Question
A small spherical ball is released from a point at a height h on a rough track shown in figure. Assuming that it does not slip anywhere, find its linear speed when it rolls on the horizontal part of the track.



Therefore potential energy it has gained w.r.t the surface will be converted to angular kinetic energy about the centre & linear kinetic energy. Therefore $\text{mgh}=\Big(\frac{1}{2}\Big)\text{l}\omega^2=\Big(\frac{1}{2}\Big)\text{mv}^2$$\Rightarrow\text{mgh}=\frac{1}{2}\times\frac{2}{5} \text{mR}^2\omega^2+\Big(\frac{1}{2}\Big)\text{mv}^2$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
