A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity v ms -1 . If it is to climb the inclined surface, then v should be

Question

A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity v ms -1 . If it is to climb the inclined surface, then v should be <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a0324f5ce2b67b4bc7339aa2aa55ba7e-.png" /> (A) ≥ √(10/7) gh (B) ≥ √2 g h (C) √(10/7) gh (D) (10/7) gh

Tardigrade Admin · Accepted Answer

According to energy conservation,
Potential energy

=

Translational kinetic energy Rotational kinetic energy

\Rightarrow m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}

= \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{5} m R^{2}) \cdot \frac{v ^{2}}{R ^{2}}

or

= \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2}

= \frac{7}{10} m v^{2}

v = \frac{10}{7} g h

Hence, to climb the inclined surface

v

should be

\geq \frac{10}{7} g h

Q. A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity $v m s^{- 1} .$ If it is to climb the inclined surface, then $v$ should be

$\geq \frac{10}{7} g h$

$\geq 2 g h$

$\frac{10}{7} g h$

$\frac{10}{7} g h$

Q. A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity vms−1. If it is to climb the inclined surface, then v should be

≥710​​gh

≥2gh​

710​​gh

710​gh

According to energy conservation, Potential energy = Translational kinetic energy Rotational kinetic energy ⇒mgh=21​mv2+21​Iω2 =21​mv2+21​(52​mR2)⋅R2v2​ or =21​mv2+51​mv2 =107​mv2 v=710​gh​ Hence, to climb the inclined surface v should be ≥710​gh​

Q. A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity $v m s^{- 1} .$ If it is to climb the inclined surface, then $v$ should be

$\geq \frac{10}{7} g h$

$\geq 2 g h$

$\frac{10}{7} g h$

$\frac{10}{7} g h$