determine the moment of inertia of flywheel experiment

Experiment: Determination of Moment of Inertia of a Fly Wheel

Experiment: Determination of moment of Inertia of a Fly Wheel

Theory: The flywheel consists of a weighty round disc/massive wheel fixed with a strong axle projecting on either side. The axle is mounted on ball bearings on two fixed supports. There is a little peg on the axle. One end of a cord is loosely looped around the peg and its other end carries the weight-hanger.

Suppose, the angular velocity of a wheel is ω and its radius r. Then lineal velocity of the wheel is, v = ωr. If the moment of inertia of a body is I and the wheel is rotating around an axle.

Then its rotational kinetic energy, E = ½ Iω 2 .

Apparatus: An iron axle, a heavy wheel, some ropes, a mass, stopwatch, meter scale, slide calipers.

Description of the apparatus:

The flywheel was set as shown with the axle of the flywheel straight or parallel. A polystyrene tile was placed on the floor to avoid the collision of the mass on the floor.

(1) First of all, let us measure the radius of the axle by a slide caliper.

(2) Then for the determination of a number of rotation a mark by chalk is put on the axle and a rope is wound on the axle. At the other end of the rope a mass m is fastened and if it is dropped from position R, the wheel after rotating a few times, the weight with the rope will fall to position S. The wheel makes m 1 number of rotation to touch the point S and time for this drop is noted from the stopwatch.

Now the rope is again wound on the axle and the mass is fastened on the other end of the rope. From position R the mass is allowed to fall to the ground and as soon as it touches the ground, the stopwatch is started. When the axle comes to rest the stop wealth is stopped. Total time and the number of rotation of the wheel before it comes to rest are noted i.e., a total number of rotation (n 2 ) as noted.

Table 1: radius (r) of the axle B

Table 2: Determination of time and number of rotation

Calculation : If the axis takes time t for n 2 number of rotation, the average angular velocity,

ω 2 = (2πn 2 )/t

The axle acquires zero velocity with uniform retardation from angular velocity ω, so its average angular velocity,

ω 2 = (ω + 0) / 2 = ω/2

or, (2πn 2 )/t = ω/2

or, ω = 4πn 2 rad S -1

Then, I = (2mgh – mω 2 r 2 ) / ω 2 (1+ n 1 / n 2 ) = ….. g.cm 2 = ….. Kg.m 2

By inserting the value of n 2 , ω can be found out. By increasing the values of m, ω, r, h, n 1 , n 2 and g in an equation; the moment of inertia of the heavy wheel can be found out.

Precautions:

In the axle, a rope is to be wounded in such a way that while unwinding from the wheel it can easily drop on the ground.

• There should be the least friction in the flywheel.
• A number of rotation n and time t is to be unwired correctly.
• The length of the string should be less than the height of axle from the floor.
• Height ‘h’ is to be measured from the mark on the axle.
• ‘h’ is to be measured correctly.
• There should be no kink in string and string should be thin and should be wound evenly.
• The stopwatch should be started just after detaching the loaded string.

Applications: The main function of a flywheel is to maintain a nearly constant angular velocity of the crankshaft.

• A small motor can accelerate the flywheel between the pulses.
• The phenomenon of precession has to be considered when using flywheels in moving vehicles.
• Flywheels are used in punching machines and riveting machines.

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• IIT JEE Study Material
• Moment Of Inertia

Moment Of Inertia Of Flywheel

Moment of inertia of a flywheel is calculated using the given formula;

Where I = moment of inertia of the flywheel.Here, the symbols denote;

m = rings’ mass.

N = flywheel rotation.

n = number of windings of the string.

h = height of the weight assembly.

g = acceleration due to gravity.

r = radius of the axle.

Or, we can also use the following expression;

Flywheels are nothing but circular disc-shaped objects which are mainly used to store energy in machines.

Determining The Moment Of Inertia Of Flywheel

To determine the moment of inertia of a flywheel we will have to consider a few important factors. First, we have to set up a flywheel along with apparatus like a weight hanger, slotted weights, metre scale and we can even keep a stopwatch.

