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A solid sphere rolls without slipping, first horizontal and then up to a point $X$ at height $h$ on an inclined plane before rolling down, as shown.
The initial horizontal speed of the sphere is
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The initial horizontal speed of the sphere is
Solution:
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Verified Answer
The correct answer is:
$\sqrt{10 \mathrm{gh} / 7}$
$m g h=\frac{1}{2} m v^{2}+\frac{1}{2} \frac{2}{5} m R^{2} \frac{v^{2}}{R^{2}}$
$m g h=\frac{7}{10} m v^{2}$
$v=\sqrt{\frac{10 g h}{7}}$
$m g h=\frac{7}{10} m v^{2}$
$v=\sqrt{\frac{10 g h}{7}}$
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