Question

A solid sphere rolls down an inclined plane of inclination 60°, the acceleration of the centre of mass of the sphere is (Assume pure rolling) ​

Answer 1

Answer:

The acceleration of the center of  mass of the sphere is 6.062 $m/s^{2}$

Explanation:

Let mass of the sphere is M

Net force of the sphere = $Mg\sin \theta = Mg\sin\ 60 = Mg\times \frac{\sqrt{3} }{2}$

Frictional force = f

So,

  $Ma=Mg\sin \theta -f$             (where a is the acceleration of the sphere)

 ⇒$f=Mg\sin \theta -Ma$

We know that in rolling motion,

              $I\alpha =fR$ ( where, $\alpha \rightarrow$ angular acceleration

                                           $I\rightarrow$ moment of inertia

                                           $f\rightarrow$ frictional force

                                           $R\rightarrow$ radius )

        ⇒$\frac{2}{5}MR^2\times \frac{a}{R}=fR$                    ( for pure rolling $\alpha = \frac{a}{R}$ )

            $\frac{2}{5}M\times a=Mg\sin \theta -Ma$

          $\frac{2a}{5}+a=g\sin \theta$

          $\frac{7a}{5}=g\sin \theta$

            $a=\frac{5}{7}g\sin \theta$ $=\frac{5}{7}\times 9\cdot 8\times \sin 60=6.062 m/s^{2}$