Question

A disk of radius 25.0cm turns about an axis through the center. The pull on the string produces a linear acceleration a(t)=At on the ball The disk starts from rest and after 3 seconds, linear a(3)=1.80m/s2. Find A and then write an expression for the angular acceleration α(t).

Answers

  1. Answer:

    The value for  A  is A= 0.6

    The angular acceleration is  \alpha (t) = 2.4 \ t \ m/s^2

    Explanation:

    From the question we are told that  

        The radius of the disk is  r = 25.0 \ cm = 0.25 \ m

         The linear acceleration is  a(t) = At

         At time   t = 3 \ s

         a(3) = 1.80 \ m/s^2

    Generally angular acceleration is  mathematically represented as  

             \alpha(t) = \frac{a(t)}{r}

    Now  at t = 3 seconds  

             a(3) =  A *  3

    =>      1.80 =  A  *  3  

    =.>       A =  0.6

    So  therefore

                 a(t) =  0.6 t  

    Now  substituting this into formula for angular acceleration

            \alpha (t) = \frac{0.6 t }{R}

    substituting for  r  

             \alpha (t) = \frac{0.6 t }{0.25}

             \alpha (t) = 2.4 \ t \ m/s^2

     

         

Leave a Comment