Module : Rotational Kinematics

Helena Dedic, Kreshnik Angoni and Kevin Lenton

Module RotationalKinematicsWorksheet.docx
Module RotationalKinematicsWorksheet.pdf

Learning Objectives

After reading this page, watching the videos and reviewing the exercises, you will be able to:

• Understand the definitions of rotational motion: angular displacement, angular velocity and angular acceleration.
• Convert between linear and rotational variables.
• Recognize the analogy between linear and angular equations of motion.
• Calculate the motion of a rotating object, in the special case of constant angular acceleration.

ROTATIONAL KINEMATICS

This module concerns Rotational Motion. In rotational motion, rigid objects rotate around an axis passing through the object or close to it. In this type of motion you cannot model the whole object as a single material point because each point that makes up the object if moving differently. You have to think of the object as made up of a system of particles that rotate together. Each particle in the object is moving on its own circular path centered on the rotation axis.

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Angular Displacement

Consider the rotating object shown in Figure 2.

In particular, consider the red point at A. It starts out at some angle, which could be called the initial angular position $\theta _{i}$. Of course you need to define some angular origin, which is most often the +x axis. After a certain time the red point at A has rotated around to point B. This angle could be called the final angular position $\theta _{f}$.
You could define the angular displacement, that is to say the change in angular position as:

$\Delta \theta =\theta _{f}-\theta _{i}$

You should be noticing that this is exactly like the definitions you used for linear position and displacement, except you are dealing with angles $\theta$ rather than linear positions $x$.

Units: rad (radians)

Note: the displacement is often described as a number of revolutions (rev)

$1rev=2\pi$ rad
Note as well that the whole object has rotated about this $\Delta \theta$. This means that each point in the object has also rotated through this angle.

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Converting between Angular and Linear Displacements

Looking at Figure 2, the red point has moved on a circular path s. This path s is a distance, not an angle.

Because you are dealing with circles, you can relate s, r (the radius) and the angle $\theta$ by
$\Delta s=r\Delta \theta$

This equation is simply a formula you already know
$Arc\ length\ of\ circle=radius\times \ angle\ subtended$

Or, for a full circle:
$C=2\pi r$

Any angular measurement can be converted to a linear measurement, by multiplying by a radius.
This means that for a small radius, a point travels a certain angle but travels a proportionally small linear path. For a big radius, a point travels through the same angle, but travels a big linear distance because it is far from the axis.

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Still Vectors!

In fact, the angular measurements are still vectors because they have a sense or direction of rotation. The direction is defined by the right hand rule (see NYB). But for this course, the angular displacement is positive if the particle moves in a counterclockwise direction and it is negative if the particle moves in a clockwise direction.

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Angular Velocity

In Figure 2, the red point rotated $\Delta \theta$ to point B in a time interval $\Delta t$.
The average angular velocity $\omega _{average}$ is defined as:

$\omega _{average}={\frac {\theta _{f}-\theta _{i}}{\Delta t}}={\frac {\Delta \theta }{\Delta t}}$

The instantaneous angular velocity $\omega$ is defined as:

$\omega =lim_{\Delta t\rightarrow 0}{\frac {\Delta \theta }{\Delta t}}={\frac {d\theta }{dt}}$

That is to say, angular velocity is the rate of change of angular displacement with respect to time. Basically, the faster the angular velocity the faster the object is rotating.

Units: rad/s and sometimes rpm= revolutions per minute

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Conversion of units for Angular Velocity

The angular velocity is sometimes expressed in rpm (revolutions per minutes, or revs per minute)
That is to say $1\ rpm={\frac {2\pi }{60}}{\frac {rad}{s}}$

Another variable that is used to describe constant angular motion is the Period $T$.
The Period $T$ is the time to complete 1 revolution. Since the speed is constant in uniform circular motion, the angular velocity $\omega$ is also constant and we can write:

$\omega ={\frac {2\pi }{T}}={\frac {Angle\ in\ 1\ rev}{Time\ for\ 1\ rev}}$

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Relationship between the angular velocity and the linear speed of the particle

Because you are dealing with circles, you can relate v, r (the radius) and the anglular velocity $\omega$ by
$v=lim_{\Delta t\rightarrow 0}{\frac {\Delta s}{\Delta t}}=lim_{\Delta t\rightarrow 0}{\frac {r\Delta \theta }{\Delta t}}=rlim_{\Delta t\rightarrow 0}{\frac {\Delta \theta }{\Delta t}}=r\omega$

Or

$v=r\omega$

Any angular measurement can be converted to a linear measurement, by multiplying by a radius. This means that for a small radius, a point can have a certain angular velocity but has a proportionally small linear velocity. For a big radius, a point can have the same angular velocity, but has a big linear velocity because it is far from the axis.

