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

• A. $\geq \sqrt{\frac{10}{7}} gh$
• B. $\geq \sqrt{2 g h}$
• C. $\sqrt{\frac{10}{7}} gh$
• D. $\frac{10}{7} gh$

Answer 1

Answer: (A)

According to energy conservation,
Potential energy $=$ Translational kinetic energy Rotational kinetic energy
$\Rightarrow mgh =\frac{1}{2} mv ^{2}+\frac{1}{2} I \omega^{2} $
$=\frac{1}{2} mv ^{2}+\frac{1}{2}\left(\frac{2}{5} mR ^{2}\right) \cdot \frac{ v ^{2}}{ R ^{2}}$
or $=\frac{1}{2} mv ^{2}+\frac{1}{5} mv ^{2} $
$=\frac{7}{10} mv ^{2}$
$ v =\sqrt{\frac{10}{7} gh }$
Hence, to climb the inclined surface $v$ should be
$\geq \sqrt{\frac{10}{7} gh }$