Question

The speed of a homogenous solid sphere after rolling down an inclined plane of vertical height $h$ from rest without sliding is

• A. $\sqrt{\left(\frac{g}{5}\right)gh}$
• B. $\sqrt{gh}$
• C. $\sqrt{\left(\frac{4}{3}\right)gh}$
• D. $\sqrt{\left(\frac{10}{7}\right)gh}$

Answer 1

The correct option is A $\sqrt{\left(\frac{g}{5}\right)gh}$

For solid sphere $I=\frac{2}{5}m{R}^2$

Work- energy theorem, $mgh=\frac{1}{2}I{w}^2+\frac{1}{2}m{v}^2$

$mgh=\frac{1}{2}\times \frac{2}{5}m{R}^2{w}^2+\frac{1}{2}m{v}^2$ where $v=Rw$ (due to pure rolling)

$⟹v=\sqrt{\frac{10}{7}gh}$