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).

Answer 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  

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

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