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

Answer 2

Answer:1:1:2

Explanation: