Moment of Inertia of Standard Geometrical Bodies

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 Expression for Moment of Inertia of Uniform Rod About a Transverse Axis Passing Through its Centre:

Moment of Inertia 31

  • Consider a thin uniform rod, of mass M, an area of cross-section A, length l, and density of material ρ. Let us consider an infinitesimal element of length dx at a distance of x from the given axis of rotation.  Then its Moment of Inertia about the given axis is given by

dI = x² . dm,  …… (1)

But,   Mass =  Volume x Density

∴   dm    =  A . dx . ρ  ……   (2)



From Equations (1) and (2)

dI  =    x² A . dx .ρ

∴ dI  = A .ρ . x² . dx

The moment of inertia I of the rod about. the given axis is given by



Moment of Inertia 32This is an expression for moment of inertia of a thin uniform rod about transverse axis passing through its centre.

Expression for Moment of Inertia of a Uniform Rod About a Transverse Axis Passing Through its End:

Method – I:

Moment of Inertia 33

  • Consider a thin uniform rod, of mass M, an area of cross-section A, length l, and density of material ρ. Let us consider an infinitesimal element of length dx at a distance of x from the given axis of rotation.  Then its Moment of Inertia about the given axis is given by
  • dI = x² . dm,  …… (1)

    But,   Mass =  Volume x Density

    ∴   dm    =  A . dx . ρ  ……   (2)

    From Equations (1) and (2)

    dI  =    x² A . dx .ρ

    ∴ dI  = A .ρ . x² . dx

    The moment of inertia I of the rod about. the given axis is given by

Moment of Inertia 34

This is an expression for moment of inertia of a thin uniform rod

about transverse axis passing through its end.



Method – II

Moment of Inertia 35

  •  We know that  moment of inertia of a thin  rod about transverse axis passing through its centre G is given  by

Moment of Inertia 36

  • We have to find the moment of Inertia about the parallel transverse axis passing through the end of the rod.  Let Io be the moment of Inertia about this axis.

Then by principle of parallel axes,

Moment of Inertia 37

This is an expression for moment of Inertia of thin uniform rod about transverse axis passing through its end.





Expression for the Moment of Inertia of an Annular Ring:

  • Consider uniform, thin annular disc of mass M having inner radius R1, outer radius R2, thickness t, and density of its material ρ. Let us assume that disc is capable of rotating about a transverse axis passing through its centre. Let us assume that the disc is made up of infinitesimally thin rings.

Moment of Inertia 38

  • Consider  one  such ring of radius r and width dr.  Moment of Inertia of such element is  given, by,

dI  =   r² . dm   ………….. (1)

But,      Mass   =    volume × density

dm  =  ( 2 π r . dr .  t) ρ   ………  (2)

From Equation (1) and (2)



dI  =   r² . ( 2 π r . dr .  t) ρ

dI  =   2 π t  ρ  r³ . dr .

The moment of Inertia I of the annular disc will be given by

Moment of Inertia 39



Where M is total mass of the annular ring.

This is an expression for moment of inertia of annular ring

about a transverse axis passing through its centre.

Expression for Moment of Inertia of a Thin Uniform Disc About a Transverse Axis Passing Through its Centre and Perpendicular to its Plane:

  • The moment of inertia of annular ring about a transverse axis passing through its centre is given by

Moment of Inertia 40For solid disc, there is no centre hole, hence R2 =   R and R1 = 0

Moment of Inertia 41

This is an expression for moment of inertia of thin uniform disc about a transverse axis passing through its centre.



Expression for Moment of Inertia of a Thin Uniform Ring About an Axis Passing through its Centre and Perpendicular to its Plane:

  • The moment of inertia of annular ring about a transverse axis passing through its centre is given by

Moment of Inertia 40

For ring, the centre hole extends up to its periphery, hence R2 =   R and R1 =R

Moment of Inertia 42

This is an expression for moment of inertia of thin uniform ring about a transverse axis passing through its centre.



Expression for Moment of Inertia  of Thin Uniform Disc About its Tangent Perpendicular to its Plane:

Moment of Inertia 43

  • The moment of inertia of thin uniform disc about a transverse axis passing through its centre is given by

Moment of Inertia 44

  • We have to find the M.I. about a tangent perpendicular to the plane of the disc. These two axes are parallel to each other. By parallel axes theorem,

Moment of Inertia 45

This is an expression for M.I. of a thin uniform disc about its tangent perpendicular to its plane.

