If the moment of inertion of a disc about an axis tangentially and parallel to its surface by I, then what will be the moment of inertia about the axis tangential but perpendicular to the surface

Options

A. `(6)/(5)I`
B. `(3)/(4)I`
C. `(3)/(2)I`
D. `(5)/(4)I`

Answer

To solve the problem of finding the moment of inertia of a disc about an axis tangential but perpendicular to its surface, we can follow these steps:

### Step 1: Understand the Given Information
We know that the moment of inertia of a disc about an axis tangentially and parallel to its surface is given as \( I \).

### Step 2: Use the Moment of Inertia Formula for a Disc
For a circular disc, the moment of inertia about an axis passing through its center and parallel to the surface is given by:
\[ I_{\text{center, parallel}} = \frac{1}{4} m r^2 \]
where \( m \) is the mass of the disc and \( r \) is its radius.

### Step 3: Apply the Parallel Axis Theorem
To find the moment of inertia about the tangential axis parallel to the surface, we use the parallel axis theorem:
\[ I_{\text{tangential, parallel}} = I_{\text{center, parallel}} + m d^2 \]
where \( d \) is the distance from the center of the disc to the tangential axis. In this case, \( d = r \) (the radius of the disc).

Substituting the values, we have:
\[ I = \frac{1}{4} m r^2 + m r^2 \]
\[ I = \frac{1}{4} m r^2 + \frac{4}{4} m r^2 = \frac{5}{4} m r^2 \]

### Step 4: Relate \( m r^2 \) to \( I \)
From the above equation, we can express \( m r^2 \):
\[ m r^2 = \frac{4}{5} I \]

### Step 5: Find the Moment of Inertia for the Perpendicular Axis
Now, we need to find the moment of inertia about the axis tangential but perpendicular to the surface. The moment of inertia about the center and perpendicular to the surface is given by:
\[ I_{\text{center, perpendicular}} = \frac{1}{2} m r^2 \]

Using the parallel axis theorem again:
\[ I_{\text{tangential, perpendicular}} = I_{\text{center, perpendicular}} + m d^2 \]
Here, \( d = r \) again, so:
\[ I' = \frac{1}{2} m r^2 + m r^2 \]
\[ I' = \frac{1}{2} m r^2 + \frac{2}{2} m r^2 = \frac{3}{2} m r^2 \]

### Step 6: Substitute \( m r^2 \) from Earlier
Now, substituting \( m r^2 = \frac{4}{5} I \) into the equation:
\[ I' = \frac{3}{2} \left(\frac{4}{5} I\right) \]
\[ I' = \frac{12}{10} I = \frac{6}{5} I \]

### Final Answer
Thus, the moment of inertia about the axis tangential but perpendicular to the surface is:
\[ I' = \frac{6}{5} I \]