question_answer

A ring, a disc and a sphere all of the same radius and same mass roll down an inclined plane from the same height h. Which of the three reaches the bottom (i) earliest (ii) latest?

Answer:

                We have already deduced in Art. 5(b).20 that acceleration of an object down an inclined plane is given by \[\text{a }=\frac{g\sin \theta }{1+I/m\,{{r}^{2}}}\] For a ring, \[\text{I }=\text{m }{{\text{r}}^{\text{2}}}\] \[{{\text{a}}_{\text{ring}}}=\frac{g\sin \theta }{1+1}=\text{ }0.\text{5 g sin}\theta \] For a disc, \[\text{I }=\frac{1}{2}\text{m }{{\text{r}}^{\text{2}}}\] \[{{\text{a}}_{\text{disc}}}=\frac{g\sin \theta }{1+\frac{1}{2}}=\frac{2}{3}\text{g sin}\theta =\text{ }0.\text{67 g sin}\theta \] For a sphere, \[\text{I }=\frac{2}{5}\text{m}{{\text{r}}^{\text{2}}}\] \[{{\text{a}}_{\text{sphere}}}=\frac{g\sin \theta }{1+\frac{2}{5}}=\frac{5}{7}\text{g sin}\theta =\text{ }0.\text{71 g sin}\theta \] As \[{{\text{a}}_{\text{sphere}}}\]is maximum it will reach the bottom at the earliest. Again, as \[{{\text{a}}_{\text{ring}}}\] is minimum, it will reach the bottom at the end.