Question

A ship's anchor weighs 5000N. It's cable passes over a roller of negligible mass and is wound around a hollow cylindrical drum of mass 380 kg and radius 1.1 m. The drum is mounted on a frictionless axle. The anchor is released and drops 16m to the water. Use energy considerations to determine the drum's rotation rate when the anchor hits the water. Neglect the mass of the cable.

Answer 1

Hi! Great first step would be to understand the scenario (in my opinion). So two great ways would be to draw a picture or rephrase it. If something else works, do that! You just need to "see" the situation so that you can take some away from it.

Then I think a good next step is to conceptualize everything. Put everything into a context like a physics book would. The anchor is pulled 5000N downward - that's weight. The roller will act like a pulley, and we can ignore it's properties except that it's part of a pulley system (we can ignore stuff because it has "negligible" mass and no other details are given). And then we have the hollow cylindrical drum with one radius measurement given; so we can think of this as a made-up shape with mass - a cylindrical soda can without a top or bottom (but no thickness) and a 380kg mass. The anchor is drops 16m. It hints at energy. The energy that the drum gets is all do to this anchor pulling on the rope (which is really just a means of transferring force, since we neglect its mass and get no details).

Feel free to pause here to make sure you can get the scenario in your head.

So, we want to know something about the barrel as it's rolling. The rotation rate. How many turns per some time. But don't worry yet, we can find a way to work that in. Since the rope pulls and spins the drum, the drum is spun, and gets energy. One way to find the kinetic energy of the spinning drum uses the radius, mass, and rate of rotation. More on that soon.

And how does having some equation with the drum's kinetic energy, radius, mass, and rate of rotation help? Well, we can find all of those except our rate of rotation and solve for the rate of rotation. The energy is the only mystery, but that all comes from the dropping anchor. Can we find that energy? Yeah, there's a way to find the energy that gravity gives our anchor based on it's the force and how far that force moves it.

So, first for the anchor. Linear work is simple:  $W=F d$
So you have your force and distance we associate with the anchor, so you have your work. We'll call that "$W_1$" when we need it.

Next the drum's situation. Thanks to http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html, we have the equation for kinetic energy.
Generally, we have $KE=\frac12I\omega^2$, and we need the "$I$," which deals with rotational inertia. That is pretty much how hard it is to rotate the drum based only on the idea that your getting the mass to move (acceleration). That site refers to our hollow drum as a "hoop," and gives says that we can consider the rotational inertia to be $I=MR^2$. Now that we know the rotational inertia, we can use good old mathematical substitution to get the kinetic energy to look like
$KE=\frac12MR^2\omega^2$
And we can rearrange that to get
$\omega=\sqrt{\frac{2KE}{MR^2}}=\sqrt{\frac{2KE}{M}}\cdot\frac1R$

Since the energy change from the anchor's fall is the energy change of the drum, this $KE$ is the "$W_1$" from before. So
$\omega=\sqrt{\frac{2W_1}{M}}\cdot\frac1R=\sqrt{\frac{2\left(F d\right)}{M}}\cdot\frac1R$

Now everything's set up. It's a matter of checking my work, carefully using a calculator, and making sure the answer makes sense (ie. this should be a lot of energy - much more than 1 Joule). Also, follow up by making sure you can do it again, alone. And feel free to ask or lookup questions you need along the way if there are missing pieces in your understanding.

Good luck! :)