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Angular Acceleration

When a particle is in non-uniform circular motion its angular velocity also varies. The rate of change of the angular velocity is called angular acceleration $\alpha$:

$\alpha =lim_{\Delta t\rightarrow 0}{\frac {\Delta \omega }{\Delta t}}={\frac {d\omega }{dt}}$

Units: $rad/s^{2}$ or rad/s/s e.g., 3 rad/s/s implies that the angular velocity increases 3 rad/s every second.

The relationship between angular acceleration and linear (tangential) acceleration:

$a_{t}=lim_{\Delta t\rightarrow 0}{\frac {\Delta v}{\Delta t}}=lim_{\Delta t\rightarrow 0}{\frac {\Delta (r\omega )}{\Delta t}}=rlim_{\Delta t\rightarrow 0}{\frac {\Delta \omega }{\Delta t}}=r\alpha$

$a_{t}=r\alpha$

Any angular measurement can be converted to a linear measurement, by multiplying by a radius. This means that for a small radius, a point can have a certain angular acceleration but has a proportionally small linear acceleration . For a big radius, a point can have the same angular acceleration , but has a big linear acceleration because it is far from the axis.

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Equations of Rotational Motion

You have learned that if linear acceleration a is constant, the v vs t graph will be a straight line with slope equal to a. The same rationals apply to rotation motion. If the angular acceleration is constant the $\omega$ vs t graph is a straight line with a slope equal to $\alpha$. We can use this graph to solve problems involving motion with constant angular acceleration.

In fact you can write equations of motion which are directly analagous to the linear motion equations:

	Linear Equations of Motion (X Axis)	Rotational Equations of Motion
Equation 1	$v_{f_{x}}=v_{i_{x}}+a_{x}t$	$\omega _{f}=\omega _{i}+\alpha t$
Equation 2	$x_{f}-x_{i}=v_{i_{x}}t+{\frac {1}{2}}a_{x}t^{2}$	$(\theta _{f}-\theta _{i})=\omega _{i}t+{\frac {1}{2}}\alpha t^{2}$
Equation 3	$v_{f_{x}}^{2}=v_{i_{x}}^{2}+2a(x_{f}-x_{i})$	$\omega _{f}^{2}=\omega _{i}^{2}+2\alpha \Delta \theta$
Equation 4	$(x_{f}-x_{i})=+{\frac {1}{2}}(v_{f_{x}}+v_{i_{x}})t$	$(\theta _{f}-\theta _{i})=+{\frac {1}{2}}(\omega _{f}+\omega _{i})t$

Video Resources

Part I ;Review of Above information

Part II; with examples

Exercises

Example 1

At t = 0 a flywheel is rotating at 50 rpm. A motor gives it a constant acceleration of $0.5\ rad/s^{2}$ until it reaches 100 rpm. The motor is then disconnected and the flywheel rotates at constant angular velocity. How many revolutions are completed at t = 20 s?

Solution:

There are a couple of ways to approach this problem. You can work with motion graphs, or you can work with Rotational Equations of Motion. For completeness, both methods are shown, but probably you will want to use the equations of motion.
First, we convert the angular velocity to rad/s (use the conversion factor found above):

$\omega _{i}=50\ rpm({\frac {\frac {2\pi \ rad}{60\ s}}{1rpm}})=5.23\ rad/s$

$\omega _{f}=100\ rpm({\frac {\frac {2\pi \ rad}{60\ s}}{1rpm}})=10.47\ rad/s$

Note that the motion is in two parts:
Part 1: It accelerates with an angular acceleration $\alpha =0.5\ rad/s^{2}$ for a certain time until Part 2: it reaches the required angular velocity, and then it continues at a constant angular velocity for the remaining time

Graphical Method:

The acceleration lasts until the angular velocity reaches 10.47 rad/s. After that the rotor continues to spin with constant angular velocity. We can sketch the $\omega$ - t graph. It begins at 5.23 rad/s. The graph is a straight line with slope 0.5 rad/s/s. When the angular velocity reaches 10.46 rad/s the graph becomes a horizontal line. See the graph below:

To determine the angle through which the rotor turns we have to determine the area under the graph. First we need to know when the rotor reaches the angular velocity of 10.46 rad/s.
You will need to use a kinematic equation here.
Make a table of what you know.

