Assertion A: Moment of inertia of a circular disc of mass M and radius R about X, Y-axes (passing th

Q: Why is $I_X$ equal to $I_Y$?

Because a uniform circular disc is perfectly symmetrical. The distribution of mass around any diameter (like the $X$ or $Y$ axis) is identical, so the resistance to rotation (moment of inertia) is the same.

Q: Can we use the Parallel Axis Theorem here?

No, the Parallel Axis Theorem is used to find the moment of inertia about an axis parallel to an axis passing through the center of mass. Here, all three axes intersect at the center of mass, so we use the Perpendicular Axis Theorem.

Q: What exactly is the radius of gyration physically?

The radius of gyration is the radial distance from the axis of rotation to a point where, if the entire mass of the body were concentrated, the moment of inertia would remain the same.

Q: Is Reason R related to Assertion A?

No, Reason R is an independent, factually correct statement defining a rigid body. It does not explain why Assertion A is true or false.

Q: Why is $I_Z = \frac{MR^2}{2}$?

This is a standard derived result for a uniform solid disc. It is calculated by integrating the moments of inertia of infinitesimally thin rings that make up the disc from the center to the radius $R$.

Question

Assertion A: Moment of inertia of a circular disc of mass $M$ and radius $R$ about $X, Y$-axes (passing through its plane) and $Z$-axis which is perpendicular to its plane were found to be $I_x, I_y$ and $I_z$, respectively. The respective radii of gyration about all the three axes will be the same.
Reason R: A rigid body making rotational motion has fixed mass and shape.
In the light of the above statements, choose the most appropriate answer from the options given below.

Options:

(A) Both A and R are correct and R is the correct explanation of A.

(B) Both A and R are correct but R is not the correct explanation of A.

(C) A is correct but R is not correct.

(D) A is not correct but R is correct.

Sigma Percentile · Accepted Answer

### Visualizing the Disc

Imagine a uniform circular disc of mass $M$ and radius $R$ lying perfectly flat in the $X-Y$ plane. The $Z$-axis passes right through its center, standing perpendicular to the plane of the disc. This is a classic setup in rotational mechanics, and understanding the distribution of mass around these axes is crucial.

### The Perpendicular Axis Theorem

To find the relationship between the moments of inertia about these three axes, we invoke the **Perpendicular Axis Theorem**. This theorem is a powerful tool for planar bodies (like our disc) and states that the moment of inertia about an axis perpendicular to the plane is equal to the sum of the moments of inertia about two mutually perpendicular axes lying in the plane.

Mathematically, this is expressed as:

$$I_Z = I_X + I_Y$$

### Utilizing Symmetry

We already know the standard formula for the moment of inertia of a uniform circular disc about its central perpendicular axis (the $Z$-axis):

$$I_Z = \frac{MR^2}{2}$$

Now, look at the disc from the top. It is perfectly symmetrical. Whether you spin it around the $X$-axis or the $Y$-axis, the mass distribution looks exactly the same. Because of this symmetry, the moment of inertia about any diameter must be equal:

$$I_X = I_Y$$

Substituting this back into our theorem gives:

$$I_Z = 2I_X$$

$$\Rightarrow I_X = I_Y = \frac{I_Z}{2} = \frac{MR^2}{4}$$

### Radius of Gyration

The radius of gyration, denoted by $k$, is a measure of how the mass of a rotating body is distributed about its axis of rotation. It is defined by the equation:

$$k = \sqrt{\frac{I}{M}}$$

This tells us that the radius of gyration is directly proportional to the square root of the moment of inertia ($k \propto \sqrt{I}$).

### Conclusion

Let's compare the moments of inertia we found:

$$I_X = I_Y = \frac{MR^2}{4}$$

$$I_Z = \frac{MR^2}{2}$$

Clearly, $I_X$ and $I_Y$ are not equal to $I_Z$. Since the moments of inertia are different, their corresponding radii of gyration must also be different ($k_X = k_Y 
eq k_Z$). Therefore, **Assertion A is incorrect**.

On the other hand, **Reason R** states that a rigid body making rotational motion has a fixed mass and shape. This is the fundamental definition of a rigid body in mechanics and is absolutely correct.

Thus, Assertion A is false, but Reason R is true.

Select Answer:

Visualized Solution

$V i s u a l i z in g t h e D i sc$

$P er p e n d i c u l a r A x i s T h eor e m$

$S y mm e t r y o f t h e D i sc$

$C a l c u l a t in g I_{X} an d I_{Y}$

$R a d i u s o f G y r a t i o n$

$C o n c l u s i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Moment of Inertia

Visualizing the Disc

The Perpendicular Axis Theorem

Utilizing Symmetry

Radius of Gyration

Conclusion

Similar Questions

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A circular disc of radius $b$ has a hole of radius $a$ at its centre (see figure). If the mass per unit area of the disc varies as $(\frac{σ _{0}}{r})$ , then the radius of gyration of the disc about its axis passing through the centre is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Disc

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Perpendicular Axis Theorem

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Symmetry of the Disc

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating IX​ and IY​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Radius of Gyration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward