Informative line

Parallel Axis Theorem

Learn how to calculate moment of inertia of ring and disk about axis passing through their circumference. Practice parallel axis theorem example.

Parallel Axis Theorem

  • Suppose, moment of inertia (MOI) of a body about a given line AB is to be calculated.
  • CZ is the line parallel to AB passing through center of mass C of the body.
  • \(I\) and \( I_{CM}\)  are the moment of inertia of the body about AB and CZ, respectively.

  • The parallel - axis theorem states :

           \( I = I_{CM} + Md^2\)

where, d is the perpendicular distance between AB and CZ and M is the mass of the body.

Note :

  • \(I_{CM}\) is the moment of inertia about an axis passing through the center of mass.
  • The two axes taken, must be parallel.

Illustration Questions

In which of the following figures, \(I\)=\(I_0 + Md^2\) can be applied?

A

B

C

D

×

Option (A) is incorrect because the axis (along which moment of inertia is \(I_0\)), is not passing through the center of mass.

image

Option (B) is incorrect because the axes (along which moment of inertia are \(I\) and \(I_0\) ), are not parallel to each other.

image

Option (C) is incorrect because the axes (along which moment of inertia are \(I\) and \(I_0\) ), are not parallel to each other

 

image

Option (D) is correct because both the axes, (along which moment of inertia are \(I\) and \(I_0\)), are parallel to each other. Also, the axis (along which moment of inertia is \(I_0\)), is passing through the center of mass of the body.

image

In which of the following figures, \(I\)=\(I_0 + Md^2\) can be applied?

A image
B image
C image
D image

Option D is Correct

Moment of Inertia of a Uniform Rod about an End

(using parallel axis theorem)

  • The moment of inertia of a uniform rod about a perpendicular axis passing through its center of mass is

            \(I_0=\dfrac {ML^2}{12}\) ....(1)

  • To calculate moment of inertia of the rod about the axis AB (perpendicular to the rod passing through an end), parallel-axis theorem is used.

  • By parallel- axis theorem, \(I=I_0+Md^2\)

The distance between the two parallel axes considered is,

\( d =\) \(\dfrac {L}{2}\)  ... (2)

 \(I=\dfrac {ML^2}{12}\)+ \(M\)\(\left(\dfrac {L}{2}\right)^2\)       ... [from (1) & (2)]

\(I\)\(\dfrac {ML^2}{3}\) 

Illustration Questions

Calculate the moment of inertia \((I)\) of a rod of length (L) 20 m and mass (M) 1 kg about the axis AB, as shown in figure. Given that the moment of inertia (\(I_0\)) about the line perpendicular to the rod and passing through the center is 26 kg-m2.

A 100 kg-m2

B 130 kg-m2

C 126 kg-m2

D 120 kg-m2

×

Length of rod, L = 20 m

Mass of rod, M = 1 kg

Moment of inertia about the perpendicular line at the center of rod, \(I_0\) = 26 kg-m2

Distance between two axes, d = 10 m

image

Using parallel-axis theorem;

Moment of inertia about AB

\(I=I_0+Md^2\)

\(I\) = 26 + 1(10)2

= 26 + 100

= 126 kg-m2

image

Calculate the moment of inertia \((I)\) of a rod of length (L) 20 m and mass (M) 1 kg about the axis AB, as shown in figure. Given that the moment of inertia (\(I_0\)) about the line perpendicular to the rod and passing through the center is 26 kg-m2.

image
A

100 kg-m2

.

B

130 kg-m2

C

126 kg-m2

D

120 kg-m2

Option C is Correct

Moment of Inertia of Ring and Disk about Axis passing through their Circumference

 

For ring

  • Moment of inertia of a ring of mass M and radius R about its axis perpendicular to the plane, passing through its center is

?            \(I_0=MR^2\) ...(1)

 

  • Then, moment of inertia \((I)\) about the axis AB passing through its circumference is given by,

using parallel-axis theorem,

\(I=I_0+Md^2\)

  • For ring, d = R  ... (2)

Thus,  \( I = MR^2 + M(R)^2 \)    [ from (1) and (2)]

         \( I = 2MR^2\)

For disk 

  • Moment of inertia of a disk of mass M and radius R about its axis perpendicular to the plane, passing through its center is 

\(I_0=\dfrac {MR^2}{2}\) ...(3)

  • Then, moment of inertia (I) about the axis AB passing through the circumference is given by,

using parallel-axis theorem,

\(I=I_0+Md^2\)

  • For disk, d = R ... (4)

Thus,  \(I=\dfrac{MR^2}{2}+M(R)^2\)... [from equation (3) and (4)]

          \(I=\dfrac {3MR^2}{2}\)

Illustration Questions

Calculate the moment of inertia of a disk of mass M = 2 kg and radius R = 1 m about the perpendicular axis passing through its circumference.  

