Relationship between angular and translational velocity on inclined surface I have been researching about rolling motion and I was calculating a way to predict the translational velocity of the object at the bottom of the incline. I know that the kinetic energy of a cylinder undergoing rolling motion is given as
$$E_k = \frac{1}{2} I \omega^2$$
Can angular velocity $\omega$ be replaced as $v/r$ even if the object is a partially filled cylinder?
 A: Linear tangential speed $v_t$ of a particle at radius $r$ from the axis of rotation is
$$v_\text{tan} = r \omega$$
The fact that the cylinder is only partially filled does not affect the above equation, it affects only moment of inertia of the body. Please note that many equations for rotational motion assume that body is rigid! Total kinetic energy of a rolling cylinder must include both translational and rotational kinetic energy
$$\boxed{K = \frac{1}{2} I \omega^2 + \frac{1}{2} m v_t^2}$$
Moment of inertia of a full cylinder is $I = \frac{1}{2} m r^2$ and the above equation is simplified into
$$K_\text{full-cyl} = \frac{3}{4} m v_t^2$$
A: You can say that $\omega = v/r$ — but only for the cylinder, not for the water.  The cylinder will be rolling without slipping, and so you can only the rotational kinetic energy equation for the cylinder.
The water inside the cylinder will be executing a different sort of motion.  The simplest assumption is that the water will not be "sloshing" around and will therefore be at rest relative to the center of the cylinder.  In that case, the water will only have translational kinetic energy, with zero rotational KE.
