Question

A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed $v_0$. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel ?

• A. $\frac{10v_0^2}{7g\sin \theta }$
• B. $\frac{v_0^2}{5g\sin \theta }$
• C. $\frac{2}{5}\frac{v_0^2}{g\sin \theta }$
• D. None of these

Answer 1

Answer: (D)

Angular momentum conservation about A
${\mathrm{mv}}_0$ a $\cos \theta +\frac{2}{5}{\mathrm{ma}}^2\omega$
$=mva+\frac{2}{5}m{a}^2{\omega }^1$
$m{v}_{0}a\left[\frac{2}{5}+\cos \theta \right]=\frac{7}{5}mva$
$v=\frac{5}{7}=v_0\left[\frac{2}{5}+\cos \theta \right]$
$\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=\frac{7}{10}m{v}^2=mgh$