Step 1

We are given the mass, force and radius is $m = 10\;{\rm{kg}}$, $F = 10\;{\rm{N}}$ and $r = 0.4\;{\rm{m}}$ respectively.

We are asked to determine the ring’s initial angular acceleration and the acceleration of its mass center G.

 
Step 2

The following is the free body diagram.

Here, ${F_f}$ is the friction force.

 
Step 3

To find the angular acceleration, we will use the relation,

\[\begin{array}{c} \Sigma {M_C} = {M_0} + M\\ \left[ \begin{array}{l} \left( {F\sin 45^\circ } \right) \times \left( {0.4\cos 30^\circ } \right) - \\ \left( {F\cos 45^\circ } \right) \times \left( {0.4\left( {1 + \sin 30^\circ } \right)} \right) \end{array} \right] = \left[ { - \left( {mar} \right) - I\alpha } \right]\\ \left[ \begin{array}{l} \left( {\left( {10\;{\rm{N}}} \right)\sin 45^\circ } \right) \times \left( {0.4\cos 30^\circ } \right) - \\ \left( {\left( {10\;{\rm{N}}} \right)\cos 45^\circ } \right) \times \left( {0.4\left( {1 + \sin 30^\circ } \right)} \right) \end{array} \right] = \left[ \begin{array}{l} - \left( {10\;{\rm{kg}} \times a \times 0.4\;{\rm{m}}} \right) - \\ \left( {10\;{\rm{kg}} \times {{\left( {0.4\;{\rm{m}}} \right)}^2}} \right)\alpha \end{array} \right]\\ 4a + 1.60\alpha = 1.793 \end{array}\] … (1)

To find the acceleration, we will use the relation,

\[\begin{array}{c} a = \alpha r\\ a = \left( {0.4\;{\rm{m}}} \right)\alpha \end{array}\] … (2)
 
Step 4

On plugging the values in equation (1), we get,

\[\begin{array}{c} 4\left( {0.4\alpha } \right) + 1.60\alpha = 1.793\\ \alpha = 0.56\;{\rm{rad/}}{{\rm{s}}^2} \end{array}\]

To find the acceleration, substitute the values in equation (2).

\[\begin{array}{l} a = \left( {0.4\;{\rm{m}}} \right)\left( {0.56\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}} \right)\\ a = 0.224\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \end{array}\]