Circular Motion

 < Day Day Up > 

Believe it or not, you already understand all the underlying concepts of circular motion. Previous chapters defined quantities such as displacement, velocity, and acceleration. Now all you have to do is apply these same concepts to rotational motion. Let's start by looking at the motion of a CD-ROM when your computer is reading it. Figure 14.1 shows a top-down view of the spinning CD.

Figure 14.1. Point P on a CD-ROM.

graphics/14fig01.gif

The center point ( C ) represents the axis of rotation. A point on the CD ( P ) therefore rotates about the center, which means that it travels along a circular path with a fixed radius ( r ). After some time has passed ( D t ), point P has moved a certain distance ( s ) along the circular path. This distance is called the arc length , and it is illustrated in Figure 14.2.

Figure 14.2. Point P has moved.

graphics/14fig02.gif

Also in Figure 14.2, you see that the angle of rotation ( q ) for this time interval is labeled. This angle must always be measured in radians. Chapter 3, "Trigonometry Snippets," discussed the conversion between degrees and radians. This angle of rotation ( q ) represents the angular displacement , and it's defined as the arc length divided by the radius.

Angular Displacement

q = s / r

where s = arc length and r = radius.


NOTE

Notice that if the arc length ( s ) is equal to the radius ( r ), q = 1 radian. This is precisely how the radian was originally defined.

Here is an example of a function that can calculate the angular displacement given the arc length and radius:

 float angDisplacement(float arc, float radius)      {        float theta;        theta = arc/radius;        return theta;      } 

Now that angular displacement has been defined, we can look at average angular velocity ( w ). As you might have guessed, angular velocity is the rate of change of the angular displacement. In other words, angular velocity is the angular displacement divided by time, just like linear velocity.

Average Angular Velocity

graphics/14equ01.gif


where q = angular displacement and t = time.


NOTE

The symbol for angular displacement ( w ) is the Greek letter omega.


If you look closely at the definition of angular velocity, you can probably guess the appropriate units. If angular displacement is measured in radians and time is measured in seconds, angular velocity must be measured in radians/second (rad/s).

Example 14.1: Average Angular Velocity of a CD-ROM

A certain manufacturer of CD-ROM drives advertises a speed of 9500rpm (revolutions per minute). What is the angular velocity in rad/s?

Solution
  1. Calculate the angular displacement. If one revolution equals 2 p radians, then

    graphics/14equ02.gif


  2. Convert the time to seconds. 1 minute equals 60 seconds.

  3. Plug those two values into the definition:

    graphics/14equ03.gif


Here is one way to turn this into a function.

 float avgAngularVelocity(float arcStart, float arcEnd, float time, float radius)      {       float initialDisplacement, endDisplacement,omega;         //calculate the angular displacement.         initialDisplacement = arcStart/radius;         endDisplacement = arcEnd/radius;         //apply the formula.         omega = (endDisplacement - initialDisplacement) / time;         return omega;      } 

This particular version will take the beginning and ending points of the arc, the total radius, and the overall time interval to calculate against. This function determines the angular displacement internally, and then applies the formula to determine the average angular velocity.

While this is a useful method, in general it is best to try to find the smallest possible chunk of work that can be done, then write a function to process it. A function like the previous one is useful, but only in the situation where the arc lengths and radius are known. Bear this in mind when designing your own functions.

At this point, we can make a distinction between linear velocity and angular velocity. Looking at the CD-ROM described in Example 14.1, you can see that any point on the CD-ROM moves with the same angular velocity because the CD is a solid (rigid body) object. However, if you look at linear velocity , points on the outside of the CD have to move faster than points close to the center in order to keep up.

So far, we've discussed only average angular velocity. Chapter 8, "Motion in One Dimension," discussed the difference between average linear velocity and instantaneous linear velocity. The same concept applies here as well. The instantaneous angular velocity is simply the average angular velocity over an extremely small time interval, as close to an instant as you can get.

Now let's redefine acceleration in terms of circular motion. Chapter 8 also said that linear acceleration is the rate of change of linear velocity. In other words, linear acceleration is equal to the change in linear velocity divided by time. The same definition carries over to circular motion: Average angular acceleration ( graphics/alpha.gif ) is equal to the rate of change of angular velocity.

