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Questions 3-22 · Q12

Q.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 the first complete revolution and the last complete revolution.
Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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The wheel starts from rest (ω0=0\omega_0=0) and reaches n=2n=2 rev/s (i.e. ω=2πn=4π\omega=2\pi n=4\pi rad/s) after t=10t=10 s, with uniform angular acceleration α=ω−ω0t=4π10=0.4π\alpha=\dfrac{\omega-\omega_0}{t}=\dfrac{4\pi}{10}=0.4\pi rad/s^2. Total angle turned in these 10 s: θ=ω0t+12αt2=0+12(0.4π)(10)2=20π rad=20π2π=10 revolutions\theta=\omega_0t+\frac{1}{2}\alpha t^2=0+\frac{1}{2}(0.4\pi)(10)^2=20\pi\text{ rad}=\frac{20\pi}{2\pi}=10\text{ revolutions} so the ant, stuck to the rim, completes exactly 10 revolutions in these 10 seconds.

Time for the FIRST complete revolution (θ1=2π\theta_1=2\pi rad): using θ1=12αt12\theta_1=\frac{1}{2}\alpha t_1^2, t1=2θ1α=2(2π)0.4π=10≈3.16 st_1=\sqrt{\frac{2\theta_1}{\alpha}}=\sqrt{\frac{2(2\pi)}{0.4\pi}}=\sqrt{10}\approx3.16\text{ s} …

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