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Moment Of Inertia For Continuous Bodies

Learn how to calculate the moment of inertia of a uniform rod about the axis perpendicular to the rod and passing through one of its ends. Practice mass moment of inertia of a rod, ring, and disk.

Moment of Inertia for Continuous Mass Distribution

  • For the calculation of moment of inertia for a continuous body, integration can be used.
  • Consider a small element of a body of mass dm.

  • Element should be chosen such that all the elements of same shape if added together, form the whole body and also the moment of inertia of the element about the given axis should be known.
  • Assume that the perpendicular distance of the element from the axis of rotation is r.
  • Then, moment of inertia of the element will be \(dI=r^2dm\) ... (1)

  • Moment of inertia of the whole body will be obtained by integrating equation (1) on both sides within proper limits,

\(\int{dI}\) = \(\int {r^2dm}\)

⇒  \(I=\int {r^2dm}\)

  • The lower and upper limits within proper limits of integration are the minimum and maximum values of the variable respectively, so as to determine the moment of inertia of whole body.

Examples :

1. Uniform rod 

  • To calculate the moment of inertia about perpendicular bisector of a uniform rod of mass 'M' and length 'L', consider an element of length 'dr' at a perpendicular distance 'r' from the axis of rotation.
  • Then, moment of inertia will be

            \(I=\int\limits_{–L/2}^{L/2}r^2dm\)

where, dm is mass of the element.

  • The lower and upper limits for integration are the minimum and maximum values of the variable, respectively to determine the moment of inertia of whole body.

2. Ring 

  • A ring of mass 'M' and radius 'R' is made up of particles, all at a distance of R from the axis.
  • An element of mass dm should be taken for the calculation of moment of inertia, as shown in figure.

3. Disk 

  • A disk is made up of rings of different radii. Once the moment of inertia for a ring is calculated, then we can find out the moment of inertia for the complete disk.
  • The radius of inner circle is 'x' & that of outer circle is 'x + dx'.
  • So, the thickness of the ring becomes dx.
  • As x varies from center to circumference (0 to R), moment of inertia of all such rings can be added together.
  • The lower limit is 'zero(0)' and upper limit is 'R'.

Illustration Questions

To calculate the moment of inertia of a uniform rod about the axis perpendicular to the rod and passing through one of its ends, the element chosen would be -

A

B

C

D

×

Option (C) is correct  because the small element will cover up the whole body and its moment of inertia is known.

 

Option (A) is incorrect because we cannot use rod to calculate the moment of inertia of rod.

 

Option (B) is incorrect because moment of inertia of element chosen is not known.

 

Option (D) is incorrect because square will not cover up the whole body.

To calculate the moment of inertia of a uniform rod about the axis perpendicular to the rod and passing through one of its ends, the element chosen would be -

A image
B image
C image
D image

Option C is Correct

Moment of Inertia of a Uniform Rod about the Axis passing through the Center of Mass

  • For the calculation of moment of inertia for a continuous body, integration can be used.
  • Consider a small element of a body of mass dm.

  • Element should be chosen such that all the elements of same shape if added together, form the whole body and also the moment of inertia of the element about the given axis should be known.
  • Assume that the perpendicular distance of the element from the axis of rotation is r.
  • Then, moment of inertia  of the element will be \(dI=r^2dm\) ... (1)

  • Moment of inertia of the whole body will be-

Integrating equation (1) on both sides within proper limits,

\(\int{dI}\) = \(\int {r^2dm}\)

  \(I=\int {r^2dm}\)

  • The lower and upper limits within proper limits of integration are the minimum and maximum values of the variable respectively, so as to determine the moment of inertia of whole body.
  • To  calculate the moment of inertia about perpendicular bisector of a uniform rod of mass 'M' and length 'L' , consider an element of length 'dr' at a perpendicular distance 'r' from the axis of rotation.
  • Then, moment of inertia will be

            \(I=\int\limits_{–L/2}^{L/2}r^2dm\)

where dm is mass of the element.

  • The lower and upper limits for integration are the minimum and maximum values of the variable respectively, to determine the moment of inertia of whole body.
  • Mass per unit length = \(\dfrac{M}{L}\)

therefore,  \(dm=\dfrac{M}{L}.{dr}\) ...(1)

\(I=\int\limits_{-L/2}^{L/2}r^2dm\)

 \(I=\dfrac{M}{L}\)\(\int\limits_{-L/2}^{L/2}r^2dr\)  ...[from (1)]

 \(I=\dfrac{M}{L}\)\(\left[\dfrac{r^3}{3}\right]_{-L/2}^{L/2}\)

 \(I=\dfrac{ML^2}{12}\)

 

Illustration Questions

Calculate the moment of inertia of a uniform rod of mass M =12 kg and length L = 12m about its perpendicular bisector.

