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.
You need to login to perform this action.
You will be redirected in
3 sec