sign in sign up
<<
Home , Velocity

Circular

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 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 , 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 (or dot) product and the vector (or cross) product. In this chapter we will make use, for the first , 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, the cosine of the 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 , 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 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 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 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 makes with the positive x-axis, or the 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 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 of the angle versus time is 휃called the 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 per time, or per . 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 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 푟Equation⃗ ∙ 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 to the circular trajectory. This property is sometimes expressed as

= = , [19]

푧 where 푣⃗ is푣 the휃� generic푅휔 휃� unit vector (magnitude 1) tangent to the circle at the position of the particle (see Fig. 3), and we made use of Eq. [15]. 휃�

FIG. 3

4. Uniform circular motion

Before moving to the more general description of circular motion, let’s review the simpler case of uniform circular motion, that is, the motion of an object moving at constant angular velocity , or equivalently, at constant speed v. First of all, if the particle moves at constant speed, we can easily calculate the time it takes for the particle to complete a full 푧 circle, called the period of the motion. We can do that in two different ways,휔 using either the speed or the angular 4 Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden. velocity. To complete a full circle, the particle must travel a distance equal to the circumference of the circle 2 . The period T is therefore given by 휋푅 = . [20] 2휋푅 Similarly,푇 푣 we can say that to complete a full circle the particle must cover an angle, expressed in radians, equal to 2 , traveling at constant angular velocity . The period is therefore given by the angle divided by the angular velocity:

푧 휋 = . 휔 [21] 2휋 푇 휔푧 Notice how Eqs. [20] and [21] are related, and can be derived from each other using Eq. [15]. Another quantity used to describe uniform circular motion is the f of the motion, that is, the number of full circles (revolutions) completed by the particle per unit time. It is easy to see that the frequency is just the inverse of the period, e.g., if a particle takes ½ s to compete a full circle, then it will make 2 revolutions per second.

= . [22] 1 Another푓 푇 way of looking at the frequency is through the angular velocity . In fact, the angular velocity represents the number of radians completed by the particle per time, while the frequency represents the number of revolutions per 휔푧 time, and since there are 2 per revolution, it follows that

= 2 . 휋 푟푎푑푖푎푛푠 [23]

휔Notice푧 휋푓that, indeed, Eq. [23] can be derived easily by combining Eqs. [21] and [22]. Finally, it is important to remember that Eqs. [20] through [23] are not general, but ARE ONLY VALID FOR UNIFORM CIRCULAR MOTION, which we defined as motion at constant speed (or constant angular velocity).

Having discussed the properties of uniform circular motion, let’s now review how it can be generated. Even if the speed of the particle is constant, the direction of the velocity, and thus the velocity itself, change constantly as the particle moves around the circle. This implies that the particle is subject to a non-zero . To find the value of such acceleration we can differentiate Eq. [14], keeping into account that, in this specific case, the angular velocity is constant: 휔푧 = = cos sin = cos sin . [24] 푑푣�⃗ 푑휃 푑휃 2 2 푧 푧 푧 푧 By푎⃗ comparing푑푡 −푅휔 Eq. [24휃]푑푡 with횤̂ − Eq.푅휔 [10] we휃 푑푡obtain횥̂ −: 푅휔 휃 횤̂ − 푅휔 휃 횥̂

= = , [25] 2 2 that is,푎⃗ the −acceleration휔푧 푟⃗ −휔 푧responsi푅푟̂ ble for the circular trajectory of the particle is aligned with the radius of the circle at the particle position (unit vector ), is directed toward the center of the circle (the negative ), and has constant magnitude equal to: 푟̂ = = = , [22] 2 2 2 푣 푣 푧 where 푎we made휔 푅 use� 푅of� Eq.푅 [15].푅 Due to its direction toward the center of the circular trajectory, such acceleration is often called centripetal acceleration.