A 144 kgm2

B 140 kgm2

C 130 kgm2

D 150 kgm2

×

Consider an element of length 'dr' and mass 'dm' at a distance 'r' from the axis.

image

Mass of element,

since,   \(dm=\dfrac{M}{L}.{dr}=\dfrac{12}{12}.{dr}\)... (1)

image

choosing limits, upper limit = 6

lower limit = –6

image

Moment of Inertia,

 \(I=\int\limits_{-6}^{6}r^2dm\)

     =\(\int\limits_{-6}^{6}r^2dr\) ...[from (1)]

     =\(\left[\dfrac{r^3}{3}\right]_{-6}^{6}\)

     = \(\left[\dfrac{(+6)^3}{3}–\dfrac{(–6)^3}{3}\right]\)

     =144 kg-m2

image

Calculate the moment of inertia of a uniform rod of mass M =12 kg and length L = 12m about its perpendicular bisector.

A

144 kgm2

.

B

140 kgm2

C

130 kgm2

D

150 kgm2

Option A is Correct

Moment of Inertia of Rod at Different Locations of Axis

  • For the calculation of moment of inertia for a continuous body, integration can be used.
  • Consider a small element of a body of mass dm.

  • Element should be chosen such that all the elements of same shape if added together, form the whole body and also the moment of inertia of the element about the given axis should be known.
  • Assume that the perpendicular distance of the element from the axis of rotation is r.
  • Then, moment of inertia of the element will be \(dI=r^2dm\) ... (1)

  • Moment of inertia of the whole body will be-
  • Integrating equation (1) on both sides within proper limits,
  • \(\int{dI}\) = \(\int {r^2dm}\)

    ⇒  \(I=\int {r^2dm}\)

  • The lower and upper limits within proper limits of integration are the minimum and maximum values of the variable respectively, so as to determine the moment of inertia of whole body.
  • To calculate the moment of inertia about perpendicular bisector of a uniform rod of mass 'M' and length 'L', consider an element of length 'dr' at a perpendicular distance 'r' from the axis of rotation.

  • Then, moment of inertia will be

            \(I=\int\limits_{–L/2}^{L/2}r^2dm\)

where, dm is mass of the element.

  • The lower and upper limits for integration are the minimum and maximum values of the variable, respectively to determine the moment of inertia of whole body.

Illustration Questions

Consider a uniform rod of mass 'M' and length 'L'. Calculate the moment of inertia of the rod about an axis perpendicular to the rod at a distance \(\dfrac{L}{3}\) from one end.

A \(\dfrac {ML^2}{9}\)

B \(\dfrac {ML^2}{6}\)

C \(\dfrac {ML^2}{3}\)

D \(\dfrac {ML^2}{5}\)

×

Consider an element of length 'dr' and mass 'dm' at a distance 'r' from the axis.

image

Mass of element,

since,  \(dm=\dfrac{M}{L}.{dr}\)  ... (1)

image

choosing limits, upper limit = \(\dfrac{2L}{3}\)

                         lower limit = \(\dfrac{-L}{3}\)

 

image

Moment of inertia,

 \(I=\int\limits_{-L/3}^{2L/3}r^2dm\)

\(\dfrac{M}{L}\)\(\int\limits_{-L/3}^{2L/3}r^2dr\) ...[from (1)]

\(\dfrac{M}{L}\)\(\left[\dfrac{r^3}{3}\right]_{-L/3}^{2L/3}\)

\(\dfrac{M}{3L}\) \(\left[\left(\dfrac{2L}{3}\right)^3–\left(\dfrac{–L}{3}\right)^3\right]\)

\(\dfrac{ML^2}{9}\)

 

 

image

Consider a uniform rod of mass 'M' and length 'L'. Calculate the moment of inertia of the rod about an axis perpendicular to the rod at a distance \(\dfrac{L}{3}\) from one end.

image
A

\(\dfrac {ML^2}{9}\)

.

B

\(\dfrac {ML^2}{6}\)

C

\(\dfrac {ML^2}{3}\)

D

\(\dfrac {ML^2}{5}\)

Option A is Correct

Moment of Inertia of a Circular Ring about the Line Perpendicular to the Plane of the Ring passing through its Center

  • Consider a ring of mass 'M' and radius 'R'.
  • As all the elements of the ring are at the same perpendicular distance 'R' from the axis, the moment of inertia of the ring will be-

\(I=\int{r^2}dm\)

 \(I=\int{R^2}dm\)

\(I=R^2\int{dm}\)

   \(= MR^2\)

Illustration Questions

Calculate the moment of inertia of a ring of mass M = 2 kg and radius R = 1 m about the line perpendicular to the plane of the ring through its center.

