$A$ thin uniform circular ring is rolling down an inclined plane of inclination $30^o$ without slipping. Its linear acceleration along the inclined plane will be

$A$ solid sphere,a disc,and a solid cylinder start rolling down an inclined plane from rest. All three objects are made of the same material and have the same mass. Then,

Two uniform thin spherical shells are made from different materials. Both shells have a mass of $2 \,kg$ and an outer radius of $20 \,cm$. When they are both rolled down the same inclined plane without slipping,the times they take to cover equal distances differ by $1 \%$. If the thickness of the thinner shell is $0.5 \,cm$,the thickness of the other one is closest to ........... $\,cm$.

$A$ solid sphere of mass $m$ rolls down an inclined plane without slipping,starting from rest at the top of an inclined plane. The linear speed of the sphere at the bottom of the inclined plane is $v$. The kinetic energy of the sphere at the bottom is

$A$ solid cylinder $P$ rolls without slipping from rest down an inclined plane,attaining a speed $v_p$ at the bottom. Another smooth solid cylinder $Q$ of the same mass and dimensions slides without friction from rest down the inclined plane,attaining a speed $v_q$ at the bottom. The ratio of the speeds $\frac{v_q}{v_p}$ is

Prove that the velocity $v$ of translation of a rolling body (like a ring,disc,cylinder,or sphere) at the bottom of an inclined plane of height $h$ is given by $v^{2} = \frac{2gh}{1 + k^{2}/R^{2}}$ using dynamical considerations (i.e.,by considering forces and torques). Note: $k$ is the radius of gyration of the body about its symmetry axis,and $R$ is the radius of the body. The body starts from rest at the top of the plane.