Question

12. 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.

​

Answer 1

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