Carpet Rolling 101

This problem is easily solved using Conservation of Energy.
$\displaystyle E_{init} = E_{final}$

When the Carpet is in the state where its radius is halved, it can be easily proven that the mass of the remaining rolled part of the cylinder is $\displaystyle \frac{M}{4}$ .

Thus,
Using $\displaystyle PE_i + KE_i + RE_i = PE_f + KE_f + RE_f$ , we get,

$\displaystyle MgR + 0 + 0 = (\frac{M}{4})g(\frac{R}{2}) + \frac{1}{2}(\frac{M}{4})v_{cm}^2 + \frac{1}{2} I_{cm} w^2$

Also, note that,

$\displaystyle I_{cm} = \frac{(\frac{M}{4})(\frac{R}{2})^2}{2}$ , and,

$\displaystyle w = (\frac{v_{cm}}{\frac{R}{2}})^2$

Therefore,

$\displaystyle MgR = \frac{MgR}{8} + \frac{Mv^2_{cm}}{8} + \frac{Mv^2_{cm}}{16}$

$\displaystyle \Rightarrow v_{cm} = \sqrt{\frac{14gR}{3}} = \sqrt{9} = \boxed{3m/s}$

I loved this problem! So exciting!! Ok, let's go :). Oh, wait- first, a few comments to Ruchir Mehta: thanks for such a cool problem. Also, little typo- you ask for the horizontal velocity in m/s^2, so you should probably fix that. Ok, now on to my solution.

This problem just screams conservation of energy, doesn't it :)? All you need to look at are the beginning and ending situations. At the beginning, we have a rolled up carpet with a decent amount of gravitational potential energy. Afterwards, however, our carpet has very little gravitational potential energy in comparison. It must make up for this loss with some (linear and rotational) kinetic energy. So, we just apply conservation of energy and solve :). One thing, though- for me it was troublesome to calculate the potential energy of the cylinder before and after. The answer is very intuitive, but I actually did some integration using densities and differential angles and pesky little things like that. That's not really necessary to do on here (you can try it if you want); we'll just do a hand-wavy proof for the potential energy of the cylinder: the potential energy of a point at the bottom of the cylinder is zero. At the top it's 2MgR. We take the average for the whole cylinder, and get that for a general cylinder of uniform density (cylinder must be small enough so g is pretty much constant), the potential energy is MgR.

For the cylinder at the beginning: $r_1=R$ and $m_1=M$ . For the cylinder at the end: $r_1=R/2$ (given). And $m_2$ ? Well, we can find it like this: the density of the cylinder remains constant, so its mass at the beginning can be written as $pL\pi r_1^2=pL\pi R^2$ where $p$ is density and $L$ is length. At the end, it can be written like $pL\pi r_2^2=pL\pi R^2/4$ . So $m_2$ is a quarter of $m_1$ .

$m_2=M/4$ . Thus, the potential energy at the beginning $U_1$ is equal to $MgR$ while at the end $U_2=MgR/8$ . The beginning potential energy is the total energy of the cylinder. At the end, it has potential, rotational kinetic, and linear kinetic. The final energy is equal to the initial, of course:

$U_1=U_2+K_R+K_L$

$MgR=MgR/8+1/2I\omega^2+1/2m_2V^2$

$7MgR/8=1/2*1/2*m_2*r_2^2\omega^2+1/2m_2V^2$

$7MgR/4=1/2*m_2*r_2^2\omega^2+m_2V^2$

$7MgR/4=1/2*m_2*r_2^2(V/r_2)^2+m_2V^2=1/8*M*V^2+1/4*M*V^2$

$V=\sqrt{\frac{7MgR/4}{3M/8}}$

$V=\sqrt{\frac{14gR}{3}}$

Plugging in the beautiful numbers given by Ruchir, we find that V=3m/s.