by the particle is nothing more than the arc subtended by the angle , we can also write: = . 휃 [11] From푙 푅 휃Eq. [10] we can also easily find an expression for the particle velocity. By using the chain rule for derivatives and the fact that in circular motion R is constant, i.e., the distance from the origin does not change, we get: = = ( sin ) + ( cos ) , [12] 푑푟⃗ 푑휃 푑휃 where 푣⃗we mad푑푡 e use−푅 of the휃 fact푑푡 that,횤̂ indeed,푅 휃 the푑푡 angle횥̂ is the only quantity in Eq. [10] that changes with time. The first derivative of the angle versus time is 휃called the angular velocity of the particle and is indicated with the symbol (the Greek letter omega): 푧 휔= , [13] 푑휃 푧 Where휔 the index푑푡 “z” is used to indicate that, as mentioned before, the rotation is around the z-axis. We note that, in general, the angular velocity is a vector, with each component quantifying the rotation around a different axis. However, in this chapter (and in most of the book) we will only consider rotations around the z-axis, and thus only use the z- component of the vector. We also point out that the angular velocity measures angles per time, or radians per second. However, radians are usually treated as if they are not a real unit, and the SI unit for angular velocity is simply . Using 휔푧 Eq. [13] we can rewrite Eq. [12] as: −1 푠 = sin + cos . [14] 푣⃗ −푅휔푧 휃 횤̂ 푅휔푧 휃 횥̂ We can also use Eq. [12] to find the speed of the particle = | | = + . However, it is easier to calculate it 2 2 using the fact that the speed is the first derivative of the distance 푣l traveled푣⃗ in� Eq.푣푥 [11]:푣푦 ( ) = = = = , or = , [15] 푑푙 푑 푅휃 푑휃 푣 푧 푧 where 푣we used푑푡 the푑푡 fact that푅 푑푡 the 푅radius휔 R휔 is constant.푅 Equation [15] represents a very simple and straightforward relation between the particle speed and its angular velocity. 3 Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden. 3. Position vs. Velocity in circular motion Before continuing with our analysis, let’s take a quick look at the properties of the vectors position and velocity associated with circular motion. The vector position is the vector going from the origin of the coordinate system to the particle position. However, since in our choice the origin of the coordinate system coincides with the center of the circular path, the magnitude of the vector position will always푟⃗ be constant and equal to the radius R of the circle and will always have a “radial” direction, i.e., it will always be oriented with the radius of the circle at the position of the particle. This property can be expressed as = , [16] where 푟⃗ is 푅the푟̂ generic unit vector (magnitude 1) aligned with the radius of the circle at the position of the particle. Using푟̂ Eqs. [10] and [14] we can also demonstrate that, in circular motion, the velocity is perpendicular to the position . The simplest way to demonstrate it is by calculating the dot-product between the two vectors, using the 푣⃗ definition of dot-product with components derived in Eq. [7] and the fact that the two vectors only have x and y components:푟⃗ = ( + ) + = + = ( cos )( sin ) + ( sin )( cos ) , [17] 푥 푦 푥 푦 푧 푧 which 푟⃗simplifies∙ 푣⃗ 푥횤 ̂to: 푦횥̂ ∙ �푣 횤̂ 푣 횥̂� 푥푣 푦푣 푅 휃 −푅휔 휃 푅 휃 푅휔 휃 = cos sin + sin cos = 0. [18] 2 2 Equation푟⃗ ∙ 푣⃗ − 푅[18휔] 푧is valid휃 in 휃the most푅 휔 푧general휃 case휃 of circular motion, and implies that, if a particle can only move on a circular path, then its velocity is always perpendicular to its vector position. Moreover, since the vector position is always aligned with a radius of the circular path, this means that the velocity is always perpendicular to the radius, i.e., it is always tangent to the circular trajectory.