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Perpendicular Axis Theorem

Practice perpendicular axis theorem example. Learn how to calculate moment of inertia of a ring of mass and radius about the given tangent.

Perpendicular Axis Theorem

  • For a planar body, consider that X and Y-axes are in the plane of the body and Z-axis is perpendicular to the plane of the body. The three axes chosen should be mutually perpendicular.
  • Then, the theorem states that -

Moment of inertia along Z-axis = Sum of Moment of inertia along X and Y-axes

          i.e., \(I_Z = I_X + I_Y\)

Note :  This theorem is applicable only to the planar bodies

 

Illustration Questions

In which one of the following figures, \(I_Z = I_X + I_Y\) can be applied?

A A cuboid

B A sphere

C A disk

D A disk

×

Options (A) and (B) are incorrect because given bodies are non-planar.

Option (C) is correct because given body is planar and lies in X – Y plane.

Thus, \(I_Z = I_X + I_Y\) holds true.

Option (D) is incorrect because given body lies in Z–X plane and not in X –Y plane.

Thus, here \(I_Y=I_Z+I_X\) holds true.

In which one of the following figures, \(I_Z = I_X + I_Y\) can be applied?

A

A cuboid

.

image
B

A sphere

image
C

A disk

image
D

A disk

image

Option C is Correct

Application of Perpendicular Axis Theorem

  • For a planar body, consider that X and Y-axes are in the plane of the body and Z-axis is perpendicular to the plane of the body. The three axes chosen should be mutually perpendicular.
  • Then, the theorem states that -

Moment of inertia along Z-axis = Sum of Moment of inertia along X and Y-axes

          i.e., \(I_Z = I_X + I_Y\)

Note :  This theorem is applicable only to the planar bodies

Illustration Questions

A plate lies in X –Y plane as shown. Find moment of inertia of the plate about OZ axis. Given that moment of inertia of the body about OX and OY are \(I_1\) and \(I_2\) respectively.

A \(I_1-I_2\)

B \(I_1+I_2\)

C \(2I_1+I_2\)

D \(I_1-2I_2\)

×

Moment of inertia about  OX, \(I_X=I_1\)

Moment of inertia about  OY, \(I_Y=I_2\)

Moment of inertia about  OZ, \(I_Z=I\)

image

Applying perpendicular -axis theorem

\(I_Z=I_X+I_Y\)

\(\therefore\) \(I=I_1+I_2\)

image

A plate lies in X –Y plane as shown. Find moment of inertia of the plate about OZ axis. Given that moment of inertia of the body about OX and OY are \(I_1\) and \(I_2\) respectively.

image
A

\(I_1-I_2\)

.

B

\(I_1+I_2\)

C

\(2I_1+I_2\)

D

\(I_1-2I_2\)

Option B is Correct

Moment of Inertia For a Ring about its Diameter

  • Consider a ring of radius R and mass  M.

  • Moment of inertia of the ring about OZ, \(I_Z = MR^2\)....(1)
  • The mass distribution about the axes OX and OY is same. Thus, moment of inertia about OX and OY will also be same.

              Hence, \(I_X=I_Y=I\) ....(2)

  • Applying perpendicular axis theorem-

   \(I_Z=I_X+I_Y\)

\(MR^2=2I\)                 [ From equation (1) and (2) ]

       \(\therefore\) \(I=\dfrac {MR^2}{2}\)

  • Consider a disk of mass M and radius R.

  • Moment of inertia of the disk about OZ,  \(I_Z=\dfrac {MR^2}{2}\)...(1)
  • The mass distribution about the axes OX and OY is same. Thus, moment of inertia about OX and OY will also be same.

          Hence, \(I_X=I_Y=I\) ....(2)

  • Applying perpendicular axis theorem-

           \(I_Z=I_X+I_Y\)

          \(\dfrac {MR^2}{2}=2I\)

          \(I=\dfrac {MR^2}{4}\)

Illustration Questions

A ring of mass (M) 4 kg and  radius (R) 3 m is lying in X-Y plane with center O. Calculate its moment of inertia about X-axis.

A 14 kg m2

B 16 kg m2

C 12 kg m2

D 18 kg m2

×

Ring of Mass M = 4 kg and

Radius R = 3 m

image

Moment of inertia about X-axis

\(I=\dfrac {MR^2}{2}\)

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

\(I=18\) kg m2

A ring of mass (M) 4 kg and  radius (R) 3 m is lying in X-Y plane with center O. Calculate its moment of inertia about X-axis.

image
A

14 kg m2

.

