Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

System of Particle and Rotational Motion

Question:

What is the radius of gyration of a uniform rod whose length is L and passes through the centre of mass ?

Options:

\(\frac{L}{\sqrt{3}}\)

\(\frac{L}{2\sqrt{3}}\)

\(\frac{L}{\sqrt{2}}\)

\(\frac{L^2}{12}\)

Correct Answer:

\(\frac{L}{2\sqrt{3}}\)

Explanation:

Radius of gyration : k

\(I = \frac{1}{12} mL^2\) 

Now, as mk2 = I

\(\Rightarrow mk^2 = \frac{1}{12} mL^2\)

\(\Rightarrow k = \frac{1}{\sqrt{12}}\)

and \(\Rightarrow k = \frac{1}{2\sqrt{3}}\)