*Q.23.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
i. Number of revolutions completed by the ant
in these 10 seconds.
ii. Time taken by it for first complete revolution
and the last complete revolution.
Answers
Answer:
Explanation:
Given :
Diameter = 1m ∴ Radius (r) = 0.5 m
After 10 seconds, frequency (n) = 2 rev/s
To find :
Number of revolutions completed by ant in 10 seconds
Time taken for first revolution
Time taken for last revolution
Solution :
The ant is stuck to the rim of the wheel. Hence revolutions completed by wheel will be same as that completed by the ant.
n = 2 rev/s
∴ ω = 4π rad/s ....(angular velocity)
Initially, the wheel is at rest. ∴ ω₀ = 0
It is then uniformly accelerated. Hence using laws of motion,
ω = ω₀ + αt where α is the angular acceleration
∴ 4π = 0 + α(10)
∴ α = 0.4 π rad/s²
Number of revolutions completed will be given by :
θ = ω₀t + (1/2)αt²
= 0 + (1/2)(0.4π)(10²)
∴ θ = 20π rad
Hence, number of revolutions will be,
θ ÷ 2π = 20π 2π
= 10 revolutions.
For first revolution, distance covered will be 2π.
∴ 2π = 0 + (1/2)(0.4π)t₁²
∴ t₁² = 10
∴ t₁ = √10 sec.
= 3.16 seconds
It takes 10 seconds to complete 10 revolutions.
Hence,
time taken for 10th revolution = 10 - time taken for 9 revolutions
To calculate time taken for 9 revolutions,
9×2π = 0 + (1/2)(0.4π)t₉²
∴ t₉² = 90
∴ t₉ = √90
= 9.49 seconds
Hence, time taken for 10th revolution = 10 - 9.49 = 0.11 seconds
Answers:
The ant will complete 10 revolutions in 10 seconds.
Time taken for first complete revolution is 3.16 seconds.
Time taken for last complete rotation is 0.11 seconds.