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
Generally the torque experienced by the spool is
[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]
=> [tex]I = \frac{1}{2} * 25 *0.30^2[/tex]
=> [tex]I = 1.125 \ kg \cdot m^2 [/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]
=> [tex]\alpha =14.67 \ rad/s^2[/tex]