Using ’s second law, we can move one step further and say that, to move on a circular trajectory at constant speed, a particle requires a net force directed toward the center of the circle with constant magnitude 5 Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden. = = = 2. [23] 2 푚푣 푧 Let’s now퐹 review푚푎 푚how휔 푅we can푅 apply the concepts just learned to real problems, with a few examples.

Example 1

A car with mass m moves at constant speed v on a vertical circular track of radius R (Fig. 4a). Find the normal force exerted by the track on the car at [a] the highest and [b] the lowest point, and [c] the minimum speed needed by the car not to fall from the track.

FIG. 4

[a] To find the normal force we simply use Newton’s second law. The only two acting on the car at the top are the and the normal force. To stay on the track, the acceleration, for uniform circular motion, must be equal to the centripetal acceleration or / . The free body diagram of the car at the highest point is show in Fig. 4b. Using a reference frame with the y-axis2 vertical, only the y-component of the forces is non-zero, thus we can write: 푣 푅 = = 2. [24] 푣 푦 Solving− 푚푔for the− 푛 magnitude푚푎 of− 푚the푅 normal force n we get

= = = 2 2 therefore 2 . [25] 푣 푣 푣 [b] The푛 procedure푚 퐿 − 푚푔 is identical푚 � 퐿 to− 푔what� we did for푛�⃗ part−푚 [a],� 퐿 however,− 푔� 횥̂ this time the normal force and the acceleration are directed toward the positive y-axis. The free body diagram is shown in Fig. 4c, and Newton’s second law for the y- direction is:

= = + = + 2 2 2 . [26] 푣 푣 푣 Comparing푛 − 푚푔 Eqs. [25]푚 퐿 and∴ [26]푛 it푚 is� 퐿clear푔 that� ∴ not푛�⃗ only푚 �the퐿 direction푔� 횥̂ of the normal force changes from the highest to the lowest point, but also its magnitude is bigger at the bottom than at the top.

[c] If the car is not fast enough, then it will fall from the track without completing a full circle. A practical experiment to show this is by filling a bucket with water and spinning it in a vertical circle with your arm. If the speed of the bucket is high enough, then the water stays in the bucket, otherwise… well, I will let you do the experiment 

The reason for it is evident in Eq. [24]. If the speed is high enough, then there is a normal force that, combined with the weight, generates the right acceleration for circular motion. However, if the speed is too small, then we would need a negative normal force to maintain circular motion. Unless the car is strapped to the track, it will fall before reaching the maximum height. To find the minimum speed necessary to avoid it, all we need to do is set the normal force equal to zero in Eq. [25]. If the normal force is zero, then alone is responsible for the centripetal acceleration. Any speed smaller than that, gravity is too big and thus the acceleration of the car will be bigger than the centripetal acceleration making it

6 Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden. fall. For any speed bigger than that, then the normal force will be able to balance the gravity to provide the proper centripetal acceleration. Setting = 0 in Eq. [25] we get:

= 0 = 푛 2 . [27] 푣 Example퐿 −2 푔 ∴ 푣 �퐿푔

A car with mass m, traveling at constant speed v, moves around a flat curve with radius R. How much force must be exerted on the car to stay on the road and what provides it?

FIG. 5

Solution: For the car to move of uniform circular motion, there must be a force directed toward the center of the circular trajectory, capable of generating the . Figure 5 shows the car with the free body diagram. Notice that in the figure the car is moving on a plane perpendicular to the page, and the instant velocity is straight out of the page. In the figure we also show the center of the circle and the vector position. Gravity is perpendicular to direction of the acceleration, so it cannot be responsible for it, and it is simply “balanced” by the normal force of the road. We thus added the “unknown” force f. Using a reference frame with the x-axis horizontal pointing to the right, and the y-axis vertical, pointing up, Newton’s second law in the x and y directions gives:

= 2 [28] 푣 = 0 = 푓 푚 푅 The x-푛com− 푚푔ponent of∴ Newton’s푛 푚푔 law gives the magnitude of the force responsible for the circular motion, while the y- component gives the magnitude of the normal force.