Then we make some assumptions. We will take the mass as (m) for the weight hanger as well as the hanging ring. The height will be (h). Now we consider an instance where the mass will descend to a new height. There will be some loss in potential energy and for which we write the equation as;

P loss = mgh

Meanwhile, there is a gain in kinetic energy when the flywheel and axle are rotating. We express it as;

K flywheel = (½) Iω 2

I = moment of inertia

ω = angular velocity

Similarly, the kinetic energy for descending weight assembly is expressed as;

K weight = (½) Iv 2

Here, v = veocity

We also have to take into account the work that is done in overcoming the friction. This can be found out by;

W friction = nW f

In this case,

n = number of windings of the string

W f = work done in overcoming frictional torque

If we state the law of conservation of energy then we obtain;

P loss = K flywheel + K weight + W friction

We will substitute the values and the equation will now become;

mgh = (½)Iω 2 + (½) mv 2 + nW f

Moving on to the next phase, we look at the flywheel assembly’s kinetic energy that is used in rotating (N) number of times against the frictional torque. We get;

NW f = (½ ) Iω 2 and W f = (1 / 2N) Iω 2

Further, we establish a relation between the velocity (v) of the weight assembly and the radius (r) of the axle. The equation is given as;

We have to substitute the values for W f and v.

mgh = (½) Iω 2 + (½ )mr 2 ω 2 + (n / N) x ½ Iω 2

If we solve the equation for finding the moment of inertia, we obtain;

\(\begin{array}{l}I = \frac{Nm}{N+n}(\frac{2gh}{\omega ^{2}}-r^{2})\end{array} \)

⇒ Check Other Object’s Moment of Inertia:

• Moment Of Inertia Of Circle
• Moment Of Inertia Of A Quarter Circle
• Moment Of Inertia Of Semicircle
• Moment Of Inertia Of A Sphere
• Moment Of Inertia Of A Disc

Parallel Axis Theorem

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MOMENT OF INERTIA OF FLYWHEEL

This experiment is an introduction to some basic components of rotational dynamics to develop an understanding of Moment of Inertia

Torque and Mass Moment of Inertia

If a body is free to rotate about a fixed axis, then a torque is required to initiate or change the rotational motion of the body.

The torque τ ⃗ of a force about an axis is given by the cross-product of the force F ⃗ and the distance from the axis of rotation

The net torque is proportional to the angular acceleration α ⃗ of the body and shall exist during the entire time the torque acts. The equation is given as

Where I is the constant of the body known as the Mass Moment of Inertia about the specified axis of rotation. Mass moment of inertia (also known as rotational inertia) is a measure of a body’s resistance to a change in its rotation direction or angular momentum. The moment of inertia depends not only on the mass but also the distribution of the mass around the axis. Just as the mass is a measure of resistance of linear acceleration, mass moment of inertia is a measure of resistance to angular acceleration.

The experiment consists of estimating the mass moment of inertia of the flywheel system. A flywheel is a heavy thick circular discs designed for storing rotational energy. It is generally made of cast iron or steel along is mounted on an axle free to rotate on ball bearings. In other words, it’s a kind of system that needs a large force to start or stop spinning. The capacity of storing / shedding of kinetic energy depend on the rotational inertia of the flywheel.

In real life, flywheels come in all shapes and sizes. For obtaining the maximum moment of inertia per volume, most flywheels have a heavy outer circular rim with spokes. They may be mounted on the crankshaft of machines such as turbines, steam engines, diesel engines etc. This makes the engine run smoothly by storing kinetic energy when the machines are on higher loads and maintains that constant angular velocity during idle conditions.

Mass Moment of Inertia of Flywheel

The Mass Moment of Inertia of cylindrical objects about an axis passing through the centre can be given by the equation

I = (mr^2)/2.

The flywheel in this experiment is a solid disc of mass M1 and radius R attached to a shaft of mass M2 and radius r. So the moment of inertia of the flywheel system is given as

I = Σ (mr^2)/2= (M_1 R^2)/2+(M_2 r^2)/2

For complex geometries, the mass moment of Inertia of the flywheel can be estimated by measuring the approximate mass of different simplified geometrical components and adding the Mass Moment of Inertia about the central axis (from the known equations of MI of rings, cylinders, rods, etc).

Experimental Setup and Theory

In the experiment, a hanging mass m attached to the end of a spring, the remainder of which is wrapped around the axle is allowed to fall initiating the necessary torque τ ⃗ to the flywheel system initially at rest. Suppose that the string is wrapped around the axle n times and that a mass m is suspended from its free end and the system is released at time t = 0. As the mass accelerates downward, the flywheel attains an angular acceleration α ⃗. Because of the friction in the bearings, there will be an additional torque in the direction opposite to the motion of the flywheel. This frictional torque (α_f ) ⃗ depends upon a number of factors such as speed of rotation, coefficient of friction, etc but shall be assumed to be a constant value for simplicity.

