Balance and Rotation - Lesson 1 - Rotational Kinematics

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Discovering the Links

A chain is made up of a series of links.  What do these links do?  In general, they connect things.  In this part of our lesson, we are going to discover three very helpful links.  What will our links do?  They will help us connect rotational quantities that describe an object’s motion to their corresponding translational quantities. 

In the previous section we saw that when a point on the rim of a spinning wheel rotates, we described its angular velocity with the variable ω.  We learned that we often measure ω in units of radians/sec. But doesn’t this point also have a translational (linear) velocity v that we might measure in m/sec?  Indeed, it does.  It turns out that these two velocities—the angular velocity ω and the translational velocity v—are related.  We can find one if we know the other.

Similarly, during an interval of time, the point on the rim of the wheel rotated through an angle ∆θ.  We called this the angular displacement and measured it in units of radians.  But doesn’t this black dot on the rim of the wheel below also travel some distance on a curved path along the circumference of the circle that we could measure in meters?  It does.  We’ll call this distance ∆s. Once again, the angular displacement ∆θ and the distance along the curved arc ∆s are related.

An image showing the relationship between linear acceleration velocity and distance to rotational acceleration velocity and angle.  A circle is shown with the initial position at 0 degrees, and a final position aprox 45 degrees.  There is a red Delta Theta degree on the arc and a red Delta X to show the distance.  A Green letter a acceleration arrow is pointing to the top left, and a curved alpha acceleration arrow is showing the angular acceleration.  Lastly a larger blue V arrow is pointing to the top left to show the velocity, and a curved blue line with omega showing the angular velocity.

We might also say that, if the spinning dot experiences an angular acceleration (that we could measure in radians/s2), then it will also have a translational acceleration (measured in m/s2).  In every case, the rotational quantity of a given point that is rotating is related to its corresponding translational quantity.

We sometimes call the translational velocity of the rotating object its tangential velocity.  This makes sense because this velocity is always tangent to the circular orbit.  Similarly, we sometimes call the translational acceleration of the rotating object its tangential acceleration.  You can see from the picture that it also points tangent to the circle.

The 'Link' Equations

How do we connect these rotational and translational quantities?  We do so with the radius—how far the rotating point is from the axis.  We can now write three ‘link equations’ that will take us from the rotational quantity to its corresponding translational quantity.

The three equations to link linear to rotational motion.  For Distance (or displacement), Delta s (change in distance) equals r (radius) times delta theta (change in angle).  For Velocity, v (velocity) equals radius time omega (angular velocity).  For Acceleration, a (acceleration) equals radius times alpha (angular acceleration).

In our previous section we noted that physicists often prefer radians to degrees because using radians offers a more convenient way to move between the rotational and translational quantities.  We can now see why by considering the distance link equation above.  Let’s say the blue dot, which is 1 m from the axis of rotation, has rotated through an angle of ∆θ =1 radian.  Then the distance it traveled (∆s) is 1 m—the same as the radius.  This makes sense since this was our definition of a radian from the previous section.   Now let’s look at the red dot that also rotated through an angle ∆θ =1 radian.  Let’s say the red dot is 2 m from the axis of rotation.  We could say that r = 2 m, but more precisely we mean r = 2 m/radius or 2 m/rad.  Because of how we define a radian, the ‘radius’ can be replaced with ‘radians.’  Thus, the distance traveled (∆s) by the red dot is 2 m—again, the same as the radius.  The radius essentially ‘scales up’ the angular quantity to allow us to determine the translational quantity.  The bigger the radius, the bigger the distance traveled along the arc.  And since we had measured the angle of rotation in radians, we see how the units conveniently cancel out.

A circle with 1 meter and 2 meter radius markers and the calculation to find the distance traveled for a 1 radian angle.  The equation shows change in distance = radius times change in angle.  The change in angle is either 1 meter per radian or 2 meters per radian depending on the radius, and since it moves 1 radian, we are left with 1 meter for the 1 meter radius and 2 meters for the 2 meter radius distance traveled.

Let’s look at three examples—one for each of the link equations—to see how this works.

Example 1: Finding Arc Length

Problem:  Three colored dots (green, blue, and red) are fixed to a disk that is allowed to rotate as shown.  Each colored dot is a different distance from the axis of rotation. Complete the table below by determining the angular displacement and arc length (or distance) for each.

A circle with 3 radius outlined, green 0.20 meters, blue 0.4 meters, and red 0.6 meters.  Each start at 0 and make an arc of pi radians (180 degrees).  A table is shown to solve, with rows of Green Blue and Red, and Colums of Angular Displacement in Radians and Arc Length in Meters.

Solution:  All three dots had the same angular displacement as they all rotated 3.14 radians.  Using ∆s = r ∆θ, we find that the greater the radius, the greater the arc length traveled. 

Solution equations for the 3 radius.  Each show the equation delta s (change in distance) = radius time delta theta (change in angular displacement), which for each is 3.14 radian times the radius (0.2, 0.4, and 0.6) to give the answer 0.63 meters, 1.26 meteres, and 1.89 meters respectively.The table from the example's image filled out.  All 3 have a angular displacement of 3.14 radians.  The lengths are 0.63 meters for green, 1.26 meters for blue, and 1.89 meters for red.

