What is the downward acceleration of a cylinder as it unwinds from a tape?

[Ahwleung]

Homework Statement

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.

Homework 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)

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)

 

[Bob Loblaw]

This is not too disimilar from a problem I did a little while ago: https://www.physicsforums.com/showthread.php?t=195028

 

[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?

 

[Roball93]

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

 

[rwisz]

because of the tension of the tape.

I would like to clarify a few points about this problem. First, we need to define the direction of positive and negative acceleration. In this case, we will define the downward direction as positive and the upward direction as negative. This is because the cylinder is accelerating downwards due to the force of gravity.

a. The free body diagram for this problem would include the tension force acting upwards and the weight (mg) acting downwards. The tension force is provided by the tape and is equal to the weight of the cylinder.

b. To find the downward acceleration of the cylinder, we can use the equation for gravitational potential energy (mgh) and equate it to the sum of kinetic energy (translational and angular). This gives us:

mgh = 1/2mv^2 + 1/2Iω^2

Where I is the moment of inertia and ω is the angular velocity. Since the cylinder is unrolling without slipping, the angular velocity is related to the linear velocity (v) by ω = v/R, where R is the radius of the cylinder.

Substituting this into the equation and solving for a, we get:

a = g/(1+I/mR^2)

This gives us the downward acceleration of the center of the cylinder as it unrolls from the tape.

c. As mentioned earlier, the cylinder is accelerating downwards due to the force of gravity. As it unrolls from the tape, the center of the cylinder will continue to move straight down, since there are no horizontal forces acting on it. However, once it reaches the end of the tape, there will be a change in direction as the tension force from the tape will cause the cylinder to move towards the tape. This is because the tension force acts in the direction of the tape, causing a change in the direction of motion of the center of the cylinder.