$A$ solid sphere rolls down without slipping on a $30^{\circ}$ inclined plane. If $g = 10\,m/s^2$,the acceleration of the rolling sphere is

$A$ ring of radius $3a$ is fixed rigidly on a table. $A$ small ring of mass $m$ and radius $a$ rolls inside it without slipping,as shown in the figure. The small ring is released from position $A$ (the horizontal position). What will be the speed of the center of the small ring when it reaches the lowest point?

The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height $h$,from rest without sliding,is

$A$ ring takes times $t_1$ and $t_2$ to reach the bottom of an inclined plane of length $L$ by sliding and rolling,respectively. What is the ratio of $t_1$ to $t_2$?

$A$ ball is rolling without slipping. The radius of gyration of the ball about an axis passing through its center of mass is $K$. If the radius of the ball is $R$,what fraction of the total energy is rotational kinetic energy?

$A$ ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is $\frac{7}{x}$ where $x$ is . . . . . . .