Expression for Moment of Inertia of Thin Uniform Ring About its Tangent Perpendicular to its Plane:

Moment of Inertia 46

  • The moment of inertia of thin uniform ring about a transverse axis passing through its centre is given by

I = MR² = IG



  • We have to find the M.I. about a tangent perpendicular to the plane of the rimg. These two axes are parallel to each other. By parallel axes theorem

IO =  IG +  Mh²

∴   IO =  MR² +  MR²

∴   IO =  2 MR²

This is an expression for M.I. of a thin uniform ring about its tangent perpendicular to its plane.



Expression for Moment of Inertia of Thin Uniform Disc About its  Diameter:

Moment of Inertia 47

  • The moment of inertia of thin uniform disc about a transverse axis passing through its centre is given by

Moment of Inertia 48

  • We have to find the M.I. about a diameter of the disc. Let us consider a system of three axes, such that the z-axis is along the geometrical axis of the disc and x-axis and y-axis lie in the plane of the disc such that the centre of mass G lies at the origin of the system of axes.

Moment of Inertia 49

By perpendicular axes theorem, we have

lz    =    lx  +   ly   …………. (2)

Due to symmetry M.I. of the disc about diameter is

l=  lx  +   ly   …………. (3)

Substituting values of equations (1) and (3) in (2)

Moment of Inertia 50

This is an expression for M.I. of a thin uniform disc about its diameter.

Expression for Moment of Inertia  of Thin Uniform Ring About its  Diameter:

Moment of Inertia 51

  • The moment of inertia of thin uniform ring about a transverse axis passing through its centre is given by

I = MR²

  • We have to find the M.I. about a diameter the ring. Let us consider a system of three axes, such that the z-axis is along the geometrical axis of the ring and  x-axis and y-axis lie in the plane of the ring such that the centre of mass G lies at the origin of the system of axes.

l = MR²  ………(1)

By perpendicular axes theorem, we have

lz    =    lx  +   ly   …………. (2)

Due to symmetry M.I. of the disc about diameter is

l=  lx  +   ly   …………. (3)

Substituting values of equations (1) and (3) in (2)

Moment of Inertia 53

This is an expression for M.I. of thin uniform ring about its diameter.

Expression for Moment of Inertia of a Thin Uniform Disc About an Axis Tangent to the Disc and in the Plane of the Disc:

Moment of Inertia 54

  • The moment of inertia of thin uniform disc about a transverse axis passing through its centre is given by

Moment of Inertia 48

  • We have to find the M.I. about a diameter the disc first. Let us consider a system of three axes, such that the z-axis is along the geometrical axis of the disc and x-axis and y-axis lie in the plane of the disc such that the centre of mass G lies at the origin of the system of axes.

Moment of Inertia 49

By perpendicular axes theorem, we have

lz    =    lx  +   ly   …………. (2)

Due to symmetry M.I. of the disc about diameter is

l=  lx  +   ly   …………. (3)

Substituting values of equations (1) and (3) in (2)

Moment of Inertia 55

Now y axis passes through the centre of mass G of the disc

Moment of Inertia 56

Now the tangent to the disc in the plane of the disc is parallel to the y-axis.

By parallel axes theorem.

Moment of Inertia 57

This is an expression for M.I. of a thin uniform disc about its tangent in its plane and in the plane of the disc.



Expression for Moment of Inertia of a Thin Uniform Ring About an Axis Tangent to the Ring and in the Plane of the Ring:

Moment of Inertia 58

  • The moment of inertia of thin uniform ring about a transverse axis passing through its centre is given by

l  = MR²

  • We have to find the M.I. about a diameter the ring first. Let us consider a system of three axes, such that the z-axis is along the geometrical axis of the ring and x-axis and y-axis lie in the plane of the ring such that the centre of mass G lies at the origin of the system of axes.

l = MR²  ………(1)

By perpendicular axes theorem, we have

lz    =    lx  +   ly   …………. (2)

Due to symmetry M.I. of the disc about diameter is

l=  lx  +   ly   …………. (3)

Substituting values of equations (1) and (3) in (2)

Moment of Inertia 59

Now the tangent to the ring in the plane of the ring is parallel to the y-axis.

By parallel axes theorem.

Moment of Inertia 60

This is an expression for M.I. of a thin uniform ring about its tangent in its plane.

Expression for Moment of Inertia  of a Solid Cylinder About its Geometrical Axis:

Moment of Inertia 61

  • Consider a solid cylinder of mass M, length ‘’ and radius ‘r’ capable of rotating about its geometrical axis. Let ‘m be its mass per unit length.

m = M/l      Hence M = m . l

  • A solid cylinder can be regarded as a number of thin uniform discs of infinitesimal thickness piled on top of one another. Let us consider one such disc of thickness ‘dx’ at a distance of ‘x’ from the centre C of the cylinder.

