Answer to Question #140162 in Molecular Physics | Thermodynamics for Sridhar

Question

A solid sphere and a solid cylinder each of mass M and radius R are rolling with a linear speed on a flat surface without slipping.Let L(1) be magnitude of the angular momentum of the sphere with respect to a fixed point along the path of the sphere. Likewise L(2) be the magnitude of angular momentum of the cylinder with respect to the same fixed point along its path.The ratio (L(1))/(L(2)) is
Ans 14/15

Answer 1

Solution

Angular momentum of sphere about the point where it touches sorface

$L_1=I\omega$

Moment of inertia about the point where where it touches the surface

$I=\frac{2}{5}MR^2+MR^2=\frac{7}{5}MR^2$

$L_1=\frac{7}{5}MR^2\omega\\and\\\omega=\frac{v}{R}\\so\\L_1=\frac{7}{5}MvR$

Angur momentum of cylinder

$L_2=I\omega$

And moment of inertia of cylinder about the point where it touches the path surface

$I=\frac{1}{2}MR^2+MR^2=\frac{3}{2}MR^2$

$L_2=\frac{3}{2}MR^2\omega$

$And \\\omega=\frac{v}{R}\\so\\L_2=\frac{3}{2}MvR$

So $\frac{L_1}{L_2}$ Can be calculated as

$\frac{L_1}{L_2}=\frac{\frac{7}{5}MvR}{\frac{3}{2}MvR}\\$

So

$\frac{L_1}{L_2}=\frac{14}{15}$