A uniform rod of length 1.5 m is suspended from one end and performs small angular oscillations in a vertical plane. Take g = 10 m/s². If the square of its angular frequency is N s⁻², find N.
Physics · Simple Harmonic Motion · angular shm and its time period
JEE Mains NEET JEE Advanced

Question

A uniform rod of length 1.5 m is suspended from one end and performs small angular oscillations in a vertical plane. Take g = 10 m/s². If the square of its angular frequency is N s⁻², find N.

G1 Physics · Accepted Answer

For a rod pivoted at one end, I = ML²/3. The centre of mass lies at d = L/2. For a physical pendulum, ω² = Mgd/I. Therefore, ω² = Mg(L/2)/ (ML²/3) = 3g/(2L). Substitute g = 10, L = 1.5. Then, ω² = 30/3 = 10 s⁻². Hence, N = 10.