# Hollow cylinder

1. Apr 11, 2012

### playg

1. The problem statement, all variables and given/known data
Hi everyone! I did an experiment recently and I've come really far but I have some difficulties with the final equation. In this experiment you have to explain hollow cylinders rolling down an incline without using advanced physics like moment of inertia or something like that. Just simple physics.

I got the equation for solid cylinders:

t=sqrt(3s/a) from s=v0t+(at^2/2) where v0t=0 and I know the acceleration is 2/3gsinα

put them together and you get t=sqrt(3s/a)

But, then the promblems occur... I cannot figure out the equation of the hollow cylinder. I know one variable is d/D^2 (inner/outer diameter), by doing a diagram.

2. Relevant equations
t=sqrt(3s/a)+(1+k(d/D)^2) <----- t=sqrt(3s/a)+[(sqrt(3s/a))x(k2(d/D^2))]
they're the same but you need to shorten the second equation so you can get the first one with k(constant). k should be 1/6.

3. The attempt at a solution
Look at 1.

Sorry for my bad english but I really need help with this. Just how to move on. Any tips would be nice. I need to get to this equation: t=sqrt(3s/a)+[(sqrt(3s/a))x(k2(d/D^2))]

Data you might need:
t(s)) | d | D | The time for solid cylinders is constant: 1.7 with
1.95 | 5.4 | 5.8 | the angle 7.11° , that means that :
1.85 | 2.5 | 3.5 | sqrt(3s/a)=1.7=t
1.79 | 3.5 | 5.5|

2. Apr 30, 2012

### playg

Rolling Cylinders on an incline plane

1. The problem statement, all variables and given/known data
I recently did an experiment using a cylinder and letting it roll down an incline plane, frictionless. When I did the experiment it was known to me that the factors involved was gravity, acceleration, inner & outer radius and distance (g, a, r1,r2,s). The goal was to find an equation to explain how long it will take for the cylinder to reach the bottom. for example t=....

The problem is that ther were hollow cylinders too. You cannot use moment of inertia.

2. Relevant equations
s=v0t+(at^2)/2 is one equation to explain the motion of the solid cylinder. You'll get the equation: t=sqrt(3s/gsin¤). But how should I do when I use hollow cylinders? I know that the relationship (r1/r2) should be used.

3. The attempt at a solution
I explained how I got the equation for solid cylinders on 2). And I know the final equation should be: t = sqrt(3s/gsin¤)(1 + k3 (r1/r2)^2) k3 is a constant. But I don't know why the equation is t = sqrt(3s/gsin¤)(1 + k3 (r1/r2)^2). Can someone explain this to me?

3. Apr 30, 2012

### LawrenceC

Re: Rolling Cylinders on an incline plane

"I recently did an experiment using a cylinder and letting it roll down an incline plane, frictionless."

If it is frictionless, then how can it roll?

4. Apr 30, 2012

### tiny-tim

welcome to pf!

hi playg! welcome to pf!

(try using the X2 and X2 buttons just above the Reply box )
yes, the solid cylinder formula is correct

for a rolling solid cylinder, the acceleration is a constant, 2/3gsinα

for a rolling hollow cylinder, the acceleration is also constant, a different multiple of gsinα

so shouldn't your formula for t have a √ round (1 + k3 (r1/r2)2) ?

(and how are we to "explain" this without using moment of inertia? )

5. Apr 30, 2012

### playg

What do you mean "different multiple of gsina ? And yeah you can explain this with mOment of inertia But not use the equations, just using simple physics. I know that the equation should be
t(x) = √((3 + (r1/r2)²)x/(gsin(α))).
If you use moment of inertia equations, but the intention of this experiment is not to use it, just simple physics. I can explain it with moment of inertia but not calculate with it.

Im sorry, but my english isn't the best.

Last edited: Apr 30, 2012
6. May 1, 2012

### tiny-tim

hi playg!

(just got up :zzz:)
i'm still not sure what you mean

my guess is that, in the experiment, you're supposed to measure the accelerations (a) for different values of d, and then plot them on a graph with axes chosen to make it easy to find the constant k

7. May 1, 2012

### playg

I solved it! The constant should be k=c•(r1/r2)^2 and I got the equation :
t = √((2+2k)x/(gsin(α))) without using moment. Thanks anyway :D