Step 1

We are given a $m = 20{\rm{ kg}}$ sphere rolling down an inclined plane without slipping.

We are asked to determine the angular acceleration of the sphere and the acceleration of its mass center.

 
Step 2

The free-body diagram of the sphere is given below:

Here, $r$ is the radius of the sphere, $\theta $ is the angle of inclined plane and $P$ is the contact point between the sphere and the inclined plane.

 
Step 3

The moment of inertia of the sphere about its center of mass is given by:

\[{I_G} = \frac{2}{5}m{r^2}\]

On substituting the values in the above expression, we get,

\[\begin{array}{c} {I_G} = \frac{2}{5} \times \left( {20{\rm{ kg}}} \right) \times {\left( {0.15{\rm{ m}}} \right)^2}\\ {I_G} = 0.18{\rm{ kg}} \cdot {{\rm{m}}^2} \end{array}\]
 
Step 4

At equilibrium condition, taking moments about the contact point between the sphere and the inclined plane, we have,

\[mg\sin \theta \times r = {I_G} \times \alpha + m{a_G} \times r\]

Here, ${a_G}$ is the acceleration of the center of mass of the sphere and its value is given by,

\[\begin{array}{c} {a_G} = \alpha r\\ {a_G} = \alpha \times \left( {0.15{\rm{ m}}} \right) \end{array}\]

On substituting the values in the above expression, we get,

\[\begin{array}{c} \left( {20{\rm{ kg}}} \right) \times \left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\sin 30^\circ \times \left( {0.15{\rm{ m}}} \right) = \left( {0.18{\rm{ kg}} \cdot {{\rm{m}}^2}} \right) \times \alpha + \left( {20{\rm{ kg}}} \right) \times \alpha \times \left( {0.15{\rm{ m}}} \right) \times \left( {0.15{\rm{ m}}} \right)\\ \left( {0.18{\rm{ kg}} \cdot {{\rm{m}}^2}} \right) \times \alpha + \left( {0.45{\rm{ kg}} \cdot {{\rm{m}}^2}} \right) \times \alpha = \left( {14.715{\rm{ N}} \cdot {\rm{m}}} \right)\\ \left( {0.63{\rm{ kg}} \cdot {{\rm{m}}^2}} \right) \times \alpha = \left( {14.715{\rm{ N}} \cdot {\rm{m}}} \right)\\ \alpha = 23.36{\rm{ rad/}}{{\rm{s}}^2} \end{array}\]
 
Step 5

Then, the acceleration of the center of mass of the sphere and its value can be calculated as,

\[\begin{array}{c} {a_G} = \alpha r\\ {a_G} = \left( {23.36{\rm{ rad/}}{{\rm{s}}^2}} \right) \times \left( {0.15{\rm{ m}}} \right)\\ {a_G} = 3.504{\rm{ m/}}{{\rm{s}}^2} \end{array}\]