Question

A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill. If it takes 3.60 s for it to make the first complete revolution, how long will it take to make the next complete revolution?

Answer 1

Answer:

The time taken is  $\Delta t = 1.5 \ s$

Explanation:

From the question we are told that

   The mass of the ball is  $m = 1.25 \ kg$

    The time taken to make the first complete revolution is  t= 3.60 s

    The displacement of the first complete revolution is  $\theta = 1 rev = 2 \pi \ radian$

Generally the displacement for one  complete revolution is mathematically represented as

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

Now given that the stone started from rest $w_i = 0 \ rad / s$

     $2 \pi =0 + 0.5* \alpha *(3.60)^2$

     $\alpha = 0.9698 \ s$

Now the displacement for two  complete revolution is

         $\theta_2 = 2 * 2\pi$

         $\theta_2 = 4\pi$

Generally the displacement for two complete revolution is mathematically represented as  

     $4 \pi = 0 + 0.5 * 0.9698 * t^2$

=>   $t^2 = 25.9187$

=>   $t= 5.1 \ s$

So

 The  time taken to complete the next oscillation is mathematically evaluated as

     $\Delta t = t_2 - t$

substituting values

      $\Delta t = 5.1 - 3.60$

     $\Delta t = 1.5 \ s$