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 it's 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^(-2 π)] s, where α= _________.

Solution

Physics sir · 4 steps

Equal components give dv/ds = v/R, so speed grows exponentially with distance along the circle.

1. Set up the condition. Normal acceleration is v²/R and tangential acceleration is dv/dt. Setting them equal, and writing dv/dt = vdv/ds, gives dv/ds = v/R.
2. Speed versus distance. Integrate from v₀ at s = 0: ln v/v₀ = s/R, so v = v₀ e^(s/R).

   t = ∫₀^(2π R) ds/(v₀ e^(s/R)) = R/v₀[1 - e^(-2π)]

3. Put in numbers. One revolution is s = 2π R. With R = 0.5 m and v₀ = 4 m s⁻¹, R/v₀ = 0.5/4 = 1/8 s.
4. So. Comparing with 1/α[1 - e^(-2π)] gives α = v₀/R = 8.

The common slip: 0.08. Leaving the radius as 50 cm instead of 0.5 m gives R/v₀ = 12.5, so α = 0.08, not 8.

The idea to keep. When a condition ties acceleration to position, switch dv/dt to vdv/ds. Then get time from dt = ds/v.

Answer from the NTA final answer key. Solution by PrepWiser.