A 1 kg m2

B 2 kg m2

C 3 kg m2

D  4 kg m2

×

Mass of disk (M) = 2 kg

Radius (R)  = 1 m

image

Moment of Inertia

 \(I=\dfrac {3}{2}MR^2\)

    =\(\dfrac {3}{2}(2)(1)^2\) 

    = 3 kg m2

image

Calculate the moment of inertia of a disk of mass M = 2 kg and radius R = 1 m about the perpendicular axis passing through its circumference.  

image
A

1 kg m2

.

B

2 kg m2

C

3 kg m2

D

 4 kg m2

Option C is Correct

Moment of Inertia of a Square Plate

 

  • Consider an element as a rod of thickness \(dx\) of a square sheet of mass M and length L. Mass of the element is given by :

 \(dm=\dfrac {Mass \;of \;body} {Area \;of \;body}{(Area\; of \;element)}\) 

 \(dm=\dfrac{M}{L^2}(Ldx)\)

 \(dm=\dfrac{M}{L}(dx)\)  ...(1)

  • Moment of inertia of the element about an axis passing through its center of mass (z'-axis) is

 \(dI_0=\dfrac{dmL^2}{12}\)

  • If the moment of inertia of all the elements about the axis of rotation of the body is known, then on integrating it, the moment of inertia of the whole body about the axis is obtained.
  • For the calculation of moment of inertia of element about z-axis, parallel- axis theorem can be used.
  • Thus, \(I_z=dI_0+Md^2\)  

       ⇒  \(dI=\dfrac{dmL^2}{12}+dmx^2\) 

  • Moment of inertia of square sheet,

 \(I=\int dI\)

=\(\int \left(\dfrac {L^2}{12}+ x^2\right)dm\)

\(\dfrac{M}{L}\) \(\int\limits_{–L/2}^{L/2}\)\(\left(\dfrac {L^2}{12}+ x^2\right)dx\) ...[from(1)]

\(I\) = \(\dfrac{M}{L}\) \(\left[\dfrac {L^2x}{12}+ \dfrac{x^3}{3}\right]_{-L/2}^{L/2}\)

 \(I=\dfrac{ML^2}{6}\)

Illustration Questions

Calculate the moment of inertia of a square plate of mass M = 4 kg and length L = 3 m about the axis AB, as shown in figure.

A 16 kg m2

B 8 kg m2

C 6 kg m2

D 10 kg m2

×

 Mass of square plate (M) =  4 kg

Length (L) = 3 m

image

Moment of inertia

 \(I=\dfrac{ML^2}{6}\)

\(I=\dfrac{4(3)^2}{6}\)

\(I = 6\; kg\,\,m^2\)

image

Calculate the moment of inertia of a square plate of mass M = 4 kg and length L = 3 m about the axis AB, as shown in figure.

image
A

16 kg m2

.

B

8 kg m2

C

6 kg m2

D

10 kg m2

Option C is Correct

Moment of Inertia of Combination of Ring and Disk Attached to Each Other about a Line AB

  •  Consider a disk of mass Mand radius R1 and a ring of mass M2 and radius R2.

  • The moment of inertia of a disk about the perpendicular axis (X1Y1) passing through center of mass is

\(I_{o_1}=\dfrac{M_1R_1^2}{2}\)

  • By using parallel axis theorem, moment of inertia (\(I_1\)) of a disk about axis AB is

\(I=I_{o_1}+Md^2\)

 \(I_1=\dfrac{M_1R_1^2}{2}\) + \(M_1R_1^2\)

\(I_1=\dfrac {3}{2}M_1R_1^2\) ...(1)

  • The moment of inertia  of a ring about the perpendicular axis (X2Y2) passing through the center of mass is

  \(I_{o_2}=M_2R_2^2\)

  • By using parallel axis theorem, the moment of inertia (\(I_2\)) of a ring about axis AB is

\(I_2=I_{o_2}+Md^2\)

  \(I_2=M_2R_2^2\) + \(M_2R_2^2\)

\(I_2=2M_2R_2^2\)  ... (2)

  • Thus, the moment of inertia of the combination of ring and disk about the given axis AB is the sum of moment of inertia of individual bodies about AB.