If T is the tension in the string, then the net torque exerted by the wheel is

The net force on the mass m is

If the frictional torque is constant, then the angular acceleration of the system, (α_f ) ⃗, is also constant The flywheel will achieve a maximum angular velocity at the instant when the string detaches from the axle. The axle will continue to rotate until all the work is used to overcome the friction in bearings. Finally, the axle will stop rotating against the frictional forces.

Theoretical Calculations

As the slotted weight falls a particular height, it loses its potential energy. The loss in potential energy during unwinding is converted into its translation kinetic energy and rotational kinetic energy of flywheel. Some of the energy is lost in overcoming frictional forces in the bearings. Applying the law of conservation of energy at the instant the mass hits the ground.

(P.E)m = (R.K.E.)F + (L.K.E.)m + Frictional losses

The loss of potential energy (P.E)m of the slotted weights as it hits the ground is given as

(P.E)m = m g h = mg (2 π r n)

Note that we have neglected the thickness of the cord since radius of elastic cord cannot be determined experimentally. Another source of error is the slipping of the cord from the axle during unwinding.

The rotational kinetic energy of the flywheel (R.K.E)f can be given by

(R.K.E.)m = ½ Iω^2

The gain in linear Kinetic Energy (L.K.E)m of the slotted weights just before the mass touches the ground is given as (L.K.E.)m = ½ mv^2

If ω is the angular speed of the disc just as the mass hits the ground, then the final velocity of slotted weights is given by

The frictional losses are mainly due to friction in the axle and bearing assembly of the apparatus. We assume that the bearing frictional losses per unit rotation to be a constant value Wf. The total bearing friction depends on the number of wounds of cord around the axle

Bearing friction at the end of n1 rotations = n Wf

It is worth mentioning that air friction acting on the surface of the rotating disc as well as the moving weights may also result in losses which are ignored here.

Applying the individual equations in the law of conservation of energy, we obtain

Now, even after the mass detaches from the axle, the flywheel will continue to rotate. The angular velocity of the flywheel would decline gradually and finally come to a rest when all is rotational kinetic energy of flywheel (R.K.E)f is spent to overcome the frictional forces. If N is the number of rotation made by the flywheel after the string has left the axle then

Substituting the values of v and Wf in the equation,

Solving the equation for I, we obtain the following equation for mass moment of inertia of a flywheel which is,

The maximum angular velocity ω in the above equation can be found out by calculating the average velocity ωa as the flywheel comes to a final stop.

If N revolutions take a time t, then the average angular is given by

The above two equations give us a direct relationship between maximum angular velocity, number of rotations after detachment and the time required to complete that revolution

The experimental moment of inertia calculated by the equation may be slightly different from the theoretical moment of inertia because of the following criteria The thickness of the cord is assumed to be negligible. The bearing friction per rotation was assumed to be a constant value throughout the rotation. The air frictional losses are ignored. Any slip between the cords and the axle during unwinding is ignored

Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!

Flywheels - Kinetic Energy

The kinetic energy stored in flywheels - the moment of inertia..

A flywheel can be used to smooth energy fluctuations and make the energy flow intermittent operating machine more uniform. Flywheels are used in most combustion piston engines.

Energy is stored mechanically in a flywheel as kinetic energy.

Kinetic Energy

Kinetic energy in a flywheel can be expressed as

E f = 1/2 I ω 2 (1) where E f = flywheel kinetic energy (Nm, Joule, ft lb) I = moment of inertia (kg m 2 , lb ft 2 ) ω = angular velocity ( rad /s)

Angular Velocity - Convert Units

• 1 rad = 360 o / 2 π =~ 57.29578 o
• 1 rad/s = 9.55 rev/min (rpm) = 0.159 rev/s (rps)

Moment of Inertia

Moment of inertia quantifies the rotational inertia of a rigid body and can be expressed as

I = k m r 2 (2) where k = inertial constant - depends on the shape of the flywheel m = mass of flywheel (kg, lb m ) r = radius (m, ft)

