Question

A thin walled hollow cylinder is rolling down an incline , without slipping. At any instant the ratio , "rotational K.E.: Translational K. E.: total K. E.:" is

Answer 1

Answer:

Rotational kinetic energy, K_{R}=\frac{1}{2} l \omega^{2}

K_{R}=\frac{1}{2} \times \frac{M R^{2}}{2} \omega^{2}=\frac{1}{4} M v^{2} \quad(\because v=R \omega)

Transnational kinetic energy

K_{T}=\frac{1}{2} M v^{2}

Total kinetic energy =K_{T}+K_{R}

\begin{array}{l}=\frac{1}{2} M v^{2}+\frac{1}{4} M v^{2} \\=\frac{3}{4} M v^{2}\end{array}

therefore, “Rotational K.E. : Transnational K.E. : Total K.E.”=

\frac{1}{4} :\frac{1}{2} :\frac{3}{4} = 1:2:3