Average Angular Acceleration

graphics/14equ04.gif


where w = angular displacement and t = time.


NOTE

The symbol for angular acceleration ( a ) is the Greek letter alpha.


If you look closely at the definition, you can see that the units for angular acceleration are rad/s 2 , which is very similar to the m/s 2 that you're used to seeing for linear acceleration. Also similar to linear acceleration, the instantaneous angular acceleration is the average angular acceleration over a tiny time interval, as close to an instant as you can get.

Example 14.2: Average Angular Acceleration of a Wheel

Suppose you're coding a racing game like Need for Speed Under Ground , and the car's wheels have an angular velocity of 5rad/s when the player pushes the acceleration button. If 2 seconds later the angular velocity is 15rad/s, what was the wheel's average angular acceleration?

Solution

Let's go straight to the definition:

graphics/14equ05.gif


Just as you saw with the CD-ROM, notice that the angular acceleration is the same for every point on the wheel even though the linear acceleration is different depending on how far the point is from the axis of rotation.

To calculate alpha in code, we can use a function that takes the angular displacement and divides by time:

 float avgAngAcceleration(float angVelBegin, float angVelEnd, float time)      {        float alpha;        alpha = (angVelEnd - angVelBegin)/time;        return alpha;      } 

Note here that again, the change in time has been pre-computed and put through as a final value instead of performing the subtraction of start and end times. This function could also calculate the instantaneous angular acceleration if it were passed a sufficiently small time interval. If you need to modify this function to take degree measurements, don't forget to convert!

Now that we've redefined displacement, velocity, and acceleration for circular motion, you can see that basically the same concepts carry over from linear motion. As you might have guessed, the five equations of motion (discussed in Chapter 8) also carry over from linear motion. The only difference is that the variables change slightly.

Displacement goes from D x to D q , velocity changes from v to , and acceleration changes from a to a . The revised equations are listed in Table 14.1.

Table 14.1. The Five Equations of Motion Revisited

Equation 1

w f = w i + a t

Equation 2

graphics/14inl01.gif

Equation 3

D q = ( w i + w f ) t

Equation 4

D q = w i t + a t 2

Equation 5

graphics/14inl09.gif

The process is the same as before. The only difference is that you plug in angular variables instead of linear variables.

Example 14.3: Wheel of Fortune

Suppose you're coding a Wheel of Fortune game. If the player gives the wheel an initial angular velocity of 8rad/s, and the pegs decelerate it at a rate of 2rad/s 2 , what will its angular displacement be after 3 seconds?

Solution
  1. Make a list of everything you know and what you're looking for, just like you did for linear motion:

    Given

    Find

    w i = 8rad/s

    D q = ?

    t = 3s

     

    a = 2rad/s 2

     
  2. Based on this list, you can choose an equation. In this case, equation 4 will work.

  3. Plug in the known values, and solve for D q :

    D q = w i t + a t 2

    D q = (8rad/s)(3s) + (2rad/s 2 )(3s) 2

    D q = 24rad 3rad

    D q = 21rad

Until now, we've focused on the similarities between linear motion and circular motion. The only difference that has been mentioned is the idea that angular velocity and acceleration are the same for all points on a fixed object where linear velocity and acceleration differ based on how far the point is from the center or axis of rotation. Let's take that one step further and look for a relationship between the linear and angular values, starting with linear and angular velocity.

At the beginning of this chapter, we said that angular displacement is equal to the arc length divided by the radius:

graphics/14equ06.gif


To get angular velocity, we must divide by time, so let's divide both sides of that equation by time:

graphics/14equ07.gif


Here's where we take a big jump. Remember that to calculate instantaneous angular velocity, we must use a very small time interval. If we use smaller and smaller time intervals, you'll see that the arc length approaches a straight line. As soon as it becomes that tiny straight line, the linear velocity equals ( D s / D t ). The equation then becomes

graphics/14equ08.gif


If you look back at the CD-ROM shown in Figure 14.2 and picture smaller and smaller time intervals, q becomes smaller, and s approaches a straight line that is tangent to the circular path. That is why the linear velocity at this point is called the tangential velocity , v t . This is shown in Figure 14.3.