Inertial constants of some common types of flywheels

• wheel loaded at rim like a bicycle tire - k =1
• flat solid disk of uniform thickness - k = 0.606
• flat disk with center hole - k = ~0.3
• solid sphere - k = 2/5
• thin rim - k = 0.5
• radial rod - k = 1/3
• circular brush - k = 1/3
• thin-walled hollow sphere - k = 2/3
• thin rectangular rod - k = 1/2

Moment of Inertia - Convert Units

• 1 kg m 2 = 10000 kg cm 2 = 54675 ounce in 2 = 3417.2 lb in 2 = 23.73 lb ft 2

Flywheel Rotor Materials

• 1 MPa = 10 6 Pa = 10 6 N/m 2 = 145 psi
• Maraging steels are carbon free iron-nickel alloys with additions of cobalt, molybdenum, titanium and aluminum. The term maraging is derived from the strengthening mechanism, which is transforming the alloy to martensite with subsequent age hardening.

Example - Energy in a Rotating Bicycle Wheel

A typical 26-inch bicycle wheel rim has a diameter of 559 mm (22.0") and an outside tire diameter of about 26.2" (665 mm) . For our calculation we approximate the radius - r - of the wheel to

r = ((665 mm) + (559 mm) / 2) / 2

=  306 mm

The weight of the wheel with the tire is 2.3 kg and the inertial constant is k = 1 .

The Moment of Inertia for the wheel can be calculated

I = (1) (2.3 kg) (0.306 m) 2

= 0.22 kg m 2

The speed of the bicycle is 25 km/h ( 6.94 m/s) . The wheel circular velocity (rps, revolutions/s) - n rps - can be calculated as

n rps = (6.94 m/s) / (2 π (0.665 m) / 2)

= 3.32 revolutions /s

The angular velocity of the wheel can be calculated as

ω = (3.32 revolutions /s) (2 π rad/ revolution )

= 20.9 rad/s

The kinetic energy of the rotating bicycle wheel can then be calculated to

E f = 0.5 (0.22 kg m 2 ) ( 20.9 rad/s ) 2

Related Topics

Related documents, angular motion - power and torque, belt transmissions - speed and length of belts, conn-rod mechanism, energy storage density, formulas of motion - linear and circular, impulse and impulse force, mass moment of inertia, rotating bodies - stress, rotating shafts - torque, salt hydrates - melting points and latent melting energy.

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Flywheel experiment

1. INTRODUCTION

A flywheel is a mechanical device with a significant moment of inertia used as a storage device for rotational energy 1 . The rotational energy stored enables the flywheel to accelerate at very high velocities, and also to maintain that sort of velocity for a given period of time. The force that enables the flywheel to attain such velocities also produces energy to slow down the flywheel’s motion.

The objectives of the experiment are;

• To determine the friction torque due to the bearings, T f  
• To determine, experimentally, the moment of inertia, I, for the flywheel.
• To estimate the moment of inertia, using simple equations.
• To compare the experimental value of I with the estimate and suggest reasons for any discrepancies.

To calculate friction torque, it is assumed that the energy lost due to bearing friction is equal to the potential energy lost by the mass during unwinding and rewinding:

               Mg(H 1 -H 2 ) = T f  θ                                                                   . . . . . (1)

Where, m        = applied mass (kg)

               H 1         = original height of mass above some arbitrary datum (m)

               H 2         = final height of mass above the same datum (m)

               T f            = friction torque (Nm)

               θ                    = total angle turned through during unwinding and rewinding (rads)

To calculate the angular acceleration, (α),

              S = u t + a t 2 /2                                                                  . . . . . . . .(2)

This is a preview of the whole essay

and       α =a/r                                                                                . . . . . . . . (3)

where, s             = distance travelled by mass during decent (m)

             u             = initial velocity of mass (=0)

             t              = time to travel distance s (s)

             a             = linear acceleration of mass (m/s 2 )

             r              = effective radius of the flywheel axle (m)

To determine, experimentally, the moment of inertia (I exp );

     T – T f = (I + m r 2 ) α                    where T = m g r             . . . . . . . . . (4)

To calculate a theoretic value for I. The equation is;

     I = MR 2 /2                                                                              . . . . . . . . (5)

Where M            = mass of flywheel (kg)

              R            = radius of flywheel (m)

3 . EXPERIMENTAL PROCEDURES

3.1     DESCRIPTION OF THE TEST EQUIPMENT

• Flywheel (disc and axle)
• Pencil (to mark distance)
• Mass (known) and mass holder

