# Speed of a solid cylinder after unwinding

1. Dec 14, 2015

### BMZ

1. The problem statement, all variables and given/known data
The problem states the following:

One end of a string is attached to the ceiling while the other end is wrapped around a solid cylinder of mass 0.20 kg and radius 0.030m. The cylinder is released from rest and the string unwinds as the cylinder rotates and accelerates downward. Determine the a) Solid cylinder's speed after it has traveled 1.0m downward and b) its rate of acceleration.

So given variables are
mass = 0.20kg
and a displacement of 1.0m downwards.
2. Relevant equations

I can't seem to find the correct one for this. I have a sheet and my Physics book with all the formulas yet all the ones I've found up to now require me to have the time in order to find the speed. I have been trying for a while now, and can't seem to find anything relevant to this. Only closest thing I could find would be rotational Kinetic Energy which would be

KEr = ½Iω2, but then there would be two unknowns at the end of this which would be ω and the KEr

Θ = ω0t + ½αt2 could be used I think. Maybe there's an error in the question, which is quite possible since my teacher makes mistakes, but unlikely.
3. The attempt at a solution

I haven't been able to attempt it at all since all the formulas that I can remember off the top of my head and the sheets seem to be missing variables. A formula would push me in the right direction, but I've looked everywhere for formulas regarding Angular displacement, and Angular velocity or even linear velocity/displacement but I seem to be missing variables like in:

ω = Θ/t
ω = v⋅r

--
Thank you for any help.

2. Dec 14, 2015

### CWatters

I haven't tried to solve the problem but perhaps it would help to ask yourself..

Where does the KE come from?
The string unwinds so is there is a relationship between rotation and some other parameter?

3. Dec 14, 2015

### BMZ

Well, I think the KE would come from the rotation of the cylinder as it unwinds from the string. But at the beginning it doesn't have any if it's not moving right? Then maybe the left side would be zero since the rotational kinetic energy is zero at the beginning, then that would be a solvable equation but I don't think it's like that.

All the relationships I can think of relating to the rotation is the radius, the velocity and maybe the length of the string. But it says "after it has traveled 1.0m downwards", doesn't specify whether the string ends there but I don't think knowing that the string is 1m long would help much.

I thought about using Potential Energy = mgh, but I don't really know H, unless the 1m is H and it can be used there.

4. Dec 14, 2015

### haruspex

In mgh, h is the change in height. The change in height is indeed 1m.
The formula you quote for rotational KE is correct, but the cylinder is not simply rotating about its central axis. There are two ways to handle this, and I don't know which you may be familiar with. You can treat the cylinder's motion as a rotation about its centre, plus a linear motion, or as a rotation about a different axis, with no linear motion. Your choice.

5. Dec 14, 2015

### CWatters

The KE doesn't "come from the rotation of the cylinder" that's where some of it goes. The cylinder is also descending with some velocity as well so also it has linear KE.

Why not?

That's the right track. There is a relationship between the length of string unwound and the number of revolutions of the cylinder. There is also a relationship between the velocity of the string relative to the cylinder and the angular velocity of the cylinder (just as there is between the angular velocity of a car wheel and the velocity of the car/road)

Cross posted with haruspex.

6. Dec 14, 2015

### BMZ

Well, if Potential Energy fits the problem by using the 1m change in height in mgh, then what would come to mind is

Potential Energy = Linear KE + Angular KE
which would be

mgh = ½mv2 + ½Iω2

Then I = 0.5mr2 since it's a solid cylinder.

equation would then be mgh = ½mv2 + ½(½mr22

You would cancel out all the masses, and since v = r⋅ω, this would end being.

gh = ½v2 + ¼v2

which by combining like terms you end up with

gh = ¾v2

But I'm not entirely sure whether that would be correct for this.

7. Dec 14, 2015

### insightful

Work the numbers, then compare to the velocity of a simple free-falling object.