Question

A solid sphere of radius ${}^{\prime }{a}^{\prime }$ and mass ${}^{\prime }{m}^{\prime }$ rolls along the horizontal plane with constant speed $v_o$. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping, how far up the iclined plane, sphere will travel (along the incline)?

• A. $\frac{2}{5}\frac{v_o^2}{g\sin \theta }$
• B. $\frac{10}{7}\frac{v_o^2}{g\sin \theta }$
• C. $\frac{1}{5}\frac{v_o^2}{g\sin \theta }$
• D. $\frac{7}{10}\frac{v_o^2}{g\sin \theta }$

Answer 1

The correct option is D $\frac{7}{10}\frac{v_o^2}{g\sin \theta }$


From energy conservation,
$mgh=\frac{1}{2}m{v}_o^2+\frac{1}{2}I{\omega }^2$
$mgh=\frac{1}{2}m{v}_o^2+\frac{1}{2}\times \frac{2}{5}m{a}^2\times \frac{v_o^2}{a^2}$
$gh=\frac{1}{2}{v}_o^2+\frac{1}{5}{v}_o^2$
$gh=\frac{7}{10}{v}_o^2$
$h=\frac{7}{10}\frac{v_o^2}{g}$
From triangle, $\sin \theta =\frac{h}{l}$
Then, $h=l\sin \theta =\frac{7}{10}\frac{v_o^2}{g}$
$l=\frac{7}{10}\frac{v_o^2}{g\sin \theta }$