91Ó°ÊÓ

The magnitude of acceleration of the center of the sphere,${\text{a}}_{\text{center}}=0.10×\text{g}$

The change in velocity per unity time interval is known as acceleration. For small angular displacements, it can also be defined as the product of angular acceleration and the radius of the path on which it is traveling. It is given as-

$\mathbf{a}\mathbf{=}\mathbf{r}\mathbf{×}\mathbf{Î\pm }$

Here, ais the acceleration,$\mathbf{Î\pm }$ is the angular acceleration, and r is the radius of the circular path.

Let a frictional force acts on the sphere at the point of contact of sphere and the inclined plane. As the weight and normal reaction act on the center of mass, the torque produced by them is zero. So, the rotations are only caused by the frictional force. So, referring to the figure, the translational equation of motion is -

$f_s-Mg\sin \mathrm{θ}=Ma····························\left(1\right)$

The rotational equation of motion is given as-

$$\begin{gathered} R{f}_s=I_{com}\mathrm{Î\pm } \\ {f}_s=\frac{I_{com}\mathrm{Î\pm }}{R}······························\left(2\right) \end{gathered}$$

As we know,$\mathrm{Î\pm }=-\frac{\text{a}}{\text{R}}$ , equation (2) becomes-

$f_s=-\frac{I_{com}a}{R^2}$

Negative sign arises due to the fact that the acceleration is along the negative x axis.

From equation (1), we have,

$$\begin{gathered} -\frac{I_{com}a}{R^2}-Mg\sin \mathrm{θ}=Ma \\ -Mg\sin \mathrm{θ}=Ma+\frac{I_{com}a}{R^2} \\ -Mg\sin \mathrm{θ}=a\left(\frac{M{R}^2+I_{com}}{R^2}\right) \\ a=-\frac{Mg\sin \mathrm{θ}}{\left(\frac{M{R}^2+I_{com}}{R^2}\right)} \end{gathered}$$

Further solving,

$$\begin{gathered} a=-\frac{Mg\sin \mathrm{θ}}{\left(\frac{M{R}^2+I_{com}}{R^2}\right)} \\ =-\frac{g\sin \mathrm{θ}}{\left(\frac{R^2+I_{com}}{R^2}\right)} \end{gathered}$$

Now, putting the value of$I_{com}=\frac{2}{5}M{R}^2$for sphere

$a=-\frac{g\sin \mathrm{θ}}{1+\left(\frac{\frac{2}{5}M{R}^2}{M{R}^2}\right)}$

From the given data, we have a= - 0.10g so we can solve it as

$$\begin{gathered} -0.10×g=-\frac{g\sin \mathrm{θ}}{1+\left(\frac{\frac{2}{5}M{R}^2}{M{R}^2}\right)} \\ -0.10×g=-\frac{g\sin \mathrm{θ}}{7/5} \\ \sin \mathrm{θ}=\frac{7×0.1}{5} \\ \mathrm{θ}={\sin }^{-1}\left(0.14\right) \end{gathered}$$

Further solving,

$\mathrm{θ}=8.0^0$

The acceleration would be greater than that of 0.1xg

If the block is frictionless, there would be no frictional force opposing the motion. The component of the gravitational acceleration will cause net acceleration of the object. It is not resisted by the frictional force, so overall acceleration would be the same as that of the component of the gravitational acceleration. This would be greater than 0.1g