Question

Somehow, an ant is stuck to the rim of a bicycle wheel of diameter 1 m. While the bicycle is on a central stand, the wheel is set into rotation and it attains the frequency of 2 rev/s in 10 seconds, with uniform angular acceleration. Calculate:

1. The number of revolutions completed by the ant in these 10 seconds.
2. Time is taken by it for first complete revolution and the last complete revolution.

Answer 1

Given:

r = 0.5 m, `omega_0 = 0`, ω = 2rps, t = 10 s

(i) Angular acceleration (α) being constant, the average angular speed,

`omega_"av" = (omega_"o" + omega)/2`

`= (0 + 2)/2`

= 1 rps

∴ The angular displacement of the wheel in time t,

θ = ω_av, t = 1 × 10 = 10 revolutions

(ii) α = `(omega - omega_"o")/"t"`

`= (2 - 0)/10`

`= 1/5` rev/s²

θ = `omega_"o" "t" + 1/2 alpha "t"^2`

`θ = 0 + 1/2 alpha "t"^2`       ...(∵ ω_o = 0)

∴ For θ₁ = 1 rev,

`1 = 1/2(1/5)"t"2/1`

∴ `"t"2/1 = 10`

∴ t₁ = `sqrt10`s

∴ t₁ = 3.162s

For θ₂ = 9 rev,

∴ `9 = 1/2(1/5)"t"2/2`

∴ `"t"_2 = sqrt90`

∴ t₂ `= 3sqrt10`

∴ t₂ = 3(3.162)

∴ t₂ = 9.486

The time for the last, i.e., the 10th, revolution is

 t - t₂ = 10 - 9.486 = 0.514s