A uniform solid sphere rolls down in an incline (a) what must be the incline angle if the linear acceleration of the center of the sphere is tohave a magnitude of the $\mathbf{0}.\mathbf{10}\mathbf{g}$? (b) If the frictionless block were to slide down the incline at that angle, would its acceleration magnitude be more than, less than, or equal to $\mathbf{0}.\mathbf{10}\mathbf{g}$? why?

The linear acceleration of the center of the sphere is $0.10g$

The force that may cause an item to revolve along an axis is measured as torque. Here, we have to calculate torque by using position vector and the force.

(a) We substitute

$\begin{array}{c}I=\frac{2}{5}M{R}^2\\ -0.10g=-\frac{gsin\theta }{1+\left[\frac{2}{5}M{R}^2\right]/M{R}^2}\\ =-\frac{gsin\theta }{\left[\frac{7}{5}\right]}\end{array}$

$\begin{array}{c}\theta =si{n}^{-1}\left(0.14\right)\\ =8.0°\end{array}$

If the magnitude of the associated torque is zero Nm.thenangle between the directions of $\overrightarrow{r}$ and is$8.0°$

(b)

It would accelerate more quickly. This can be seen in terms of either forces or energy. Since there would be no static friction on the uphill side, the only force responsible for the downhill acceleration would be gravity.In terms of the energy the potential energy is$\frac{1}{2}m{v}^2$ (without it being ‘shared’ with another terms) resulting in a greater speed.