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प्रश्न
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:
- The number of revolutions completed by the ant in these 10 seconds.
- Time is taken by it for first complete revolution and the last complete revolution.
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उत्तर
Given:
r = 0.5 m,
ω0 = 0,
ω = 2 rps,
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 = ωav,
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`
`theta = 0 + 1/2 alpha t^2` ...(∵ ωo = 0)
∴ For θ1 = 1 rev,
`1 = 1/2(1/5) t 2/1`
∴ `t 2/1 = 10`
∴ t1 = `sqrt 10`s
∴ t1 = 3.162 s
For θ2 = 9 rev,
∴ `9 = 1/2(1/5) t 2/2`
∴ `t_2 = sqrt 90`
∴ t2 `= 3 sqrt 10`
∴ t2 = 3(3.162)
∴ t2 = 9.486
The time for the last, i.e., the 10th, revolution is
t − t2 = 10 − 9.486
= 0.514 s
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