A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t = 0 is 4 m/s, the time taken to complete the first revolution will be (1/α)[1 – e⁻²π] s, where α = ______.
Physics · Circular Motion · kinematics of circular motion
JEE Mains 2024 29 Jan Evening Shift

Question

A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t = 0 is 4 m/s, the time taken to complete the first revolution will be (1/α)[1 – e⁻²π] s, where α = ______.

Physics · Circular Motion · kinematics of circular motion

JEE Mains 2024 29 Jan Evening Shift

G1 Physics · Accepted Answer

|a(c)| = |a(t)| v²/r = dv/dt ∫₄ᵛ dv/v² = ∫₀ᵗ dt/r [–1/v]₄ᵛ = t/r –1/v + 1/4 = 2t 1/v = 1/4 – 2t v = 4/(1 – 8t) = ds/dt 4∫₀ᵗ dt/(1 – 8t) = ∫₀ˢ ds (r = 0.5 m, s = 2πr = π) 4 × [ln(1 – 8t)/–8]₀ᵗ = π ln(1 – 8t) = –2π 1 – 8t = e⁻²π t = (1/8)(1 – e⁻²π) s So, α = 8