Rolling Motion

[add image: rolling wheel with top point P moving at 2v and contact point Q at rest] For pure rolling, the lowest point Q (point of contact) is instantaneously at rest: $v_Q = v - v = 0$. The highest point P moves with speed $v_P = v + v = 2v$. So point P moves faster than point Q.

Rolling without slipping: $a_1 = \frac{g\sin\theta}{1 + \frac{K^2}{R^2}}$
Slipping without rolling: $a_2 = g\sin\theta$
$\frac{a_1}{a_2} = \frac{1}{1 + \frac{K^2}{R^2}} = \frac{1}{1 + \frac{2}{5}} = \frac{5}{7}$

Conservation of mechanical energy: $\frac{1}{2}mv^2\left(1 + \frac{K^2}{R^2}\right) = mgh$
$\frac{1}{2}mv^2\left(1 + \frac{K^2}{R^2}\right) = mg\left(\frac{3v^2}{4g}\right) \Rightarrow \frac{K^2}{R^2} = \frac{1}{2}$
So the object is a disc.

At maximum compression all the kinetic energy (translational + rotational) is stored in the spring:
$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}kx^2$, with $I = \frac{mR^2}{2}$ and $\omega = \frac{v}{R}$
$\frac{1}{2}\times3\times(4)^2 + \frac{1}{2}\times\frac{3R^2}{2}\times\left(\frac{4}{R}\right)^2 = \frac{1}{2}\times200\times x^2$
$24 + 12 = 100x^2 \Rightarrow x^2 = 0.36$
$x = 0.6$ m

Acceleration of a body rolling without slipping down an incline: $a = \frac{g\sin\theta}{1 + \frac{k^2}{R^2}}$, where k is the radius of gyration.
Time to cover a length l of the incline from rest: $t = \sqrt{\frac{2l}{a}} = \sqrt{\frac{2l\left(1 + \frac{k^2}{R^2}\right)}{g\sin\theta}}$
Solid cylinder: $I = \frac{1}{2}MR^2$, so $\frac{k^2}{R^2} = \frac{1}{2}$
Hollow cylinder: its mass is farther from the axis, so $\frac{k^2}{R^2}$ is larger (it equals 1 for a thin shell).
A smaller $\frac{k^2}{R^2}$ gives a larger acceleration and a shorter time, independent of mass and of the angle of inclination.
Hence the solid cylinder reaches the bottom first.

When a body rolls without slipping, the point of contact is momentarily at rest, so the friction acting is static friction and does no net work. Hence it does not dissipate energy as heat.
For a body rolling down an incline, friction acts up the incline.
It opposes the translational motion, reducing the linear acceleration from $g\sin\theta$ to $\dfrac{g\sin\theta}{1 + I/MR^2}$.
At the same time its torque about the centre ($fR$) produces the angular acceleration, i.e. it increases the rotational motion.
So the friction takes away part of the energy that would have gone into translation and puts it into rotation; the total mechanical energy is conserved.
Hence the frictional force converts translational energy to rotational energy.

In rolling without slipping, friction does no work, so mechanical energy is conserved. The loss in potential energy becomes translational plus rotational kinetic energy:
$Mgh = \dfrac{1}{2}Mv^2 + \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}Mv^2\left(1 + \dfrac{K^2}{R^2}\right)$ (using $I = MK^2$ and $v = R\omega$)
$\Rightarrow v = \sqrt{\dfrac{2gh}{1 + K^2/R^2}}$
For a solid cylinder $I = \dfrac{1}{2}MR^2$, so $\dfrac{K^2}{R^2} = \dfrac{1}{2}$
$v = \sqrt{\dfrac{2gh}{1 + \frac{1}{2}}} = \sqrt{\dfrac{4gh}{3}}$
The speed does not depend on the length L of the incline, only on the height h.

In rolling without slipping, each point has the velocity of the centre of mass $V_{cm}$ plus a rotational velocity $R\omega = V_{cm}$ tangent to the disc.
Top point: both are in the same direction, so $v = V_{cm} + R\omega = 2V_{cm}$.
Point of contact: they are in opposite directions, so $v = V_{cm} - R\omega = 0$ (no slipping).
So the highest point moves with $2V_{cm}$ and the point of contact is momentarily at rest.