Example 2: Finding Tangential Velocity

Problem:  A child sits 1.5 m from the center of a merry-go-round that is spinning with an angular velocity of 60o/sec.  How fast is the child traveling (in m/s) relative to the ground?

Solution:  First, let’s convert 60o/s to rad/s.  Doing so leads to an angular velocity of 1.05 rad/s.  Next, we can think of the radius as 1.5 m/radius or 1.5 m/rad.  Since we converted the angular velocity to rad/s, this will allow the ‘rad’ units to cancel.  Finally, we’ll use our velocity link equation to determine the magnitude of the tangential velocity to be 1.58 m/s.

The solution to the example.  First the angular velocity (omega) is calculated in radians by taking 60 degrees per sec times pi radians over 180 degrees to get 1.05 radians per second.  T hen multiplying that by the radius 1.5 meters per radian to get 1.58 meters per second = tangential velocity.

Example 3: Finding Tangential Acceleration

Problem:  A centrifuge is used in laboratories to separate substances (often liquids) based on density by rotating rapidly. When turned on, a centrifuge can go from zero to 350 revolutions/sec in 12 seconds.  (a) Find the average angular acceleration of the centrifuge during the start-up process, and (b) find the tangential acceleration of the liquid that is 5.0 cm from the axis of rotation during this start-up process?

Solution:  (a) First, we’ll find the final angular velocity in rad/s.  Next, we’ll use our average angular acceleration equation to find that the angular acceleration during start-up is 183 rad/s2.  (b) We can use our acceleration link equation to determine the tangential acceleration of the liquid that is 5.0 cm away from the axis to be 916 cm/s2 .

The equations and calculations to solve example 3.  First we take 300 revolutions per second multiplied by 2 pi radian per revolution to get about 2199 radians per second (final velocity).  Final velocity (2199) minus initial velocity (0) over 12 seconds gets us about 183 radians per second square.  We take that value (actually 183.26) multiplied by the radius (5 centimeters per radian) to get about 916 cm per second squared tangential acceleration.

 

We Would Like to Suggest ...

Sometimes it isn't enough to just read about it. You have to interact with it! And that's exactly what you do when you use one of The Physics Classroom's Interactives. We would like to suggest that you combine the reading of this page with the use of our Rotational Motion Interactive. You can find it in the Physics Interactives section of our website. The Rotational Motion Interactive allows a learner to explore the relationship between a rotating bug’s angular velocity and its tangential (linear) velocity.

Check Your Understanding

Use the following questions to assess your understanding. Tap the Check Answer buttons when ready.

1. A red dot on the edge of a wheel or radius 'r' makes one complete revolution.  What distance did the red dot travel?

A tan circle with radius r, and a red dot at position 0 with an arrow going counter clockwise.

Check Answer

2. At track practice, your teammate says, “I want to run in lane 4 because I get a head start and have to run less distance than those in the inner lanes.”  What is the flaw in your teammate’s reasoning?

Picture of the right section of a school track, with 4 lanes.  Each lane that goes further out has a later starting point, with 4 being the farthest ahead in the starting point.

Check Answer

3. The tub of your washing machine has a diameter of 0.70 m.  During the spin cycle, an article of clothing is pinned against the wall of the tub as it makes 5 revolutions/sec.
(A) What is the angular velocity (in rad/s) of the tub during the spin cycle?
(B) How fast (in m/s) is the article of clothing moving during this cycle?

Check Part A Answer

Check Part B Answer

4. Two people (A and B) are standing on a spinning merry-go-round.  Person A says, “I have a bigger velocity since I am further from the center.”  Person B says, “No, we have the same velocity since we both revolve around the center in the same amount of time.”  Who is correct?

 

Check Answer

5. The figure below shows a top view of a ball attached to a string traveling in a circle.  The ball orbits with an angular velocity of ω = 4.0 rad/s.  The string is 1.2 m long.  The cardinal directions (N, S, E, and W) are shown for reference.

A circle with compass directions, a line (string) with a ball (black dot) on East and a blue arrow (omega) pointing counter clockwise.

(A) If the string were to break at the moment the ball is in the position shown, what would be the direction of its subsequent motion?
(B) What would be the ball’s velocity (in m/s) immediately after the string breaks?

Check Part A Answer

Check Part B Answer

6. The radius of the Earth is 6.4 x 106 m.  If you were standing on the equator of the Earth:
(A) how long would it take for you to make one complete rotation?
(B) what is the magnitude of your angular velocity?
(C) what is the magnitude of your tangential velocity?

Check Part A Answer

Check Part B Answer

Check Part C Answer

7.  A large warehouse fan has blades that extend out from the center 1.5 m.  What angular acceleration would be necessary so that a point on the rim of the blades has a tangential acceleration equal to the acceleration of gravity on Earth (9.8 m/s2)?

Check Answer

Looking for additional practice?  Check out the Calc Pad for additional practice problems.


 
Figure 1 from MS Word
Figure 2 Picture borrowed from: https://commons.wikimedia.org/wiki/File:202407_Desktop_Centrifuge_Female.svg
Figure 3 author created with person icons from MS Word


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