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: Understanding the Concept:
The radius of gyration (K) of a body about an axis of rotation is the effective distance from the axis where the entire mass of the body could be concentrated to produce the same moment of inertia. The relationship is given by \( I = MK^2 \), where I is the moment of inertia and M is the total mass. To find K, we first need to determine the moment of inertia (I) of the disc about its diameter.
Step 2: Key Formula or Approach:
1. Find the moment of inertia of a uniform circular disc about an axis passing through its center and perpendicular to its plane. This is a standard result: \( I_z = \frac{1}{2}MR^2 \).
2. Use the Perpendicular Axis Theorem to find the moment of inertia about a diameter. The theorem states that for a planar lamina, \( I_z = I_x + I_y \), where x and y are two perpendicular axes in the plane of the lamina, and z is the axis perpendicular to the plane.
3. Relate the moment of inertia to the radius of gyration using \( I = MK^2 \) and solve for K.
Step 3: Detailed Explanation:
Let the disc lie in the x-y plane. The axis perpendicular to the disc through its center is the z-axis. The moment of inertia about this axis is:
\[ I_z = \frac{1}{2}MR^2 \] Let \( I_x \) and \( I_y \) be the moments of inertia about two perpendicular diameters (the x-axis and y-axis). Due to the symmetry of the disc, the moment of inertia about any diameter is the same, so \( I_x = I_y = I_{diameter} \).
According to the Perpendicular Axis Theorem:
\[ I_z = I_x + I_y \] \[ I_z = I_{diameter} + I_{diameter} = 2I_{diameter} \] Now, we can find \( I_{diameter} \):
\[ I_{diameter} = \frac{I_z}{2} = \frac{1}{2} \left( \frac{1}{2}MR^2 \right) = \frac{1}{4}MR^2 \] Now we use the definition of the radius of gyration, \( I_{diameter} = MK^2 \):
\[ \frac{1}{4}MR^2 = MK^2 \] Cancel M from both sides:
\[ K^2 = \frac{R^2}{4} \] Take the square root of both sides:
\[ K = \sqrt{\frac{R^2}{4}} = \frac{R}{2} \] Step 4: Final Answer:
The radius of gyration of a uniform circular disc about its diameter is \( \frac{R}{2} \). This corresponds to option (E).