Answer
a) The magnitude of $\alpha$ is $5.94rad/s^2$
b) The magnitude of the frictional force $f$ is $44N$
Work Step by Step
a) Find the angular deceleration of the cylinder
We have $\omega_0=76rad/s$, $\omega=38rad/s$ and $t=6.4s$. Therefore, $$\alpha=\frac{\omega-\omega_0}{t}=-5.94rad/s^2$$
The magnitude of $\alpha$ is $5.94rad/s^2$
b) The torque acting on the cylinder is $$\tau=I\alpha=0.615\times5.94=3.65N.m$$
This torque is created by frictional force $f$, and it has lever arm equal to the cylinder's radius $r=0.083m$. Therefore, $$f=\frac{\tau}{r}=44N$$