Figure 14.3. Tangential velocity at point P.

graphics/14fig03.gif

Tangential Velocity

v t = w r

where w = instantaneous angular velocity and r = radius.


Tangential velocity is also pretty easy to turn into a function for programming purposes. Here is an example:

 float tangVelocity(float omega, float radius)      {        float velT;        velT = omega*radius;        return velT;      } 

This function returns the finalized tangential velocity given omega and time.

One way you can visualize this idea of tangential velocity is to find a ball with a string attached to it, or even a yo-yo. Take the end of the string and swing the ball so that it ends up circling your hand, which is fixed at some location. What do you suppose would happen if the ball came loose from the string? At the instant it detached, the ball would follow a straight-line path tangent to the circular path at the point where it was released, as shown in Figure 14.4.

Figure 14.4. A ball on a string.

graphics/14fig04.gif

Example 14.4: Throwing a Baseball

Suppose you're coding a baseball game like All Star Base . If the player throwing the ball has a forearm that measures 0.5m, and the ball has an instantaneous angular acceleration of 100rad/s when he releases it, what is the ball's initial linear velocity at the beginning of the projectile motion?

Solution

If v t = w r , the tangential velocity must be 100rad/s * 0.5m, which is 50m/s in the direction that is tangent to the curve at the instant the player releases the ball.

You can also derive a relationship between linear and angular acceleration. Let's start with the equation that relates the velocities:

v t = w r

To get acceleration, divide both sides by time:

graphics/14equ09.gif


a t = a r

Tangential Acceleration

a t = a r

where a = instantaneous angular acceleration and r = radius.


Example 14.5: Wheel of Fortune Revisited

Suppose you're back to coding that Wheel of Fortune game. The player gives the wheel an initial angular velocity of 8rad/s, and it takes 10 seconds to come to rest. If the wheel has a 3-meter radius, and one of the pegs on the wheel's exterior comes loose the second the player sets the wheel in motion, what is the wheel's tangential acceleration?

Solution
  1. The first thing you need to find is the angular acceleration. Make a list of what you know:

    Given

    Find

    w i = 8rad/s

    a = ?

    t = 10s

     

    w f = 0rad/s

     
  2. Now you can choose an equation. In this case, use equation 1:

    w f = w i + a t

    0rad/s = 8rad/s + a (10s)

    a = 0.8rad/s 2

  3. Now that you know the angular acceleration, you can find the tangential acceleration:

    a t = a r

    a t = 0.8rad/s 2 (3m)

    a t = 2.4m/s 2

As you can see, the basic definitions of displacement, velocity, and acceleration carry over from linear motion to angular motion. Linear motion is based on translation, whereas angular motion is based on rotation. Remember that for rigid bodies the angular velocity and acceleration are consistent for all points on the object, but linear velocity and acceleration vary depending on the distance from the center. The next section uses these definitions to investigate what causes rotational motion.

Self-Assessment

1.

If a merry-go-round completes 10 revolutions in 30 seconds, what is its average angular velocity in rad/s?

2.

If a merry-go-round has an initial angular velocity of 5rad/s when a parent gives it a good push to speed it up to 15rad/s in 3 seconds, what is the angular acceleration due to the push?

3.

As soon as the merry-go-round hits its top angular velocity of 15rad/s, it starts to slow down at a rate of 3rad/s 2 . At that rate, how long will it take to come to a complete stop?

4.

As soon as the merry-go-round hits its top angular velocity of 15rad/s, it starts to slow down at a rate of 3rad/s 2 . What is its total angular displacement when it finally comes to rest?

5.

One of the children on the merry-go-round is not holding on tight enough. If the radius of the merry-go-round is 2.5m, and the child falls off the edge at the maximum angular velocity of 15rad/s, what is his new linear (tangential) velocity?


 < Day Day Up > 


Beginning Math and Physics for Game Programmers
Beginning Math and Physics for Game Programmers
ISBN: 0735713901
EAN: 2147483647
Year: 2004
Pages: 143
Authors: Wendy Stahler

flylib.com © 2008-2017.
If you may any questions please contact us: flylib@qtcs.net