To answer to question of what is responsible for the force f, we note that the only thing, other than gravity, acting on the car is the surface of the road and that the force is parallel to such surface. We have already seen in previous chapters that the parallel force between an object and the surface over which it is placed is the between the two. The force responsible for the circular path of the car is, indeed, the static friction, i.e., = .

We should also remember that there is a maximum value the static friction may푓 have,푓푠 proportional to the magnitude of the normal force:

, [29]

Where푓 푠 ≤ is휇 the푠푛 coefficient of static friction. Using Eqs. [28] and [29] we get:

푠 휇 = 2 . [30] 푣 푠 푠 푠 That is,푚 there푅 ≤ 휇is 푛a maximum휇 푚푔 value∴ 푣 ≤for� 휇the푅푔 speed of the car, which depends on the coefficient of stating friction and the radius of the curve (but not the mass of the car), and if the speed of the car exceeds this value we all know what happens to the car. To understand the meaning of Eq. [30] we notice that, if the coefficient of static friction is small, then the maximum speed at which the car can travel to stay on the road is small as well. For example, if the road is wet, or icy, 7 Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden. then the speed at which the car may slip off the road is much smaller. Also, the maximum speed depends on the radius, that is, the maximum speed at which the car can travel staying on the road is smaller for sharp curves, as we well know from everyday experience. However, the maximum speed does not depend on the mass of the car, so a light car can travel at the same maximum speed of a heavy one.

Problem 1

The conical : An amusement park ride consists of small cars of mass m attached to steel cables of length L. The cables have negligible mass and are attached to a spinning post (see Fig. 6). When at maximum speed, an empty car forms an angle with the vertical. What is the angular velocity of the ride at maximum speed?

훼 휔푧

FIG. 6 FIG. 7 Strategy

At maximum speed the cars move of uniform circular motion around the spinning post and the forces acting on them are the weight and the in the steel cables. We can therefore use Newton’s second law to find the value of the acceleration and solve for the angular velocity.

Setup

Figure 6 already shows us one car at an instant in which it is in the x-y plane. We can thus simply choose a coordinate system with the x-axis horizontal and pointing to the right of the page, and the y-axis vertical, pointing to the top of the page. While the choice of the origin is not important in this problem, we can set it at the fixed point at the top of the steel cable. Notice that, with this choice, the car motion is actually around the y-axis, in a plane parallel to the x-z plane (horizontal and perpendicular to the page). We will therefore change the index for the angular velocity accordingly, but the physics stays the same.

Solution

Figure 7 the free body diagram of the car. In the figure we have also marked the radius of the circular motion which, with simple trigonometry, can be written as = sin . Based on the free body diagram, we can write Newton’s second law for the x and y direction as: 푅 퐿 훼

sin = = = sin = 2 2 푇 [31] 푇 cos훼 푚푎푥= 0푚 휔푦 푅 = 푚휔푦퐿 훼 ∴ 휔푦 �푚퐿 푚푔 푇 훼 − 푚푔 ∴ 푇 cos 훼 8 Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden. Combining the two equations we get:

= = . [31] 푚푔 1 푔 휔푦 �cos 훼 푚퐿 �퐿 cos 훼

Check dimensions: The units of angular velocity are simply , therefore = 푚 = = OK 2 −1 −1 푠 −2 −1 � Notice that, if instead of the angular velocity we are interested푠 in the speed 푠of the particle,푚 √ we푠 can 푠simply use Eq. [15]:

= = sin = . [32] 2 푔 퐿푔 sin 훼 푣 휔푦푅 �퐿 cos 훼 퐿 훼 � cos 훼 5. Non-uniform circular motion

We will briefly mention here the case of non-uniform circular motion, that is, when the speed of the particle is not constant. The simplest way to calculate the acceleration in non-uniform circular motion is by starting from Eq. [19]:

= ,

and since푣⃗ the푣휃� acceleration is the first derivative of the velocity, we get

= = + . [33] 푑푣�⃗ 푑푣 푑휃� � To푎⃗ find푑푡 an푑푡 expression휃 푣 푑푡 for the second term of the Eq. [33] we can simply use what we have already done for uniform circular motion. In fact, for uniform circular motion