                                 Flywheel

                         (disc + Axle)

         A        

             C          H 1

         String

        H 2

A known mass          B

Figure 3.1a

3.2   PROCEDURE

• The string is wrapped around the flywheel in a clockwise direction, which in turn lifts the known mass that is attached to the bottom of the string to a point close to the flywheel (point A on fig 3.1).
• The string, with the mass attached to it, is then allowed to wind down the flywheel until the mass reaches its lowest point (point B on fig 3.1), which is timed with a stop watch.
• The distance between points A and B is measured as H 1 .
• After reaching its lowest point, the mass then bounces back and starts to travel in the opposite direction, but then stops at a particular point (point C on fig 3.1).
• The distance between points B and C is measured as H 2 .
• The experiment is then repeated again, so as to improve reliability and accuracy of the supposed result.

3.3   RESULT                                                                                                        Table 3.3

H 1  = Original height of mass after it unwinds from the flywheel

H 2  = Final height of mass after bouncing back in opposite direction

 θ = Total angular displacement (rads)

r   = Effective radius of the axle = 13.75x10 -3 m.

Radius of shaft and rope (r) = 0.01375m

Mass of flywheel = 6.859kg

Radius of flywheel = 0.1m

Radius of axle = 0.0125m

4.  ANALYSIS OF RESULTS

To calculate the moment of inertia of the flywheel;

             T – T f = (I + m r 2 ) α                    where T = m g r  

Make ‘I’ the subject of the formula;

            I exp = (T – T f  )/α – (m r)

then, the value of T(applied torque) is;

   = 0.1 x 9.81 x (13.75x10 -3 )

   = 13.49x10 -3  Nm

To calculate T f (frictional torque);

T f  = mg (H 1  – H 2 )/θ

    = (0.1 x 9.81 x 0.77)/ 56

    = 1.35x10 -2  Nm

To calculate the angular acceleration (α);

α = 2H 1 / (r x t 2 )

    = (2 x 0.98)/ (13.75x10 -3  x 22.88 2 )

    = 0.27ms -2

I exp  = (13.49x10 -3  – 1.35x10 -2 )/0.27 - (0.1 x [13.75x10 -3 ] 2 )

       = 3.7x10 -5  – 1.89x10 -5

       = 1.81x10 -5  kgm 2

To calculate the theoretic value for the moment of inertia;

I theory  = MR 2 / 2

            = 6.859 x (0.1) 2  / 2

            = 3.43 x 10 -2  kgm 2

% error  = [(Expected Value – Actual value)/ Expected Value] x 100

               = [3.4x10 -2 / 3.43x10 -2 ] x 100 = 99.13%

5. DISCUSSIONS/CONCLUSION

Following the analysis of my results, the values of I experiment  and I theory  differ by fairly a significant amount i.e. (a percentage error of 99.13%). The errors that led to the difference in the two values can be categorize into two sub-groups called “Measurement errors” and “Procedural errors”.

Measurement Errors.

• Errors may perhaps have crept up while measuring the distances of H 1  and H 2 . These distances could have possibly been marked incorrect if the points were not marked at eye level, which could have lead to errors in the final value. However, these errors could have been minimised by taking more repeated readings, or even recording the experiment with the use of a video camera in order to help in checking for these kind of errors.
• Furthermore, another error that could have affected the final value was the timing of the stopwatch while measuring H 1  and H 2 . This human error can be significantly reduced via total concentration of everyone involved in the experiment.

Procedural Errors.

The motion of the mass that was attached to the spring could have been affected by factors, such as the air resistance and friction, which would lead to easy energy loss during the experiment. This could have also led to some errors in the final value.                                                                                

This error could have been minimised by doing the experiment in a closed system, which would have not just minimised errors, but also increase the accuracy and reliability of the result.

• Lynn White, Jr., “Theophilus Redivivus”, Technology and Culture , Vol. 5, No. 2. (Spring, 1964), Review, pp. 224-233 (233) 1  
•  Lynn White, Jr., “Medieval Engineering and the Sociology of Knowledge”, The Pacific Historical Review , Vol. 44, No. 1. (Feb., 1975), pp. 1-21 (6)

Document Details

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10.4 Moment of Inertia and Rotational Kinetic Energy

Learning objectives.

