Answer to Question #140152 in Classical Mechanics 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

The angular momentum of the solid sphere is given by,

$L(1)=MVR+I\omega\\$

where V and I represent the linear velocity and moment of inertia about an axis through the centre of mass respectively,

$\begin{aligned} L(1)&=MVR + (\frac{2}{5}MR²)(\frac{V}{R})\\&=MVR+\frac{2}{5}MVR\\ &=\frac{7}{5}MVR\\ \\ \end{aligned}$

The angular momentum of the solid cylinder is also given by,

$L(2)=MVR+I\omega$

where V and I represent the linear velocity and moment of inertia about an axis through the centre of mass respectively,

$L(2)=MVR+(\frac{1}2MR²)(\frac{V}R)\\ L(2)=MVR+\frac{1}2MVR\\ L(2)=\frac{3}2MVR.\\\hspace{2cm}\\ \textsf{the ratio L(1)/L(2) is therefore,}\\ \begin {aligned}L(1)/L(2)&= \frac{\frac{7}5MVR}{\frac{3}2MVR}\\ L(1)/L(2)&=\frac{14}{15}\end{aligned}$