Question

A thin walled hollow cylinder is rolling down an incline without sliping .at any instant the ratio Rotatioal K.E.is and translational K.E is

Answer 1

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

R = 21 lω 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)K

R = 21 × 2MR 2 ω 2 = 41 Mv 2 (∵v=Rω)

Transnational kinetic energy

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

T = 21Mv 2

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

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

= 21 Mv 2 + 41 Mv 2= 4 3 Mv2

therefore, “Rotational K.E. : Transnational K.E. : Total K.E.”=\frac{1}{4} :\frac{1}{2} :\frac{3}{4}

41: 21 :43 = 1:2:3