Question

A rectangular rigid fixed block has a long horizontal edge. A solid homogeneous cylinder of radius $R$ is placed horizontally at rest with its length parallel to the edge such that the axis of the cylinder and the edge of the block are in the same vertical plane as shown in figure. There is sufficient friction present at the edge, so that a very small displacement causes the cylinder to roll of the edge without slipping. Determine:
(i) The angle ${\theta }_C$ through which the cylinder rotates before it leaves contact with the edge.
(ii) The speed of the centre of mass of the cylinder before leaving contact with the edge and
(iii) The ratio of the translational to rotational kinetic energies of the cylinder when its centre of mass is in horizontal line with the edge.

Answer 1

for cylinder

$mgcos\theta =N+\frac{m{v}^2}{r}atleavingN=0mgcos\theta =\frac{m{v}^2}{r}\to eq1Energyequation{I}_{cylinder}aboutP=\frac{M{R}^2}{2}+M{R}^2=\frac{3}{2}{MR}^{2}mgR(1-cos\theta )=\frac{1}{2}\left(\frac{3}{2}{MR}^2\right)W^{2}mgR(1-cos\theta )=\frac{1}{2}\left(\frac{3}{2}{MR}^2\right)\times \frac{V^2}{R^2}mg(1-cos\theta )=\frac{3}{4}m\frac{V^2}{R}\to eq2Fromeq1mg(1-cos\theta )=\frac{3}{4}mgcos\theta 1=\frac{7}{4}cos\theta \theta ={cos}^{-1}\left(\frac{4}{7}\right)$