A uniform vertical rod of length 1 m is held at rest in vertical plane. At t = 0 it is released and simultaneously a sharp horizontal impulse is imparted to it at point B so that the displacement of centre of mass at t = 1 s is 5√2 m. After hit, angular velocity of rod is 10n rad/s. Find n. (g = 10 m s⁻²)
Physics · Mixed Chapter
JEE Mains

Question

A uniform vertical rod of length 1 m is held at rest in vertical plane. At t = 0 it is released and simultaneously a sharp horizontal impulse is imparted to it at point B so that the displacement of centre of mass at t = 1 s is 5√2 m. After hit, angular velocity of rod is 10n rad/s. Find n. (g = 10 m s⁻²)

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Gravity moves the centre of mass vertically by ½gt²=5 m in 1 s. The total displacement is 5√2 m, so its horizontal displacement is √[(5√2)²−5²]=5 m. Step 2: Thus the centre-of-mass horizontal speed immediately after impulse is 5 m s⁻¹, and the impulse magnitude is J=5m. Step 3: Angular impulse about the centre is J(L/2). With I_cm=mL²/12, ω=J(L/2)/I_cm=6J/(mL)=30 rad s⁻¹ for L=1 m. Step 4: ω=10n gives n=3.