# Homework Help: Dropping a cylinder

1. Nov 19, 2007

### Ahwleung

1. The problem statement, all variables and given/known data
A cloth tape is wound around the outside of a uniform solid cylnder (mass M, radius R) and fastened to the ceiling. The cylinder is held with the tape vertical and then released from rest. As the cylinder descends, it unwinds from the tape without slipping. The moment of intertia of a uniform solid cylinder about its center is .5MR^2.

a. Draw a FBD
b. In terms of g, find the downward acceleration of the center of the cylinder as it unrolls from the tape.
c. While descending, does the center of the cylinder move toward the left, toward the right, or straight down? Explain.

2. Relevant equations
Well we just finished the chapter on angular momentum (and have thus completed basic angular motion), so I suppose all the angular stuff applies (torque, angular momentum, angular kinetic energy/potential energy, and the U(alpha)M equations)

I'm pretty sure this question should be solved using potential energy (mgh is converted into kinetic + angular kinetic energy, solve for a)

3. The attempt at a solution

a. The only relevant forces would be Tension going up (from the side of the cylinder) and Gravity going down (from the center of the cylinder), am I right?

b. This is the hard one. Am I right in saying that this is a conservation of energy question? All forces are conserved (no F applied, no friction) and so it should just be Gravitational Potential Energy converted to both angular kinetic and translational kinetic energy. And because you can convert angular acc. to translational acc., you should be able to solve for regular acceleration...? I don't think the problem should be that simple. (he asks for it in terms of g, no h involved I guess?)

c. Based on my answer to A, if there are no horizontal forces then the center of the cylinder should just go straight down (and not to the right or left). However, once it reaches the end of the tape it would probably (based on intuition) move toward the tape (or toward your hand)

2. Nov 19, 2007

### Bob Loblaw

3. Nov 19, 2007

### Ahwleung

Sorry, I think I got part B wrong. You can't do it using energy because it gives you velocity squared (and you can't derive that to get acceleration).

Instead, I just summed to torques (which made it a lot easier).

Its just torque due to tension = I*(alpha)

And because the force of tension = force of gravity, you just plug in variables.

so it becomes

r * m * g * sin(90) = (1/2) * m * r^2 * a / r

Which simplifies to linear acceleration equaling 2g.

But is that right? How could acceleration double just by dropping something attatched to a string? did I make a mistake anywhere?

4. Apr 18, 2011

### Roball93

You have to use the parallel axis theorem for part b to find total inertia for the cylinder as it falls.