Question

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

Answer 1

Hint : To solve this question, we need to find out the angular acceleration of the ant, which will be equal to that of the wheel of the bicycle. Then, applying the rotational equation of motion we can get the final answer.

Formula used: The formula which is used to solve this question is given by,
 $ {\omega _f} = {\omega _i} + \alpha t $ , here $ {\omega _f} $ and $ {\omega _i} $ are the initial and the final angular velocities of a particle moving with a uniform angular acceleration of $ \alpha $ .
 $ \theta = {\omega _i}t + \dfrac{1}{2}\alpha {t^2} $ , here $ \theta $ is the angular displacement covered by a particle in $ t $ amount of time which has an initial angular velocity of $ {\omega _i} $ and a uniform angular acceleration of $ \alpha $ .

Complete step by step answer
As the ant is stuck to the rim of the bicycle, so the angular velocity and the angular acceleration of the ant will be equal to that of the wheel of the bicycle at each instant of time.
Now, since the motion of the wheel is with uniform angular acceleration, so the angular acceleration is given by
 $ {\omega _f} = {\omega _i} + \alpha t $
According to the question, we have $ {\omega _i} = 0 $ , $ {\omega _f} = 2rev/s $ , and $ t = 10s $ . Substituting the above we get
 $ 2 = 0 + 10\alpha $
 $ \Rightarrow \alpha = \dfrac{1}{5}rev/{s^2} $ …………………...(1)
Now, for determining the number of revolutions, we use the following equation
 $ \theta = {\omega _i}t + \dfrac{1}{2}\alpha {t^2} $ ……………………….(2)
Substituting $ {\omega _i} = 0 $ , $ \alpha = \dfrac{1}{5}rev/{s^2} $ , and $ t = 10s $ above we get
 $ \theta = 0 \times 10 + \dfrac{1}{2} \times \dfrac{1}{5} \times {10^2} $
 $ \Rightarrow \theta = 10rev $
Hence, the total number of revolutions completed by the ant in the $ 10 $ seconds is equal to $ 10 $ .
Let $ {t_1} $ be the time taken by the ant to complete the first revolution.
For the first revolution, we have $ \theta = 1rev $ , and $ {\omega _i} = 0 $ . Also from (1) $ \alpha = \dfrac{1}{5}rev/{s^2} $ . Substituting these in (2) we get
 $ 1 = 0t + \dfrac{1}{2} \times \dfrac{1}{5}{t_1}^2 $
 $ \Rightarrow {t_1}^2 = 10 $
Taking square root we get
 $ {t_1} = \sqrt {10} s $
 $ \Rightarrow {t_1} = 3.16s $ …………………...(3)
Let $ {t_2} $ be the time taken by the ant to complete the last revolution.
As a total $ 10 $ revolutions are covered by the ant in $ 10 $ seconds, so time taken to cover the first $ 9 $ revolutions is equal to $ \left( {10 - {t_2}} \right) $ . Substituting $ \theta = 9 $ , $ {\omega _i} = 0 $ , $ \alpha = \dfrac{1}{5}rev/{s^2} $ , and $ t = \left( {10 - {t_2}} \right) $ in (2) we get
 $ 9 = 0 \times \left( {10 - {t_2}} \right) + \dfrac{1}{2} \times \dfrac{1}{5} \times {\left( {10 - {t_2}} \right)^2} $
  $ \Rightarrow {\left( {10 - {t_2}} \right)^2} = 90 $
Taking square root we get
 $ \left( {10 - {t_2}} \right) = \sqrt {90} $
 $ \Rightarrow \left( {10 - {t_2}} \right) = 9.48 $
On rearranging we get
 $ {t_2} = 10 - 9.48 $
 $ \Rightarrow {t_2} = 0.52s $
Hence, the time taken by the ant to complete the first revolution is equal to $ 3.16s $ and that for the last revolution is equal to $ 0.52s $ .

Note
We should not convert the unit of angular velocity given in the question as revolutions per second into radians per second. This is because we have been asked to find out the number of revolutions only. So we can use the unit given in the question as it is.