Question

Radius of gyration of a uniform circular disc of radius \(R\) about its diameter is

• A. \(\frac{R}{8}\)
• B. \(2R\)
• C. \(R\)
• D. \(\frac{R}{4}\)
• E. \(\frac{R}{2}\)

Hint:

For a circular disc, first remember \(I=\frac{1}{2}MR^2\) about the perpendicular central axis, then use the perpendicular axis theorem to get the moment of inertia about a diameter.

Answer 1

Answer: (E)

Step 1: Recall the definition of radius of gyration.
The radius of gyration \(k\) about a given axis is defined by the relation:
\[ I=Mk^2 \] where \(I\) is the moment of inertia of the body about that axis and \(M\) is its mass. Hence, to find \(k\), we first need the moment of inertia of the disc about its diameter.

Step 2: Recall the standard moment of inertia of a disc about an axis perpendicular to its plane through the center.
For a uniform circular disc of radius \(R\), the moment of inertia about the axis through its center and perpendicular to its plane is:
\[ I_z=\frac{1}{2}MR^2 \]

Step 3: Use the perpendicular axis theorem.
For a plane lamina such as a circular disc, the perpendicular axis theorem states:
\[ I_z=I_x+I_y \] where \(I_x\) and \(I_y\) are the moments of inertia about two mutually perpendicular diameters in the plane of the disc.

Step 4: Use symmetry of the disc.
Since the disc is uniform and perfectly circular, all diameters through the center are equivalent by symmetry. Therefore:
\[ I_x=I_y \] Let each of them be \(I_d\), the moment of inertia about any diameter. Then:
\[ I_z=I_d+I_d=2I_d \]

Step 5: Find the moment of inertia about the diameter.
Substituting \[ I_z=\frac{1}{2}MR^2 \] into \[ I_z=2I_d, \] we get:
\[ 2I_d=\frac{1}{2}MR^2 \] \[ I_d=\frac{1}{4}MR^2 \]

Step 6: Use \(I=Mk^2\) to find the radius of gyration.
Now, \[ I_d=Mk^2 \] So, \[ Mk^2=\frac{1}{4}MR^2 \] Cancelling \(M\), we get:
\[ k^2=\frac{R^2}{4} \] \[ k=\frac{R}{2} \]

Step 7: State the final answer.
Hence, the radius of gyration of the uniform circular disc about its diameter is:
\[ \boxed{\frac{R}{2}} \] which matches option \((5)\).