Topic 1  Rotational Dynamics

In the exam, you are expected to know about:

• Concept of the moment of inertia;

•  

• Use

• Factors affecting the energy storage capacity of a flywheel
  

• Use of flywheels in machines

• Use of Equations for uniformly accelerated motion:

• Torque and angular acceleration.

• Use of

Introduction to Rotational Dynamics

When we looked at objects moving in a circle in Module 4 we only considered small objects moving around a central point.  The objects themselves were small compared to the radius of their orbit.  Examples included model aeroplanes tethered by a string, or planets moving around a star.  These objects were held to their paths by centripetal force and if that force were stopped, the objects would fly off tangentially in a straight line.

These ideas don't work for rotating systems in which an object is spinning on its axis.  Examples of these include:

• rotors of electric motors;

• flywheels.

You will come across quite a lot of odd looking symbols.  Don't worry; they are Greek letters which are used as Physics codes.  A lot of the equations are identical in concept to the equations of linear motion (motion in a straight line) which you came across in Module 2.

Let's look at the odd symbols and what they mean:

We will now have a look at some of the terms used in rotational dynamics:

Angular velocity is an important quantity.  Suppose we had a record deck turntable turning at 33 rpm (revolutions per minute).  Clearly an object at the rim of the turntable moves with a higher linear speed that an object towards the centre.  However both move through the same angle every second.  We call this the angular velocity, which has the physics code w.  We don't use degrees per second, but radians per second.  Go back to Module 4 if you are not sure what a radian is. (1 rad » 57^o)  Radians are dimensionless units, so are ignored in unit analysis.  Some purists say they should be left out altogether.  In these notes I will always include them.

Every revolution, all parts of the object turn through 2p radians.

A common way of expressing the rate of turning is revolutions per minute.  All the equations we will use need the rate of turning (angular velocity) to be in radians per second.  Therefore be a good chap and convert the revolutions per minute to radians per second.

Angular velocity = 2p(rpm ÷ 60)

Question 1 What is the angular velocity of a hi-fi record deck playing at 33 rpm? ANSWER

Angular displacement is simply the angle turned in any given direction.  It is given the code q and is measured in radians. 

 If the angular velocity is changing, it is of course accelerating.  So we have a term angular acceleration, given the code a.

Question 2  What do you think the units are for angular acceleration?  ANSWER

So, just like (linear) velocity = displacement ÷ time, we can write:

angular velocity = angular displacement ÷ time

And just like acceleration = change in velocity ÷ time interval we can write

angular acceleration = change in angular velocity ÷ time interval

Question 3  It takes a motor 5 seconds to accelerate from rest to 3000 rpm.  What is its angular acceleration?  ANSWER

All these are vector quantities, but the directions are pretty easy, clockwise or anticlockwise.  No horrible sines or cosines.  So far pretty easy, what?  Now it gets a bit harder...

Moment of Inertia

You will know that any mass has a reluctance to move, which we call inertia.  In linear dynamics, we say that all objects are point masses.  We have an analogous situation in rotational dynamics which we call the moment of inertia. The moment of inertia is the measure of the opposition of a rotating body to angular acceleration.  It is given the physics code I and its units are kg m². 

Suppose we have a disc radius r spinning with angular velocity w rad/s.  We can think of it as made up of lots of little point masses.  We know that the linear speed of each point is v, where v = wr.  We also know that kinetic energy = 1/2 mv².  Each little point mass m therefore has a kinetic energy = 1/2 m(wr)².

The total kinetic energy can be found by adding up all the kinetic energies of the little point masses:

We can rewrite this as:

And we can rewrite this as:

The term Smr² is the sum of all the terms mr².  This term is described as the moment of inertia and is given the code I.

In the exam you are NOT expected to then go on to integrate this for particular conditions.  The appropriate relationship will be given to you, or the moment of inertia will be given to you already worked out.

For a circular disc of mass M and radius r:

For a solid cylinder, the relationship is as above.  For a hollow cylinder open at both ends, the moment of inertia is:

For a solid sphere, mass M and radius r the moment of inertia is:

Question 4 Each of the objects above has a mass of 2.5 kg and a radius of 0.2 m.  What is the moment of inertia for each one?  ANSWER

In the past papers I have not yet seem the first or third of these formulae, but I have seen a question that got you to use I = Mr², ignoring the sigma bit.  Other times you have been asked to find I from either the torque or the kinetic energy.

The key thing to remember is that the moment of inertia is the rotational equivalent of mass.

Kinetic Energy

The kinetic energy of a spinning object is given by:

[I - moment of inertia (kg m²); w - angular velocity (rad/s)]

Like kinetic energy in linear motion, rotational kinetic energy is a scalar, even though the angular velocity is a vector.

Question 5 

A flywheel battery can be used in place of lead-acid batteries to provide a short-term electrical power supply when mains power fails. Energy is stored as rotational kinetic energy in a rapidly spinning rotor, which is driven up to speed by a mains-powered motor. To recover the energy, the motor is operated in reverse as a generator driven by the spinning rotor.