= = 0 = , [34] 푑푣 푑휃� But we푣 have푐표푛푠푡푎푛푡 already calculated∴ 푑푡 the∴ value푎⃗ of푣 푑푡the acceleration for uniform circular motion is Eq. [25]:

= = [35] � 2 푑휃 2 푧 푧 We푎⃗ can−휔 therefore푅푟̂ ∴ 푣write:푑푡 −휔 푅푟̂

= + = + , [36] 2 푑푣 푎⃗ −휔푧 푅푟̂ 푑푡 휃� 푎⃗푟푎푑 푎⃗푡푎푛 with | | = and | | = . 2 푑푣 푟푎푑 푧 푡푎푛 푎That⃗ is, 휔the푅 acceleration푎⃗ for푑푡 non-uniform circular motion can be divided into two terms, one directed along the radius of the circle, equal to the centripetal acceleration, the other tangent to the circle, with magnitude equal to the first derivative of the speed versus time. As in the case of uniform circular motion, the centripetal acceleration is responsible for the circular trajectory of the particle, however, if the speed is not uniform, there is an additional term of the acceleration, tangent to the particle trajectory, responsible for the change is speed.

We can go one step further by using Eq. [15] or = in Eq. [36] and considering the fact that the radius is constant in circular motion. The tangent component of the acceleration then becomes: 푣 푅휔푧 푅 ( ) | | = = = = [37] 푑푣 푑 푅휔푧 푑휔푧 푎⃗푡푎푛 푑푡 푑푡 푅 푑푡 푅훼푧 9 Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden. where = , is called the . 푑휔푧 푧 In훼 terms푑푡 of problem solving, we can solve most problems using the same strategy shown for uniform circular motion even when the speed is not constant. For example, in examples 1 and 2 and problem 1 the solution would be the same even if the speed is not constant, as long as we use the instant speed to calculate the centripetal acceleration. In general, however, there may be additional forces, tangent to the particle trajectory, responsible for the change in speed.

Example 3

FIG. 8

Let’s consider the same situation of example #1, but this time we will consider the normal force when the car is 90 degrees from the top of the circle. Figure 8 shows the free body diagram associated with the particle. Using Newton’s second law in the x and y directions we get:

= = 2 푣 [38] 푛 푚=푎푟푎푑 푚=푅 = 푑푣 푑푣 푡푎푛 Equation−푚푔 [38] shows푚푎 that,푚 in푑푡 the∴ horizontal푑푡 −푔 position only the normal force is responsible for the circular trajectory of the particle, while the weight, which is tangent to the particle position at that point, is going to increase the particle speed on the downward leg, and decrease it in the upward one. For the car to move of uniform circular motion there must therefore be an additional force (given by the car’s engine) that makes = 0 (note that, in example #1, since he weight is radial at the top and bottom, then the additional force would be zero)푡푎푛. However, even if that is not the case, as long as we know the instant speed of the car, then we can solve the problem.푎 For example, we will see later this semester that in many situations the speed of the particle at any point can be “easily” calculated using conservation of , even if it changes with time and position. In those problems, we can still calculate the forces acting on the particle using the same procedure used here.

6. Equation Sheet

= 푑휃 휔푧 = 푑푡 푑휔푧 훼푧= 푑푡 = ( + ) = 푣 휔푧푅 푟⃗ = 푅 푐표푠휃( 횤̂ 푠푖푛휃+ 횥̂ 푅)푟=̂ 푣⃗ = 휔푧푅 +−푠푖푛휃횤;̂ | 푐표푠휃|횥=̂ 휔푧;푅 휃|� | = = 2 푑푣 푎 ⃗ 푎⃗푟푎푑 푎⃗푡푎푛 푎⃗푟푎푑 휔푧 푅 푎⃗푡푎푛 푑푡 푅훼푧

10 Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden.