calculation of moment of inertia

NAME:SYED.M.ABBAS ZAIDI.

ENROLLMENT NO:01-133102-090

Submitted to:Sir IMTIAZ ALI KHAN

CALCULATIONS OF MOMENT OF INERTIA

REFERNCE BOOKS

1)Physics for engineering and science (page no131-134).

by DR MICHEAL BROWN.

2)University physics (11th edition)

page 347-350.

By YOUNG AND FREDNAN.

REFERENCE BOOKS.

3)Physics(4th Edition)

(page no 262-265)

By Paul A Tripler.

4)Physics for scientist and engineers.

(page no 301-305)

By Serway Beiunner.

5)Advanced physics(2nd edition).

By Weith Gibbs.

REFERENCE BOOKS.

6)Fundamentals of physics .

By Halliday and Resnick.

COMMON POINTS :

REFERENCE BOOKS.

1)University pysics.

2)Physics (4th edition).

3)Physics for scientist and engineers.

4)Fundamentals of physics.

Consider a point P on a rotating object that is a distance r away from the axis of rotation. As the object turns through an angle the point covers a distance given by s = r

In the above expression the angle must be in radians

If this expression is differentiated with respect to time then the left hand side will become the linear speed of particle

This speed corresponds to the velocity of the point P which is tangential to the circular arc traced out by the point. When differentiating the right hand side, we notice that r is constant and the rate of change of angular position is the angular velocity. This gives:

dt

dr

dt

ds rv Relation between linear

and angular speed


Differentiating once again gives a relationship between the tangential acceleration of the point, atan, and the angular acceleration of the rotation object:

r

dt

dr

dt

dva tan

Tangential acceleration of a point on a rotating body

Finally, recall that any object that is undergoing circular motion experiences an inwardly directed radial acceleration given by the speed squared divided by the radius. If we replace v=r we have:

rr

varad

22

Centripetal acceleration of a point on a rotating body


r

dt

dr

dt

dva tan

These equations apply to any particle that has the same tangential velocity as a point in a rotating rigid body

o Rope wound around a circular cylinder unwraps without stretching or slipping, its speed and acceleration at any instant are equal to the speed and tangential acceleration of the point at which it is tangent to the cylinder

o Bicycle chains and sprockets, belts, pulleys, …

rv rs


The rotational inertia of an object is a measure of the resistance of the object to changes in its rotational motion

For a system of particles of masses mi at distances ri from an axis passing through a point P the rotational inertia of the system about the axis is given by:

i

iirmrmrmI 2222

211 ...

Definition of moment of inertia

SI unit of moment of inertia is the kgm2 For a solid object the rotational inertia is found by evaluating an integral

as we will see later In a rigid body the distances ri are constant, and I is independent of

how the body is rotating around a given axis. The rotational inertia of some common shapes about some of their symmetry axes is given in Table 9.2 of your textbook


Rotational Inertia (Moment of Inertia)

The rotational kinetic energy of a solid object rotating about an axis for which its rotational inertia is I with angular velocity is expressed as

2

2

1 IK Rotational kinetic energy of a rigid body

Notice the similarity between this formula and the formula for the kinetic energy of a point mass m moving with speed v

This kinetic energy is the sum of kinetic energies of the individual particles that make up the rigid body

is in rad/s (NOT in rev or degrees per second ! K will be in Joules) The greater is the moment of inertia, the greater the kinetic energy of a

rigid body rotating with a given angular speed



2

2

1 IK

Greater a body’s moment of inertia, the harder it is to start the body rotating if it’s at rest and the harder it is to stop its rotation if it’s already rotating

Moments of inertia for different rotation axes One-piece machine part consists of three heavy connectors linked by light

molded struts.

A. What is the moment of inertia of this body about an axis through point A, to the plane of the slide?

B. What is the moment of inertia of this body about an axis coinciding the rod BC?

C. If the body rotates about an axis through A to the plane of the slide with angular speed 4.0 rad/s, what is its kinetic energy?


CALCULATIONS OF MOMENT OF INERTIA. Problems

Problem-Solving Strategy IDENTIFY the relevant concepts: You can use work–energy relations

and conservation of energy to find relations involving position and motion of a rigid body rotating around a fixed axis. As we saw before, the energy method is usually not helpful for problems that involve elapsed time. Later we will see how to approach rotational problems of this kind.

SET UP the problem using the following steps:

1. First decide what the initial and final states (the positions and velocities) of the system are. Use the subscript 1 for the initial state and the subscript 2 for the final state. It helps to draw sketches showing the initial and final states.

2. Define your coordinate system, particularly the level at which y=0. You will use it to compute gravitational potential energies. Equations assume that the positive direction for y is upward; use this choice consistently.

3. Identify all non-gravitational forces that do work. A free-body diagram is always helpful. If some of the quantities you need are unknown, represent them by algebraic symbols.

Rotational Energy. Problems

Problem-Solving Strategy EXECUTE the solution: Write expressions for the initial and final kinetic and potential energies (K1,

K2, U1 and U2) and the non-conservative work Wother (if any).

The new feature is rotational kinetic energy, which is expressed in terms of the body’s moment of inertia I for the given axis and its angular speed instead of its mass m and speed v.

Substitute these expressions into K1+ U1+Wother =K2+U2 (if nonconservative work is done) or K1+ U1=K2+U2 (if only conservative work is done) and solve for the target variable(s).

It’s helpful to draw bar graphs showing the initial and final values of K, U, and E=K+U.

EVALUATE your answer: As always, check whether your answer makes physical sense.

COMMON FEATURES.

Inertia Calculations

For a continuous distribution of mass the sum of the masses times the square of the distances to the axis of rotation which defines the moment of inertia become an integral.

If the object is divided into small mass elements dm in such a manner that all of the points in a particular mass element are the same perpendicular distance r from the axis of rotation then the moment of inertia is given by . dmrI 2

To evaluate the integral, you need to represent r and dm in terms of the same integration variable. 1-D object, slender rod: use coordinate x along the length and relate dm to an increment dx. 3-D object: express dm in terms of element of volume dV and density .

dVVrIdV

dm)(, 2 dVrIconst 2

dzdydxdV Limits of integral are determined by the shape and dimensions of the body


Uniform thin rod, axis to length

Slender uniform rod with mass M and length L.

Compute its moment of inertia about an axis through O, at an arbitrary distance h from the end.

L

dx

M

dm

Choose as an element of mass a short section of rod with length dx at a distance x from O. The ratio of the mass dm of this element to the total mass M is equal to the ratio of its length dx to the total length L:

dxL

Mdm

)33(3

1

322

322 hLhLM

x

L

Mdxx

L

Mdmx

hL

h

hL

h

Evaluate this general expression about an axis through the left end; the right end; through the center. Compare with Table 9.2.


Hollow or solid cylinder, rotating about axis of symmetry

Hollow, uniform cylinder with length L, inner radius R1, outer radius R2. Compute its moment of inertia about the axis of symmetry.

)2( rLdrdVdm

Choose as a volume element a thin cylindrical shell of radius r, thickness dr, and length L. All parts of this element are at very nearly the same distance from the axis. The volume of this element:

)(4

22)2( 4

142

3222

1

2

1

RRL

drrLrLdrrdmrR

R

R

R

))((2

21

22

21

22 RRRR

L

)( 2

122 RRLV )(

2

1 21

22 RRMI


Hollow or solid cylinder, rotating about axis of symmetry

If cylinder is solid, R1=0, R2=R:

If cylinder has a very thin wall, R1 and R2 are very nearly equal:

Note: moment of inertia of a cylinder about an axis of symmetry depends on its mass and radii, but not on its length!

)(2

1 21

22 RRMI

2

2

1MRI

2MRI

Parallel Axis Theorem

Different Features:

PHYSICS FOR ENGINEERING AND SCIENCES:

FUNDAMENTALS OF PHYSICS:

Parallel-Axis Theorem

To find the rotational inertia of an object about an axis that is different from one listed in Table 9.2 in your textbook you may be able to use the parallel axis theorem.

This theorem gives the rotational inertia of an object of mass M about an axis, P, that is parallel to and a distance d away from an axis that passes through the object's center of mass.

2MdII cmp Parallel-Axis Theorem


Consider two axes, both parallel to z-axis, one through the center of mass and the other through a point P.

Mass element mi has coordinates (xi, yi) with respect to an axis of rotation through the center of mass and to the plane of the slide. The mass element has coordinates (xi-a, yi-b) with respect to the parallel axis through point P.

Let’s take origin at the CM of the body: xcm= ycm= zcm=0

The axis through the CM passes through this thin slice at point O, and parallel axis passes through point P with coordinates (a, b). Then the distance of this axis from axis through CM is d: d2=a2+b2

Moment of inertia Icm about axis through O:

i

iiicm yxmI )( 22


Moment of inertia Icm about axis through P:

i

iiiP byaxmI )])()[( 22

These expressions don’t involve the coordinates zi measured to the slices. Let’s extend the sums to include all particles in all slices. Ip then becomes the moment of inertia of the entire body for an axis through P:

i i i

iiii

iiiiiP mbaymbxmayxmI )(22)( 2222

cmI 0cmx2d M0cmy

2MdII cmP

Parallel-Axis Theorem. Example

A part of a mechanical linkage has a mass of 3.6 kg. We measure its moment of inertia about an axis 0.15 m from its center of mass to be Ip=0.132 kg·m2.

What is the moment of inertia Icm about a parallel axis through the center of mass?

222

2

051.0)15.0)(6.3(132.0 mkgmkgmkg

MdII pcm

Result show that Icm is less than Ip. This is as it should be: the moment of inertia for an axis through the center of mass is lower than for any other parallel axis.

UNIQUE FEATURES:

UNIVERSITY PHYSICS BY YOUNG AND FREEDNAN.


Uniform sphere, axis through center

Uniform sphere with radius R. the axis is through its center. Find the moment of inertia about the axis is through the center of this sphere.

Divide sphere into thin disks of thickness dx, whose moment of inertia we already know. The radius r of the disk is

22 xRr

The volume is

The mass is

The moment of inertia for the disk of radius r and mass dm is

dxxRdxrdV )( 222 dxxRdxrdVdm )( 222

dxxRdxxRxRdmrdI 222222

222 )(2

])([2

1

2

1


Uniform sphere, axis through center

Integrating from x=0 to x=R gives the moment of inertia of the right hemisphere.

From symmetry, the total I for the entire sphere is just twice this:

R

dxxRI0

222 )(2

)2(

5

15

8RI

Volume of the sphere

The mass M of the sphere

3

4 3RV

3

3

4RVM

2

5

2MRI

Note: moment of inertia of a solid sphere is less than the moment of inertia of a solid cylinder of the same mass and radius! (Reason is that more of the sphere’s mass is located close to the axis)

calculation of moment of inertia

Documents