Answer:
The value is  [tex]\alpha =14.67 \  rad/s^2[/tex]
Explanation:
From the question we are told that
 The radius is  r =  0.30 m
 The mass of the spool is  m =  25 kg
  The force is  F  =  55 N
 Â
Generally the torque experienced by the spool is Â
  [tex]\tau =  I  *  \alpha[/tex]
Here  [tex]\alpha[/tex] is the angular acceleration
[tex]I[/tex] is the moment of inertia of the spool which is mathematically represented as
   [tex]I  =  \frac{1}{2} *  m * r^2[/tex]
=> Â [tex]I Â = Â \frac{1}{2} * 25 *0.30^2[/tex]
=> Â [tex]I Â = 1.125 \ kg \cdot m^2 Â [/tex]
Generally the torque experienced by the spool is  also mathematically represented as
   [tex]\tau =  F *  r[/tex]
So
   [tex] F *  r   =  1.125 *  \alpha  [/tex]
=>   [tex] 55  * 0.30  =  1.125 *  \alpha  [/tex]
=> Â [tex]\alpha =14.67 \ Â rad/s^2[/tex]