Mass of such disc is given by

Mass, dm = m.dx   =  (M /l). dx

The M.I. of such disc about a transverse axis (passing through C) is given by

Moment of Inertia 62

Integrating above expression in limits

Moment of Inertia 63

This is an expression for M. I. of a solid cylinder about its geometrical axis.

Expression for Moment of Inertia of a Hollow Cylinder About its Geometrical Axis:

Moment of Inertia 64

  • Consider a hollow cylinder of mass M, length ‘’ and radius ‘r’ capable of rotating about its geometrical axis. Let ‘m be its mass per unit length.

m = M/l      Hence M = m . l

  • A hollow cylinder can be regarded as a number of thin uniform rings of infinitesimal thickness piled on top of one another. Let us consider one such ring of thickness ‘dx’ at a distance of ‘x’ from the centre C of the cylinder.

Mass of such ring is given by

Mass, dm = m.dx  = (M/ l) dx

The M.I. of such ring about a transverse axis (passing through C) is given by

Moment of Inertia 65

This is an expression for M. I. of a solid cylinder about its geometrical axis.

Expression for Moment of Inertia  of a Solid Cylinder About a Transverse Axis Passing Through its Centre:

Expression for Moment of Inertia 66

  • Consider a solid cylinder of mass M, length ‘’ and radius ‘r’ capable of rotating about its geometrical axis. Let ‘m be its mass per unit length.

m = M/l      Hence M = m . l

  • A solid cylinder can be regarded as a number of thin uniform discs of infinitesimal thickness piled on top of one another. Let us consider one such disc of thickness ‘dx’ at a distance of ‘x’ from the centre C of the cylinder.

Mass of such disc is given by

Mass, dm = m.dx  = (M/l) dx

The M.I. of such disc about a transverse axis (passing through C) is given by

Moment of Inertia 67

The M.I. of a disc about a diameter is given by

Moment of Inertia 68

Integrating above expression in limits

Moment of Inertia 69

Moment of Inertia 70

This is an expression for M. I. of a solid cylinder about a transverse axis passing through its centre.



Expression for Moment of Inertia of a Hollow Cylinder About a Transverse Axis Passing Through its Centre:

Moment of Inertia 71

  • Consider a hollow cylinder of mass M, length ‘’ and radius ‘r’ capable of rotating about its geometrical axis. Let ‘m be its mass per unit length.

m = M/l      Hence M = m . l

  • A hollow cylinder can be regarded as a number of thin uniform rings of infinitesimal thickness piled on top of one another. Let us consider one such ring of thickness ‘dx’ at a distance of ‘x’ from the centre C of the cylinder.

Mass of such ring is given by

Mass, dm = m.dx  = (M/l) dx

The M.I. of such ring about a transverse axis (passing through C) is given by

dI = dm . R²

The M.I. of a ring about a diameter is given by

Moment of Inertia 72

Integrating above expression in limits

Moment of Inertia 73

Moment of Inertia 74

This is an expression for M. I. of a hollow cylinder about a transverse axis passing through its centre.

Expression for Moment of Inertia  of a Solid Sphere About its Diameter (Geometrical axis):

Moment of Inertia 75

  • Let us consider a solid homogeneous sphere of radius ‘R’ and mass ‘M’, capable of rotating about its diameter. Let us consider a circular strip of infinitesimal thickness ‘dx’ at a distance of x from centre ‘O’. The radius of this circular strip is PM, which is given by

Moment of Inertia 76

This circular strip can be treated as thin disc rotating about a transverse axis passing through its centre.

The M.I. of the disc about a transverse axis passing through its centre is given by

Moment of Inertia 77

The M.I. of the whole sphere about diameter can be obtained by integrating above expression.

Moment of Inertia 78

Moment of Inertia 79

The mass of the sphere = M. Hence, the M.I. of the solid homogeneous sphere is given by

Moment of Inertia 80

This is an expression for M.I. of a solid sphere about its diameter (Geometrical axis).



Expression for Moment of Inertia  of a Solid Sphere About its Tangent.

Moment of Inertia 81

 The moment of inertia of solid sphere about its geometrical  axis (diameter) is given by

Moment of Inertia 82

  • We have to find the M.I. about a tangent to the sphere. These two axes are parallel to each other. By parallel axes theorem

Moment of Inertia 83

This is an expression for M.I. of solid sphere

Science > Physics > Rotational Motion > You are Here
Physics Chemistry Biology Mathematics

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