Answer:
[tex]-0.25 rad/s^2[/tex]
Explanation:
The equivalent of Newton's second law for rotational motions is:
[tex]\tau = I \alpha[/tex]
where
[tex]\tau[/tex] is the net torque applied to the object
I is the moment of inertia
[tex]\alpha[/tex] is the angular acceleration
In this problem we have:
[tex]\tau = -12.5 Nm[/tex] (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)
[tex]I=50.0 kg m^2[/tex] is the moment of inertia
Solving for [tex]\alpha[/tex], we find the angular acceleration:
[tex]\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2[/tex]
