Question

A long cylindrical rod is welded to a thin circular disc of diameter $0.5\,\text{m}$ at a point on its circumference. The rod is in the same plane as that of the disc and forms a tangent to the disc. The radius of gyration of the disc about the rod (in m) is

• A. $\dfrac{1}{4}$
• B. $\dfrac{\sqrt{5}}{8}$
• C. $\dfrac{1}{2}$
• D. $2^2$

Hint:

When an axis is tangent and parallel to a known axis, first find the moment of inertia about the known axis and then apply the parallel axis theorem.

Answer 1

The Correct Option is B

Step 1: Find the radius of the disc. Given diameter $=0.5\,\text{m}$, \[ R=\frac{0.5}{2}=0.25\,\text{m} \]
Step 2: The rod forms a tangent in the plane of the disc. Hence, the axis is parallel to a diameter of the disc and is at a distance $R$ from the centre.
Step 3: Moment of inertia of a thin disc about a diameter: \[ I_{\text{diameter}}=\frac{1}{4}MR^2 \]
Step 4: Use the parallel axis theorem to find the moment of inertia about the tangent: \[ I = I_{\text{diameter}} + MR^2 = \left(\frac{1}{4}+1\right)MR^2 = \frac{5}{4}MR^2 \]
Step 5: Radius of gyration $k$ is defined by: \[ I = Mk^2 \Rightarrow k^2=\frac{5}{4}R^2 \Rightarrow k=\frac{\sqrt{5}}{2}R \]
Step 6: Substitute $R=0.25$ m: \[ k=\frac{\sqrt{5}}{2}\times 0.25=\frac{\sqrt{5}}{8}\,\text{m} \]