The radius of gyration of a uniform square plate of mass M and side a about its diagonal is

Options

A. `a/sqrt(6)`
B. `a/2 sqrt(3)`
C. `a/2 sqrt(2)`
D. `a/4`

Answer

To find the radius of gyration (k) of a uniform square plate of mass M and side a about its diagonal, we can follow these steps:

### Step 1: Understand the Concept of Radius of Gyration
The radius of gyration (k) is defined by the equation:
\[ I = M k^2 \]
where \( I \) is the moment of inertia of the object about the axis of rotation.

### Step 2: Determine the Moment of Inertia about the Diagonal
For a square plate, the moment of inertia about an axis through its center and perpendicular to the plane (z-axis) is given by:
\[ I_z = \frac{1}{6} M a^2 \]
However, we need the moment of inertia about the diagonal. The moment of inertia about the diagonal can be derived using the perpendicular axis theorem, which states:
\[ I_z = I_x + I_y \]
Since the square plate is symmetric, \( I_x = I_y \). Thus, we can express it as:
\[ I_z = 2I_x \]
From the perpendicular axis theorem, we have:
\[ I_z = \frac{1}{6} M a^2 \]
So:
\[ 2I_x = \frac{1}{6} M a^2 \]
This gives:
\[ I_x = \frac{1}{12} M a^2 \]

### Step 3: Use the Parallel Axis Theorem
Now, to find the moment of inertia about the diagonal, we can use the formula:
\[ I_d = I_x + Md^2 \]
where \( d \) is the distance from the center of mass to the diagonal. For a square, this distance is:
\[ d = \frac{a}{2\sqrt{2}} \]
Thus, we can calculate \( d^2 \):
\[ d^2 = \left(\frac{a}{2\sqrt{2}}\right)^2 = \frac{a^2}{8} \]

### Step 4: Substitute into the Moment of Inertia Equation
Now substituting \( I_x \) and \( d^2 \) into the equation for \( I_d \):
\[ I_d = \frac{1}{12} M a^2 + M \cdot \frac{a^2}{8} \]
To combine these, we need a common denominator:
\[ I_d = \frac{1}{12} M a^2 + \frac{3}{24} M a^2 \]
\[ I_d = \frac{2}{24} M a^2 + \frac{3}{24} M a^2 = \frac{5}{24} M a^2 \]

### Step 5: Solve for the Radius of Gyration
Now we can find the radius of gyration \( k \) using the equation:
\[ I_d = M k^2 \]
Thus:
\[ \frac{5}{24} M a^2 = M k^2 \]
Dividing both sides by \( M \):
\[ k^2 = \frac{5}{24} a^2 \]
Taking the square root:
\[ k = a \sqrt{\frac{5}{24}} \]

### Final Result
The radius of gyration of the uniform square plate about its diagonal is:
\[ k = \frac{a}{2} \sqrt{\frac{5}{3}} \]