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A object which is rotating around a horizontal axis is placed on a surface, and starts sliding (with kinetic friction). After some time, it starts rolling without sliding through the table.

The goal is to determine the disk's angular velocity once it starts rolling and the time it takes to do so.

(Notes: When the disk is rolling, we have that $v = w.r$. We can also assume that the kinetic friction has the same direction as the motion, and therefore contributes to the increase in velocity of the mass center over time.)

After-solving I noticed that the expression for $w$ doesn't depend on friction at all. Why is that?

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HINT: The disk slows its rotation because of the force of kinetic friction between it and the table. If you draw out the direction of this force, you'll find that this force tends to increase the disk's linear velocity while decreasing its angular velocity.

As to how to solve the problem, it is important to note that the force of friction with the table always acts at a point in the plane of the table, and is directed parallel to its surface.

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  • $\begingroup$ It accured to me that I was picturing the problem wrong - as if it had stopped rotating as it slided, until it began rolling again. Thank you for the tip! $\endgroup$ – Rye Nov 30 '18 at 23:19
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By using both Newton's Second Law of Motion, knowing that Friction ($F$) is the only force being applied to the body which isn't perpendicular to its movement, we see that:

$d(v)/dt = F/m$.

From that we can determine $v(t)$ via basic integration. However, since there is angular momentum ($L$), we need a second equation:

$T=d(L)/dt = Id(w)/dt$.

Since $T=r$ x $F => d(w)/dt = - rF/I => w(t) = w_0 - t(rF/I)$

Since force is constant under the given conditions, by integrating these expressions and substituting them in the condition for pure rolling, $v=wr$, we get the time it takes for the body to begin rolling. Substituing that time in * gives us the final angular velocity of the body.

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  • $\begingroup$ The last step assumes that the force is constant with respect to time, but it's not actually necessary to make that assumption. $\endgroup$ – Michael Seifert Dec 1 '18 at 15:10
  • $\begingroup$ Taking into account this was an undergraduate mechanics problem, it's a safe assumption to make. $\endgroup$ – Rye Jan 21 at 21:42

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