Informative line

### 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

#### 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

. B

A sphere C

A disk D

A disk 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

#### 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, $$I_Z=I$$ Applying perpendicular -axis theorem

$$I_Z=I_X+I_Y$$

$$\therefore$$ $$I=I_1+I_2$$ ### 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

$$I_1-2I_2$$

Option B 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-

$$I = I_0 + Md^2$$

$$I_A = I_Y +MR^2$$

$$I_A=\dfrac {MR^2}{4}+MR^2$$

$$I_A=\dfrac {5}{4}MR^2$$

#### 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 Moment of inertia about the given tangent

$$I_A=\dfrac {3MR^2}{2}$$

$$=\dfrac {3(20(5)^2}{2}$$

= 750 kg m2 ### 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

Option A 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-

$$I_Z=I_X+I_Y$$

$$\dfrac {MR^2}{2}=2I$$

$$I=\dfrac {MR^2}{4}$$

#### 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 $$I=\dfrac {MR^2}{2}$$

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

$$I=18$$ 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. 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

#### 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. A square plate of mass M = 2 kg and length L = 6 m $$I =\dfrac {ML ^2} {12}$$

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

= 6 kg m2 ### 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

Option D is Correct