A 1 kg-m2

B 2 kg-m2

C 3 kg-m2

D 5 kg-m2

×

Moment of inertia, I = MR2

    I = 2(1)2

    I = 2 kg-m2

image

Calculate the moment of inertia of a ring of mass M = 2 kg and radius R = 1 m about the line perpendicular to the plane of the ring through its center.

A

1 kg-m2

.

B

2 kg-m2

C

3 kg-m2

D

5 kg-m2

Option B is Correct

Moment of Inertia of a Uniform Circular Disk (Plate) about the Line Perpendicular to its Plane and Passing through its Center

  • Consider a circular disk of mass 'M' and radius 'R'. The center of disk is at 'O' and the axis OX is perpendicular to the plane of the plate.

  • A ring of thickness 'dx' is chosen as an element.
  • The radius of inner circle is 'x' and that of outer circle is 'x + dx'.
  • Mass per unit area = \(\dfrac{M}{\pi{R^2}}\).
  • The circumference of the ring is \(2\pi{x}\) and its thickness is dx.
  • Area of the ring can be calculated by assuming it as a ribbon of length \(2\pi{x}\) and breadth dx.

Area of ring, dA = Length × Breadth

⇒ \(dA=2\pi{x}\;dx\) 

  • Mass of the element = mass per unit area × area of the element

\(dm=\dfrac{M}{\pi{R^2}}{(dA)}\)

 \(dm=\dfrac{M}{\pi{R^2}}\) × \(2\pi{x}dx\)

 \(dm=\dfrac{2Mxdx}{{R^2}}\)

  • The moment of inertia of circular ring,

\(dI=x^2dm\)

\(x^2\dfrac{2Mxdx}{{R^2}}\)

Integrating both sides within proper limits -

\(\int\limits_O^IdI\) = \(\int\limits_O^Rx^2\dfrac{2Mxdx}{R^2}\)

 \(I=\dfrac{2M}{R^2}\)\(\int\limits_O^Rx^3dx\)

 \(I=\dfrac{MR^2}{2}\)

Illustration Questions

Calculate the moment of inertia of a disk of mass M = 2 kg and radius R = 1 m about the line perpendicular to the plane and passing through the center of disk.

A 2 kg-m2

B 1 kg-m2

C 3 kg-m2

D 4 kg-m2

×

Moment of Inertia,  \(I=\dfrac {MR^2}{2}\)

 \(I=\dfrac{(2)(1)^2}{2}\)

\(I\) = 1 kg-m2

image

Calculate the moment of inertia of a disk of mass M = 2 kg and radius R = 1 m about the line perpendicular to the plane and passing through the center of disk.

A

2 kg-m2

.

B

1 kg-m2

C

3 kg-m2

D

4 kg-m2

Option B is Correct

Moment of Inertia of the Bodies Expanded Parallel to the Axis

For Hollow Cylinder

  • Consider a ring r1 of mass m and radius R whose axis of rotation is perpendicular to the plane and passing through the center.

  • If the rings of masses m1, m2, m3 ... and of same radius R, are placed below the ring r1 such that the axis of rotation of the rings is same then the moment of inertia of ring will be m1R2, m2R2, m3R,..., respectively.
  • All the rings are placed such that they constitute a hollow cylinder of mass M (= m+ m+ m3 + ....) and radius R.
  • The moment of inertia of a ring r1,

 \(I_1=mR^2\)

Moment of inertia of all the rings,

\(I=R^2(m_1+m_2+m_3+...)\)

\(I=R^2M\)

\(I=MR^2\)

  • Thus, the moment of inertia of the bodies which are expanded parallel to the axis, remains same.

For Rectangular Plate

  • Consider a rod \(l_1\) of mass m and length L whose axis of rotation is perpendicular to the rod passing through one of its ends. 

  • If the rods of masses (m1, m2, m3, ....) and of same length L are placed below the rod \(l_1\) such that axis of rotation of all the rods is same, then the moment of inertia of rods will be \(\dfrac{m_1L^2}{3}\)\(\dfrac{m_2L^2}{3}\)\(\dfrac{m_3L^2}{3}\), ......... respectively.
  • All the rods are placed such that they constitute a rectangular plate of  mass M (m+ m+ m3 + ....) and length L.
  • The moment of inertia of a rod \(l_1\),  \(I_1=\dfrac{mL^2}{3}\)

Moment of inertia of all the rods,

\(I=\dfrac{L^2}{3}(m_1+m_2+m_3+...)\)

 \(=\dfrac{ML^2}{3}\)

  • Thus, the moment of inertia of the bodies which are expanded parallel to the axis, remains same.