By the end of this section, you will be able to:

• Describe the differences between rotational and translational kinetic energy
• Define the physical concept of moment of inertia in terms of the mass distribution from the rotational axis
• Explain how the moment of inertia of rigid bodies affects their rotational kinetic energy
• Use conservation of mechanical energy to analyze systems undergoing both rotation and translation
• Calculate the angular velocity of a rotating system when there are energy losses due to nonconservative forces

So far in this chapter, we have been working with rotational kinematics: the description of motion for a rotating rigid body with a fixed axis of rotation. In this section, we define two new quantities that are helpful for analyzing properties of rotating objects: moment of inertia and rotational kinetic energy. With these properties defined, we will have two important tools we need for analyzing rotational dynamics.

Rotational Kinetic Energy

Any moving object has kinetic energy. We know how to calculate this for a body undergoing translational motion, but how about for a rigid body undergoing rotation? This might seem complicated because each point on the rigid body has a different velocity. However, we can make use of angular velocity—which is the same for the entire rigid body—to express the kinetic energy for a rotating object. Figure 10.17 shows an example of a very energetic rotating body: an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are generated as the grindstone does its work. This system has considerable energy, some of it in the form of heat, light, sound, and vibration. However, most of this energy is in the form of rotational kinetic energy .

Energy in rotational motion is not a new form of energy; rather, it is the energy associated with rotational motion, the same as kinetic energy in translational motion. However, because kinetic energy is given by K = 1 2 m v 2 K = 1 2 m v 2 , and velocity is a quantity that is different for every point on a rotating body about an axis, it makes sense to find a way to write kinetic energy in terms of the variable ω ω , which is the same for all points on a rigid rotating body. For a single particle rotating around a fixed axis, this is straightforward to calculate. We can relate the angular velocity to the magnitude of the translational velocity using the relation v t = ω r v t = ω r , where r is the distance of the particle from the axis of rotation and v t v t is its tangential speed. Substituting into the equation for kinetic energy, we find

In the case of a rigid rotating body, we can divide up any body into a large number of smaller masses, each with a mass m j m j and distance to the axis of rotation r j r j , such that the total mass of the body is equal to the sum of the individual masses: M = ∑ j m j M = ∑ j m j . Each smaller mass has tangential speed v j v j , where we have dropped the subscript t for the moment. The total kinetic energy of the rigid rotating body is

and since ω j = ω ω j = ω for all masses,

The units of Equation 10.16 are joules (J). The equation in this form is complete, but awkward; we need to find a way to generalize it.

Moment of Inertia

If we compare Equation 10.16 to the way we wrote kinetic energy in Work and Kinetic Energy , ( 1 2 m v 2 ) ( 1 2 m v 2 ) , this suggests we have a new rotational variable to add to our list of our relations between rotational and translational variables. The quantity ∑ j m j r j 2 ∑ j m j r j 2 is the counterpart for mass in the equation for rotational kinetic energy. This is an important new term for rotational motion. This quantity is called the moment of inertia I , with units of kg · m 2 kg · m 2 :

For now, we leave the expression in summation form, representing the moment of inertia of a system of point particles rotating about a fixed axis. We note that the moment of inertia of a single point particle about a fixed axis is simply m r 2 m r 2 , with r being the distance from the point particle to the axis of rotation. In the next section, we explore the integral form of this equation, which can be used to calculate the moment of inertia of some regular-shaped rigid bodies.

The moment of inertia is the quantitative measure of rotational inertia, just as in translational motion, and mass is the quantitative measure of linear inertia—that is, the more massive an object is, the more inertia it has, and the greater is its resistance to change in linear velocity. Similarly, the greater the moment of inertia of a rigid body or system of particles, the greater is its resistance to change in angular velocity about a fixed axis of rotation. It is interesting to see how the moment of inertia varies with r, the distance to the axis of rotation of the mass particles in Equation 10.17 . Rigid bodies and systems of particles with more mass concentrated at a greater distance from the axis of rotation have greater moments of inertia than bodies and systems of the same mass, but concentrated near the axis of rotation. In this way, we can see that a hollow cylinder has more rotational inertia than a solid cylinder of the same mass when rotating about an axis through the center. Substituting Equation 10.17 into Equation 10.16 , the expression for the kinetic energy of a rotating rigid body becomes

We see from this equation that the kinetic energy of a rotating rigid body is directly proportional to the moment of inertia and the square of the angular velocity. This is exploited in flywheel energy-storage devices, which are designed to store large amounts of rotational kinetic energy. Many carmakers are now testing flywheel energy storage devices in their automobiles, such as the flywheel, or kinetic energy recovery system, shown in Figure 10.18 .

