The speed of a uniform solid sphere after rolling down from rest without slipping along a fixed inclined…

1. \(\sqrt{\frac{10 g h}{7}}\)
2. \(\sqrt{g h}\)
3. \(\sqrt{\frac{6 g h}{5}}\)
4. \(\sqrt{\frac{4 g h}{3}}\)

Solution

If \(v\) is the speed of solid sphere when it reaches at point \(O\), then by the law of conservation of energy,
\(\begin{array}{ll} & m g h=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2 \\ & =\frac{1}{2} \cdot \frac{2}{5} m R^2\left(\frac{v}{R}\right)^2+\frac{1}{2} m v^2 \\ \Rightarrow \quad & m g h=\frac{1}{5} m v^2+\frac{1}{2} m v^2 \\ \Rightarrow \quad & g h=\frac{v^2}{5}+\frac{v^2}{2} \Rightarrow g h=\frac{7 v^2}{10} \\ \Rightarrow \quad & v^2=\frac{10 g h}{7} \Rightarrow v=\sqrt{\frac{10 g h}{7}} \end{array}\)

Asked in: AP EAMCET 2020 (17 Sep Shift 2)

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