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 (i) Number of revolutions completed by the ant in these 10 seconds. (i) Time taken by it for first complete revolution and the last complete revolution​

Answer 1

Given :

Diameter of bicycle wheel = 1 m

The frequency attained by the wheel in 10 s = $2 \frac{rev}{s} = 2\times 2 \pi \frac{rad}{s} = 4 \pi \frac{rad}{s}$

To Find :

i) No of rev completed by ant in 10 sec = ?

ii) Time taken for 1st complete revolution and also for last complete revolution = ?

Solution :

We know that 1 rev =2π rad

Initially the bicycle is on central  stand so initially :

$\omega _i = 0$

And also $\omega _f= 4 \pi \frac{rev}{s}$

Now we can find the uniform angular acceleration α by  the equation  :

$\omega_f = \omega_i + \alpha t$

Or,$4 \pi = 0 + \alpha \times 10$

Or,$\alpha = \frac{4\pi}{10}$

Now the angle covered $\theta$ in these 10 sec can be found as :

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

 $= 0\times t + \frac{1}{2} \times \frac{4\pi}{10} \times 10^2$

 =$20 \pi$ rad

So, no of revolutions = $\frac{20\pi}{2 \pi}$  = 10 rev

Now , for the complete  first revolution it would cover 2π rad , so time taken will be given by using equation :

$2\pi = 0 + \frac{1}{2}\times \frac{4\pi}{10} t^2$

Or,$t^2 = 10 \\t = \sqrt{10}$

So, t = 3. 16 sec

Now to find the time it take to complete last complete revolutions , we can find the time taken for 1st 9 rev and subtract it from the total time taken , i.e.  from the time taken to complete 10 rev which is 10 sec .

$t_{(last \hspace 3 rev)} = t_{(10 \hspace 3 rev)} - t _{(9 \hspace 3 rev)$   -  (1)

So, the time taken  for 1st 9 rev is :

Or,$t^2 = 90\\t =\sqrt{90}$

Or, t = 9.48 sec

So from eq 1 we get the time for last rev as :

$t_{(last \hspace 3 rev)} = 10 - 9.48$ sec

               =0.52 sec

So, the total no  of revolutions in 10 sec is 10 , time taken to complete 1st rev is 3.16 sec and  time to complete last rev is 0.52 sec .

             

Answer 2

Answer:

i)10 revolutions in 10 sec

ii )time for last revolution == 0.5131 sec

Explanation:

Given data :-  

Diameter = 1 m  

Radius (r) = 0.5 m

After 10 seconds, frequency (n) = 2 rev/s

What we have to find out :-  

i) Number of revolutions completed by an ant in  t =10 seconds

ii)  a) Time taken for first revolution

    b) Time taken for last revolution

The ant is stuck to the rim of the wheel. Hence revolutions completed by wheel is same as that completed by the ant, an ant is just like any arbitrary or reference point on rotating point other than there is no role of ant in this numerical.

since we have,

n = 2 rev/s  hence final angular velocity is given by,

    ω =2πn  

        =2 π (2 )  

        = 4π rad /s      

similarly, Initial angular velocity Initially, ω₀ = 0 rad /s because initially the wheel is at rest.  

Since wheel is uniformly accelerated. Hence an angular acceleration is produced which is given by,

             α = (ω - ω₀ ) / t  

    = ( 4π – 0 ) / 10

            α  =  0.4 π rad/s²

      Now by using formula, the total angular displacement is given by

θ = ω₀t + (1/2)αt²          

   =  0  + (1/2)(0.4π)(10²)    

   = 20 π rad

Hence, number of revolutions = (angular displacement / 2π )  

                          = θ / 2π  

                          = 20π  / 2π  

                          = 10 revolutions in 10 sec.

ii) since the wheel is accelerating uniformly means the angular velocity is increasing with respect to time hence initial speed is minimum (requires more time to complete its first revolution i.e t1 is maximum )and final i.e after 10 sec the speed is maximum (requires less time to complete its last revolution i.e t2 is minimum)

a) For the first revolution, distance covered will be 2π because one revolution takes an angle of 3600 or angular displacement of 2π rad.  

      θ1 = ω₀t + (1/2)αt1²    

∴ 2π = 0 + (1/2)(0.4π) t₁²      

∴ t₁² = 10      

∴ t₁  = √10 sec.            

              = 3.16 seconds

Now for two revolution (which includes 1st and 2nd ), θ2 = 4 π  

θ2 = ω₀t + (1/2)αt2²  

t2  = √20 sec

Hence for upto nine revolution (which includes 1st to 9th  ), θ9 = 18 π  

θ9 = ω₀t + (1/2)αt9²  

t9  = √90 sec  

 b)Hence time for last revolution = Time for all 10 revolution - Time for 9 rev.

                      = 10  - √90

                = 10 – 9.4868

                = 0.5131 sec