The rotor of a flywheel battery is a thin-walled circular cylinder of mass 160 kg and mean radius

0.34 m, which can be rotated at a maximum safe angular speed of 44 000 rev min^-1.

Calculate:

(a) the moment of inertia of the rotor about its own axis,

(b) the rotational kinetic energy stored in the rotating rotor when it spins at 44 000 rev min^-1.

(AQA past question)

ANSWER

To show you that this is not a figment of the examiner's imagination, the picture below shows a flywheel battery for a space station.

And here's another couple for good measure:

This idea is not new.  Third-rail electric locomotives had a motor-generator to help them trundle over gaps in the electric conductor rail.

Many machines have a flywheel which is a heavy lump of metal spinning on an axis. A single cylinder petrol engine needs a flywheel to keep it running smoothly.  If it didn't have one, it would stall on the compression stroke.  Four-cylinder car engines would still be jerky without a flywheel.  Twelve cylinder car engines have less need for a flywheel, but still have one because:

• It is useful to put the toothed ring for the starter motor onto it;

• It is a useful face for a clutch.

What is the best design for a flywheel?  The obvious shape is a solid disk of steel.  The diagram shows a couple of cross-sections.  Both flywheels have the same mass and mean radius.  Both have a small flange of very small mass so that they can be bolted to the shaft.

• A solid  disk flywheel:

• A flywheel that has a thin plate in the middle and most of its mass as a ring around the outside:

Question 6  (Harder)  By considering the amount of kinetic energy each one can store, which one is the better design?  State what assumptions you make.  ANSWER

The flywheel battery in question 5 would have a flywheel of the second design.

Equations of Rotational Motion

There are four of these, each of which has an equivalent of linear motion, and are used in exactly the same way.  The only difference is the code for some of the terms, and if you are at ease with angular velocity, angular displacement, and angular acceleration, you will have no difficulty at all with these.  If you have forgotten about the linear equations of motion, go back to Module 2 and revise them.

This table shows the terms and what they mean:

(1) Used to link a second angular velocity to the first, acceleration and time.

It is used just like v = u + at

Question 7  A car wheel is being tested for balance.  It is spinning at 750 rpm and the machine then accelerates it to 1500 rpm over a period of 3 s.  What is the angular acceleration?  ANSWER

(2)  Used to link the angular displacement to the angular velocity at the start , with the time, and the angular acceleration.

It is used just like s = ut + 1/2 at².

Question 8  Use your result from question 7 to work out the angular displacement through which the wheel turns as it accelerates from 750 rpm to 1500 rpm.  ANSWER

(3) Used to link the angular velocity at end with the first angular velocity , the  angular displacement and the acceleration.

It is used just like v² = u² + 2as.

Question 9  In a recent experiment to test a "crash-proof" fuel, an aeroplane was deliberately crash-landed onto spikes which were designed to slice the fuel tanks open.  The test was a failure because the fuel burst into flames.  The reason for this was because one of the engines was stopped by a spike.  The turbine was running at 30 000 rpm and was stopped within 1/3 revolution.  The kinetic energy of the engine was dissipated as intense heat which acted as an ignition source for the fuel.

What was the angular deceleration suffered by that engine?

ANSWER

(4) This equation is used to link the angular displacement to the average of two angular velocities and the time interval:

It is used just like s = (u + v)t/2

Question 10  A car wheel is being tested for balance.  It is spinning at 750 rpm and the machine then accelerates it to 1500 rpm over a period of 3 s.  Use the above equation to work out the angular displacement. How does your answer compare to the answer you worked out in Question 8?   ANSWER

Torque and Angular Acceleration

You will be familiar with F = ma.  Rotational motion has an identical relationship.  You will remember that a torque is a turning force.  

The diagram shows a cord wrapped around a pulley (like the pull-cord used to start a lawn mower).  The pulling force F causes a torque T which can easily be worked out by:

T = Fr

If we double the pull, we will double the torque which means that the angular acceleration will be doubled.  However if we look at the mass, we find that it's not the only factor.  The shape of the pulley is important, so instead of mass, we use the moment of inertia.

So if we apply a torque to a rotating body, it will undergo angular acceleration.  In other words it will spin faster.  The angular acceleration due to a torque is given by:

T = Ia

[ T - torque (Nm); I - moment of inertia (kg m²);  a - angular acceleration ( rad/s²)]

Question 11  The wheel of a large dumper truck has a moment of inertia of 10 000 kg m² and is being tested by being rotated at 60 rpm.  It is brought to rest in a time of 40 s.

(a) What is the initial angular velocity in rad/s?

(b) What is the angular acceleration?

(c) What is the angular displacement in the first 20 s?

(d) What is the applied torque?  ANSWER

Let's sum up by comparing linear motion with rotational motion:

We will compare equations:

Linear Motion	Rotational Motion

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