The rotational and translational quantities for kinetic energy and inertia are summarized in Table 10.4 . The relationship column is not included because a constant doesn’t exist by which we could multiply the rotational quantity to get the translational quantity, as can be done for the variables in Table 10.3 .

Example 10.8

Moment of inertia of a system of particles.

• We use the definition for moment of inertia for a system of particles and perform the summation to evaluate this quantity. The masses are all the same so we can pull that quantity in front of the summation symbol.
• We do a similar calculation.
• We insert the result from (a) into the expression for rotational kinetic energy.
• I = ∑ j m j r j 2 = ( 0.02 kg ) ( 2 × ( 0.25 m ) 2 + 2 × ( 0.15 m ) 2 + 2 × ( 0.05 m ) 2 ) = 0.0035 kg · m 2 I = ∑ j m j r j 2 = ( 0.02 kg ) ( 2 × ( 0.25 m ) 2 + 2 × ( 0.15 m ) 2 + 2 × ( 0.05 m ) 2 ) = 0.0035 kg · m 2 .
• I = ∑ j m j r j 2 = ( 0.02 kg ) ( 2 × ( 0.25 m ) 2 + 2 × ( 0.15 m ) 2 ) = 0.0034 kg · m 2 I = ∑ j m j r j 2 = ( 0.02 kg ) ( 2 × ( 0.25 m ) 2 + 2 × ( 0.15 m ) 2 ) = 0.0034 kg · m 2 .
• K = 1 2 I ω 2 = 1 2 ( 0.0035 kg · m 2 ) ( 5.0 × 2 π rad / s ) 2 = 1.73 J K = 1 2 I ω 2 = 1 2 ( 0.0035 kg · m 2 ) ( 5.0 × 2 π rad / s ) 2 = 1.73 J .

Significance

In the next section, we generalize the summation equation for point particles and develop a method to calculate moments of inertia for rigid bodies. For now, though, Figure 10.20 gives values of moment of inertia for common object shapes around specified axes.

Applying Rotational Kinetic Energy

Now let’s apply the ideas of rotational kinetic energy and the moment of inertia table to get a feeling for the energy associated with a few rotating objects. The following examples will also help get you comfortable using these equations. First, let’s look at a general problem-solving strategy for rotational energy.

Problem-Solving Strategy

Rotational energy.

• Determine that energy or work is involved in the rotation.
• Determine the system of interest. A sketch usually helps.
• Analyze the situation to determine the types of work and energy involved.
• If there are no losses of energy due to friction and other nonconservative forces, mechanical energy is conserved, that is, K i + U i = K f + U f K i + U i = K f + U f .
• If nonconservative forces are present, mechanical energy is not conserved, and other forms of energy, such as heat and light, may enter or leave the system. Determine what they are and calculate them as necessary.
• Eliminate terms wherever possible to simplify the algebra.
• Evaluate the numerical solution to see if it makes sense in the physical situation presented in the wording of the problem.

Example 10.9

Calculating helicopter energies.

• The rotational kinetic energy is K = 1 2 I ω 2 . K = 1 2 I ω 2 . We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find K . The angular velocity ω ω is ω = 300 rev 1.00 min 2 π rad 1 rev 1.00 min 60.0 s = 31.4 rad s . ω = 300 rev 1.00 min 2 π rad 1 rev 1.00 min 60.0 s = 31.4 rad s . The moment of inertia of one blade is that of a thin rod rotated about its end, listed in Figure 10.20 . The total I is four times this moment of inertia because there are four blades. Thus, I = 4 M L 2 3 = 4 × ( 50.0 kg ) ( 4.00 m ) 2 3 = 1067.0 kg · m 2 . I = 4 M L 2 3 = 4 × ( 50.0 kg ) ( 4.00 m ) 2 3 = 1067.0 kg · m 2 . Entering ω ω and I into the expression for rotational kinetic energy gives K = 0.5 ( 1067 kg · m 2 ) (31.4 rad/s) 2 = 5.26 × 10 5 J . K = 0.5 ( 1067 kg · m 2 ) (31.4 rad/s) 2 = 5.26 × 10 5 J .
• Entering the given values into the equation for translational kinetic energy, we obtain K = 1 2 m v 2 = ( 0.5 ) ( 1000.0 kg ) ( 20.0 m/s ) 2 = 2.00 × 10 5 J . K = 1 2 m v 2 = ( 0.5 ) ( 1000.0 kg ) ( 20.0 m/s ) 2 = 2.00 × 10 5 J . To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is 2.00 × 10 5 J 5.26 × 10 5 J = 0.380 . 2.00 × 10 5 J 5.26 × 10 5 J = 0.380 .

