A uniform disc of radius R lies in x-y plane with its centre at origin. Its moment of inertia about the axis x=2R and y=0 is equal to the moment of inertia about the axis y=d and z=0, where d is equal to

Options

A. `(4)/(3)R`
B. `(sqrt17)/(2)R`
C. `sqrt13R`
D. `(sqrt15)/(2)R`

Answer

To solve the problem of finding the distance \( d \) such that the moment of inertia of a uniform disc about the axis \( x = 2R \) and \( y = 0 \) is equal to the moment of inertia about the axis \( y = d \) and \( z = 0 \), we can follow these steps:

### Step 1: Calculate the Moment of Inertia \( I_1 \) about the axis \( x = 2R \) and \( y = 0 \)

Using the parallel axis theorem, the moment of inertia about an axis parallel to the one through the center of mass is given by:

\[ I_1 = I_{cm} + Mh^2 \]

Where:
- \( I_{cm} \) is the moment of inertia about the center of mass,
- \( M \) is the mass of the disc,
- \( h \) is the distance from the center of mass to the new axis.

For a uniform disc, the moment of inertia about its center is:

\[ I_{cm} = \frac{1}{2} MR^2 \]

The distance \( h \) from the center of the disc (at the origin) to the axis \( x = 2R \) is \( 2R \). Thus:

\[ I_1 = \frac{1}{2} MR^2 + M(2R)^2 = \frac{1}{2} MR^2 + 4MR^2 = \frac{1}{2} MR^2 + 8MR^2 = \frac{9}{2} MR^2 \]

### Step 2: Calculate the Moment of Inertia \( I_2 \) about the axis \( y = d \) and \( z = 0 \)

Again using the parallel axis theorem, we can express \( I_2 \):

\[ I_2 = I_{cm} + Md^2 \]

For the axis through the center of mass (diameter), the moment of inertia is:

\[ I_{cm} = \frac{1}{4} MR^2 \]

Thus, we have:

\[ I_2 = \frac{1}{4} MR^2 + Md^2 \]

### Step 3: Set \( I_1 \) equal to \( I_2 \)

According to the problem, we have:

\[ I_1 = I_2 \]

Substituting the expressions we derived:

\[ \frac{9}{2} MR^2 = \frac{1}{4} MR^2 + Md^2 \]

### Step 4: Simplify the Equation

We can cancel \( M \) from both sides (assuming \( M \neq 0 \)):

\[ \frac{9}{2} R^2 = \frac{1}{4} R^2 + d^2 \]

Now, we can rearrange this:

\[ d^2 = \frac{9}{2} R^2 - \frac{1}{4} R^2 \]

Finding a common denominator (which is 4):

\[ d^2 = \frac{18}{4} R^2 - \frac{1}{4} R^2 = \frac{17}{4} R^2 \]

### Step 5: Solve for \( d \)

Taking the square root of both sides:

\[ d = \sqrt{\frac{17}{4} R^2} = \frac{\sqrt{17}}{2} R \]

### Final Answer

Thus, the value of \( d \) is:

\[ d = \frac{\sqrt{17}}{2} R \]

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