Question

A small disc is set rolling with a speed v on the horizontal part of the track of the previous problem from right to left. To what height will it climb up the curved part ?

Answer 1

A disc is set rolling with a velocity v from right to left. It has attained a height, h,

therefore, $\frac{1}{2}m{V}^2+\frac{1}{2}I{\omega }^2=mgh$

$\Rightarrow \frac{1}{2}m{V}^2+\frac{1}{2}\times \frac{1}{2}m{R}^2{\omega }^2=mgh$

$\Rightarrow \frac{1}{2}{v}^2+\frac{1}{4}{V}^2=gh$

$\Rightarrow \left(\frac{3}{4}\right)V^2=gh$

$\Rightarrow \left(\frac{3}{4}\right)V^2=gh$

$\Rightarrow h=\frac{3}{4}\left(\frac{v^2}{g}\right)$