Mathematically,

\(I=I_1+I_2\)

 \(I=\dfrac {3}{2}M_1R_1^2\) + \(2M_2R_2^2\)

Illustration Questions

A disk of mass (M1) 4 kg and radius (R1) 3 m is attached to a ring of mass (M2) 3 kg and radius (R2) 1 m, as shown in figure. Calculate the moment of inertia of the combination of both the bodies about the axis AB.  

A 19 kg m2

B 60 kg m2

C 20 kg m2

D 61 kg m2

×

A disk of mass (M1) 4 kg and radius (R1) 3 m is attached to a ring of mass (M2) 3 kg and radius (R2) 1 m 

 

image

Moment of inertia

 \(I=\dfrac{3}{2} M_1R_1^2 +2\,M_2R_2^2\)

\(\dfrac{3}{2} (4) (3)^2 + 2(3) (1)^2\)

\( 54 + 6\)

\( 60\; kg\,m^2\)

image

A disk of mass (M1) 4 kg and radius (R1) 3 m is attached to a ring of mass (M2) 3 kg and radius (R2) 1 m, as shown in figure. Calculate the moment of inertia of the combination of both the bodies about the axis AB.  

image
A

19 kg m2

.

B

60 kg m2

C

20 kg m2

D

61 kg m2

Option B is Correct

Moment of Inertia of a Disk when a Smaller Disk is removed from it

  • Consider a disk of mass M and radius R.
  • As the disk is of uniform thickness, then

     mass (M) = Density \(\left(\rho\right)\) × volume (V)

 \(M=\rho\times(A\times\,t)\)

 \(M=\rho\)  \(\times\,\pi{R}^2\,t\) ... (1)

  • A small disk of mass 'm' is removed from it.

Mass , m = Density × Volume

 \(m=\rho\times(A_1\times\,t)\)

 \(m=\rho\) \(\times\,\pi{r}^2×t\) 

\(\rho\) and t for smaller disk are same as that of large disk

       \(\therefore\)  \(m=\dfrac{M}{\pi{R^2×t}} ×\pi r^2×t\)...[from(1)]

 \(m=\dfrac{Mr^2}{R^2}\) or   \(\dfrac {m}{M}\)\(\left(\dfrac{r}{R}\right)^2\)

  • If the moment of inertia of the small disk and the remaining disk is \( I'\) and \( I\), respectively

then,

Moment of inertia of the whole disk about the axis CB is 

\(I_0=I+I' \)

\(\because\,I_0=\dfrac{MR^2}{2}\)\(I'=\dfrac{mr^2}{2}+md^2\)

\(\therefore\, I=I_0-I'\)

\(\Rightarrow I=\dfrac{MR^2}{2}\)– \(\dfrac{mr^2}{2}-md^2\)

Substituting the value of m

\(I=\dfrac{MR^2}{2}\) –  \(\dfrac{Mr^2}{R^2}\)\(\left(\dfrac {r^2}{2}+d^2\right)\)

Illustration Questions

If a small disk of radius (r) 2 m is removed from a large disk of radius (R) 4 m and mass M, then calculate the moment of inertia of the remaining disk about the axis AB, as shown in figure.

A \(\dfrac{43}{2} \,M\)

B \(\dfrac{45}{2} \,M\)

C \(\dfrac{42}{2} \,M\)

D \(\dfrac{41}{2} \,M\)

×

Radius of small disk, r = 2 m 

Radius of large disk, R = 4 m

image

The mass of small disk,

 \(m=\dfrac{(Mass\; of \; large \; disk)×r^2}{R^2}\)

 \(m=\dfrac{M(2)^2}{(4)^2}\)\(\dfrac{M}{4}\) ... (1)

image

Moment of inertia of the small disk about AB, \(I'=\dfrac{3}{2}mr^2\)..(2)

Moment of inertia of the complete large disk about AB,  \(I_0=\dfrac{3}{2}MR^2\) ... (3)

image

Moment of inertia of whole disk about AB, \(I_0=I+I'\)

Moment of inertia of remaining disk, \(I=I_0-I'\)

image

  \(I=\dfrac{3}{2}MR^2-\dfrac{3}{2}mr^2\)

    \(=\dfrac{3}{2}M(4)^2-\dfrac{3}{2}(\dfrac{M}{4})2^2\)

    \(=\dfrac{45}{2}\,M\)

image

If a small disk of radius (r) 2 m is removed from a large disk of radius (R) 4 m and mass M, then calculate the moment of inertia of the remaining disk about the axis AB, as shown in figure.

image
A

\(\dfrac{43}{2} \,M\)

.

B

\(\dfrac{45}{2} \,M\)

C

\(\dfrac{42}{2} \,M\)

D

\(\dfrac{41}{2} \,M\)

Option B is Correct

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