Parameter	Values for Problem
$\Delta \theta$	$?$
$\omega _{i}$	$5.23\ rad/s$
$\omega _{f}$	$10.47\ rad/s$
$\alpha$	$0.5\ rads^{-2}$
$t$	$?$

The equation which links the known and unknown parameter (time) is:
$\omega _{f}=\omega _{i}+\alpha t\Rightarrow \ \ \ t={\frac {\omega _{f}-\omega _{i}}{\alpha }}={\frac {10.46-5.23}{0.5}}=10.46\ s$

Therefore in Part 2 the flywheel moves at constant angular velocity for 20 - 10.46 = 9.54 s. The angular displacement is equal to the total area is the area of a triangle plus the area of two rectangles: ½ × 5.23 × 10.46 + 5.23 × 10.46 + 10.47 × 9.54 = 182 rad.

Now we have to convert the angular displacement in rads to revolutions:

$\Delta \theta =182rad{\frac {1rev}{2\pi rad}}=28.9rev$. 

Equation of Motion Method

To find the angular displacement $\Delta \theta$ for the acceleration phase Part 1.
Do an updated table:

Parameter	Values for Problem
$\Delta \theta$	$?$
$\omega _{i}$	$5.23\ rad/s$
$\omega _{f}$	$10.47\ rad/s$
$\alpha$	$0.5\ rads^{-2}$
$t$	$10.46$

Find the right equation:
Pretend for a minute, you have not yet found the time. You could use:
$\omega _{f}^{2}=\omega _{i}^{2}+2\alpha \Delta \theta$
${\frac {\omega _{f}^{2}-\omega _{i}^{2}}{2\alpha }}=+\Delta \theta$
Substitute in:
$\Delta \theta ={\frac {10.47^{2}-5.23^{2}}{2\times 0.5}}=82.1rad$
Or, if you already have time use:
$(\theta _{f}-\theta _{i})=+{\frac {1}{2}}(\omega _{f}+\omega _{i})t$
Substitute:
$(\theta _{f}-\theta _{i})=+{\frac {1}{2}}(10.47+5.23)10.46\ =\ 82.1\ rad$

Now for Part 2, which has constant angular velocity.

Parameter	Values for Part 2
$\Delta \theta$	$?$
$\omega _{i}$	$10.47\ rad/s$
$\omega _{f}$	$10.47\ rad/s$
$\alpha$	$0\ rads^{-2}$
$t$	$9.54\ s$

The equation you need is (remember $\alpha =0$):
$(\theta _{f}-\theta _{i})_{Part\ 2}=\omega _{i}t+0$
$(\theta _{f}-\theta _{i})_{Part\ 2}=10.47\times 9.54=99.88rad$
Adding the Part 1 and Part 2 together gives $(\theta _{f}-\theta _{i})_{total}=99.88+82.1=182rad$
Convert to revolutions as above.

Exercise 2

A wheel starts from rest and accelerates uniformly. As the rate of rotation changes from 20 to 50 rpms the wheel makes 40 revolutions. Find the angular acceleration and the number of revolutions completed by the time the wheel reaches 20 rpms.

Solution:

The wheel starts from rest with a constant angular acceleration.

At some time the wheel rotates at $\omega _{1}=20rpm=20*2*3.14/60=2.1rad/s$.

$\Delta \theta =40rev=40*2*3.14=251.2rad$ later the wheel rotates at $\omega _{2}=50rpm=50*2*3.14/60=5.2rad/s$. What is the angular acceleration?

For this kind of problems (given two angular velocities and the angular displacement) we should use the last equation of kinematics:

$\omega _{2}^{2}=\omega _{1}^{2}+2\alpha \Delta \theta$  

Substituting into this equation we find

$5.2^{2}=2.1^{2}+2\alpha *251.2$ 

Solving for $\alpha$ we find $\alpha =0.045rad/s^{2}$.

We can use the same equation to find the angular displacement from the beginning $\omega _{0}=0$ until the angular velocity becomes 2.1 rad/s:

$2.1^{2}=0+2*0.045*\Delta \theta$ 

Solving for $\Delta \theta$ we find $\Delta \theta =48.9rad=48.9rad*1rev/(2*3.14rad)=7.8rev$.

Video of Worked Example 1

Video of Worked Example 2