B

16 kg m2

C

12 kg m2

D

18 kg m2

Option D is Correct

Moment of Inertia of A Square Plate about any Axis Passing through its Center of Mass and Perpendicular to the Plane

  • Consider a square plate of length L and mass M.

  •  The square plate is a rod extended parallel to its axis.
  • If \(I_X\) be the moment of inertia about X-axis and \(I_Y\) be the moment of inertia about Y-axis then

Applying perpendicular axis theorem,

\(I_Z=I_X+I_Y\)

\(I_Z=\) \(\dfrac {ML^2}{12}+\dfrac {ML^2}{12}\)

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

where, M = Mass of square plate

         L = length of square plate

 

  • Consider a square plate of length L. The X and Y-axes are cutting the plane, as shown in figure.

  • The plate does not seem to be symmetric about X and Y-axes, but the four parts into which X and Y-axes cut the plane are exactly same.
  • Since mass distribution is same along X and Y-axes.

         Therefore, \(I_X=I_Y=I\)

Applying perpendicular axis theorem

\(I_Z=I_X+I_Y\)

\(\Rightarrow\dfrac{ML^2}{6}=2I\)

\(\Rightarrow \;I=\dfrac {ML^2}{12}\)

where, M = Mass of square plate

L = length of square plate

Illustration Questions

Calculate moment of inertia of a square plate ABCD of mass (M) 2 kg and side (L) 6 m about the diagonal (AB), as shown in figure.

A 3 kg m2

B 2 kg m2

C 8 kg m2

D 6 kg m2

×

Moment of inertia of a square plate about any axis passing through center of mass in the plane of the plate is same. 

image

A square plate of mass M = 2 kg and length L = 6 m 

image

Moment of inertia about AB

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

\(I= \dfrac {(2) (6)^2} {12}\)

   = 6 kg m2

image

Calculate moment of inertia of a square plate ABCD of mass (M) 2 kg and side (L) 6 m about the diagonal (AB), as shown in figure.

image
A

3 kg m2

.

B

2 kg m2

C

8 kg m2

D

6 kg m2

Option D is Correct

Moment of Inertia of a Ring about a Tangent in the Plane

  • Consider a ring of mass M and radius R.

  • Moment of inertia of the ring about an axis passing through its center of mass and perpendicular to the plane is

?           \(I_Z = MR^2\) ...(1)

  • Moment of inertia of the ring about X and Y-axes are equal.

          Thus, \(I_X=I_Y=I\)

  • Applying perpendicular axis theorem

            \(I_Z=I_X+I_Y\)

          \({MR^2}=2I\)    [ From equation (1) ]

          \(\dfrac {MR^2}{2}=I=I_X=I_Y\)

  • For the calculation of \(I_A\) ;

         use parallel axis theorem,

         \(I=I_0+Md^2\)

        \(I_A=I_Y+MR^2\)

         \(I_A=\dfrac {MR^2}{2}+MR^2\)

        \(I_A=\dfrac {3MR^2}{2}\)

  • Consider a disk of mass M and radius R.

  • Moment of inertia of the disk about an axis passing through its center of mass and perpendicular to the plane is

            \(I_Z = \)\(\dfrac {MR^2}{2}\)...(1)

  • Moment of inertia of the disk about X and Y-axes are equal.

           Thus,  \( I_X = I_Y = I\)

  • Applying perpendicular axis-theorem

          \( I_Z = I_X + I_Y \)

        \(\dfrac {MR^2}{2}=2I\)        [ From equation (1) ]

        \(\dfrac {MR^2}{4}=I=I_X=I_Y\)

  • For the calculation of \(I_A\);

      use parallel-axis theorem-

          \( I = I_0 + Md^2 \)

          \( I_A = I_Y +MR^2 \)

          \(I_A=\dfrac {MR^2}{4}+MR^2\)

          \(I_A=\dfrac {5}{4}MR^2\)

Illustration Questions

Find moment of inertia of a ring of mass (M) 20 kg and radius (R) 5 m about the given tangent (AB) as shown in figure.

A 750 kg m2

B 250 kg m2

C 500 kg m2

D 225 kg m2

×

A ring of mass M = 20 kg and radius R = 5 m

image

Moment of inertia about the given tangent

      \(I_A=\dfrac {3MR^2}{2}\)

           \(=\dfrac {3(20(5)^2}{2}\)

           = 750 kg m2

image

Find moment of inertia of a ring of mass (M) 20 kg and radius (R) 5 m about the given tangent (AB) as shown in figure.

image
A

750 kg m2

.

B

250 kg m2

C

500 kg m2

D

225 kg m2

Option A is Correct

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