Example 10.10

Energy in a boomerang.

• Moment of inertia: I = 1 12 m L 2 = 1 12 ( 1.0 kg ) ( 0.7 m ) 2 = 0.041 kg · m 2 I = 1 12 m L 2 = 1 12 ( 1.0 kg ) ( 0.7 m ) 2 = 0.041 kg · m 2 . Angular velocity: ω = ( 10.0 rev / s ) ( 2 π ) = 62.83 rad / s ω = ( 10.0 rev / s ) ( 2 π ) = 62.83 rad / s . The rotational kinetic energy is therefore K R = 1 2 ( 0.041 kg · m 2 ) ( 62.83 rad / s ) 2 = 80.93 J . K R = 1 2 ( 0.041 kg · m 2 ) ( 62.83 rad / s ) 2 = 80.93 J . The translational kinetic energy is K T = 1 2 m v 2 = 1 2 ( 1.0 kg ) ( 30.0 m / s ) 2 = 450.0 J . K T = 1 2 m v 2 = 1 2 ( 1.0 kg ) ( 30.0 m / s ) 2 = 450.0 J . Thus, the total energy in the boomerang is K Total = K R + K T = 80.93 + 450.0 = 530.93 J . K Total = K R + K T = 80.93 + 450.0 = 530.93 J .
• We use conservation of mechanical energy. Since the boomerang is launched at an angle, we need to write the total energies of the system in terms of its linear kinetic energies using the velocity in the x - and y -directions. The total energy when the boomerang leaves the hand is E Before = 1 2 m v x 2 + 1 2 m v y 2 + 1 2 I ω 2 . E Before = 1 2 m v x 2 + 1 2 m v y 2 + 1 2 I ω 2 . The total energy at maximum height is E Final = 1 2 m v x 2 + 1 2 I ω 2 + m g h . E Final = 1 2 m v x 2 + 1 2 I ω 2 + m g h . By conservation of mechanical energy, E Before = E Final E Before = E Final so we have, after canceling like terms, 1 2 m v y 2 = m g h . 1 2 m v y 2 = m g h . Since v y = 30.0 m / s ( sin 40 ° ) = 19.28 m / s v y = 30.0 m / s ( sin 40 ° ) = 19.28 m / s , we find h = ( 19.28 m / s ) 2 2 ( 9.8 m / s 2 ) = 18.97 m . h = ( 19.28 m / s ) 2 2 ( 9.8 m / s 2 ) = 18.97 m .

Check Your Understanding 10.4

A nuclear submarine propeller has a moment of inertia of 800.0 kg · m 2 800.0 kg · m 2 . If the submerged propeller has a rotation rate of 4.0 rev/s when the engine is cut, what is the rotation rate of the propeller after 5.0 s when water resistance has taken 50,000 J out of the system?

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Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction

• Authors: William Moebs, Samuel J. Ling, Jeff Sanny
• Publisher/website: OpenStax
• Book title: University Physics Volume 1
• Publication date: Sep 19, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/university-physics-volume-1/pages/1-introduction
• Section URL: https://openstax.org/books/university-physics-volume-1/pages/10-4-moment-of-inertia-and-rotational-kinetic-energy

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All Lab Experiments

To determine the Moment of Inertia of a Flywheel

• Physics Practical Files

MOI of Flywheel in .pdf

Find more Mechanics Practical Files on this Link – https://alllabexperiments.com/phy_pract_files/mech/

Watch How to perform this Experiment in your laboratory in this video – https://www.youtube.com/watch?v=9mRZDexwJUM

Also Watch the Most Important Viva Questions related to Fly-Wheel Experiment – https://www.youtube.com/watch?v=lbGJKEofVs4

2 thoughts on “To determine the Moment of Inertia of a Flywheel”

thanks sir for the material , it helped a lot

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