A sphere of radius ‘a’ and mass ‘m’ rolls along a horizontal plane with constant speed $ν_{0}$ . It encounters an inclined plane at angle $θ$ and climbs upward. Assuming that it rolls without slipping, how far up, the sphere will travel ?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{2}{5} \frac{ν_{0}^{2}}{g \sin θ}$

$\frac{ν_{0}^{2}}{2 g \sin θ}$

$\frac{7 ν_{0}^{2}}{10 g \sin θ}$

$\frac{ν_{0}^{2}}{5 g \sin θ}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} Conservation of Energy, \\ U_{i} + K_{i} = U_{f} + K_{f} \\ \Rightarrow 0 + \frac{1}{2} m v_{0}^{2} + \frac{1}{2} I ω_{0}^{2} = m g h + 0 \\ \Rightarrow \frac{1}{2} m v_{0}^{2} + \frac{1}{2} (\frac{2}{5} m a^{2}) {(\frac{v_{0}}{a})}^{2} = m g h \\ \Rightarrow h = \frac{7}{10} \frac{v_{0}^{2}}{g} \\ Now, length along the plane = L = \frac{h}{\sin θ} = \frac{7 v_{0}^{2}}{10 g \sin θ} \end{array}$