Illustration Questions

Calculate the moment of inertia of a square plate ABCD of edge 'L' about an axis in the plane of the plate parallel to edge AB passing through center.

A \(\dfrac{ML^2}{3}\)

B \(\dfrac{ML^2}{12}\)

C \(\dfrac{ML^2}{6}\)

D \(\dfrac{ML^2}{9}\)

×

Assume that the square plate is divided into thin rods of masses m1, m2, m3,... and length 'L'.

image

The moment of inertia of a rod about the perpendicular bisector,

 \(dI_1=\dfrac{m_1L^2}{12}\)

image

Moment of inertia of all the rods,

\(I=dI_1+dI_2+dI_3+...\)

 \(I=\dfrac {M_1L^2}{12}\)\(\dfrac {M_2L^2}{12}\)+\(\dfrac {M_3L^2}{12}\) ....

 \(I=\dfrac {L^2}{12}(m_1+m_2+m_3+...)\) 

 \(I=\dfrac {ML^2}{12}\)

image

Moment of inertia of the bodies which are expanded parallel to the axis, remains same.

image

Calculate the moment of inertia of a square plate ABCD of edge 'L' about an axis in the plane of the plate parallel to edge AB passing through center.

A

\(\dfrac{ML^2}{3}\)

.

B

\(\dfrac{ML^2}{12}\)

C

\(\dfrac{ML^2}{6}\)

D

\(\dfrac{ML^2}{9}\)

Option B is Correct

Moment of Inertia of a Section of a Body

  • Moment of inertia of a section of any body about an axis perpendicular  to the plane of the body and passing through the center, is equal to the moment of inertia of that complete body; as mass distribution of the section is same as that of the complete body.

Moment of inertia of a section of a disk

  • The moment of inertia of a section of a disk of radius R about an axis perpendicular to the plane of disk, passing through center, is same as that of disk.

Area of section of disk =\(\dfrac{\theta_0}{2}R^2\)

Mass per unit area = \(\dfrac{M}{\dfrac{\theta_0}{2}R^2}\)

Area of the element chosen,  \(dA=r\theta_0.{dr}\)

Mass of element, \(dm=\dfrac{M}{\dfrac{\theta_0}{2}R^2}×r\theta_0.{dr}\)

 \(I=\int{dmr^2}\)

\(I=\dfrac{2M}{R^2}\)\(\int\limits_{0}^{R} r^3dr\)

 \(I=\dfrac{2M}{R^2}\) ×\(\dfrac{R^4}{4}\)

 \(I=\dfrac{MR^2}{2}\)

  • The moment of inertia remains unchanged for a section of a disk or any other body such as ring or cylinder, as the mass distribution is uniform about the axis.

Illustration Questions

Calculate the moment of inertia of a section of a hollow cylinder along the axis shown, whose radius is 2 m, mass is 4 kg and \(\theta_0\) is 90º.

A 40 kg-m2

B 25 kg-m2

C 16 kg-m2

D 20 kg-m2

×

A section of hollow cylinder of mass M = 4 kg, radius R = 2 m and angle \(\theta_0\) = 90º

image

Moment of inertia of a section of any body about an axis perpendicular  to the plane of the body and passing through the center, is equal to the moment of inertia of that complete body; as mass distribution of the section is same as that of the complete body.

image

Moment of inertia of a section of hollow cylinder, \(I=MR^2\)

\(I\) = (4) (2)2

    = 16 kg-m2

image

Calculate the moment of inertia of a section of a hollow cylinder along the axis shown, whose radius is 2 m, mass is 4 kg and \(\theta_0\) is 90º.

image
A

40 kg-m2

.

B

25 kg-m2

C

16 kg-m2

D

20 kg-m2

Option C is Correct

Moment of Inertia of a Square Sheet about any Axis in the Plane of Sheet, Passing through its Center of Mass

  • The two perpendicular axes X and Y cut the square into four exactly identical parts.
  • The mass distribution of the plate about X-axis is same as that about Y-axis.

                     Thus, \(I_X=I_Y\)

Illustration Questions

If the moment of inertia of a square sheet is 3 kg-m2 about the axis shown, then find out the moment of inertia of same sheet along the diagonal.

A 1 kg-m2

B 3 kg-m2

C 2 kg-m2

D 4 kg-m2

×

Since, the moment of inertia of a square sheet about any axis in the plane of the sheet passing through center of mass is same.

Therefore, the moment of inertia along diagonal is 3 kg-m2.

image

If the moment of inertia of a square sheet is 3 kg-m2 about the axis shown, then find out the moment of inertia of same sheet along the diagonal.

image
A

1 kg-m2

.

B

3 kg-m2

C

2 kg-m2

D

4 kg-m2

Option B is Correct

Practice Now