# Torque Problem, acceration of mass. Help!

1. Nov 6, 2010

### mmmboh

[PLAIN]http://img825.imageshack.us/img825/4091/cylindera.jpg [Broken]

I have solved problems like this before, but they have all been, well atwoods, and the disk in that case can't move translationally.

So at the top of the cylinder the tension in the string is T, and so that exerts a torque of T*r on the cylinder. Now I call the tension at the string where the mass m is t, I know t=ma. Now I believe this doesn't exert a torque on the cylinder because it is an internal force, or something, or because the mass m is what allows the tension to be T at the other side of the rope (if there was no mass m, then there wouldn't be a T at the other side). so T*r=I$$\alpha$$...I know I=0.5mr2, and that there is no slipping, so I think that since the cylinder also moves translationally, the equation is something like
(astring-acylinder)=r*$$\alpha$$...however this doesn't seem quite right to me because the string doesn't really move uniformly if the cylinder is moving, like the string near the mass m won't even move for a while if the cylinder moves a lot, but the string on the other side will...

Anyway because of this I'm not exactly sure what to do. Can anyone help please?

Last edited by a moderator: May 5, 2017
2. Nov 6, 2010

### tiny-tim

hi mmmboh!

(have an alpha: α and a tau: √ )
no, t is external both to the cylinder and to the little mass m …

(it is only internal if you regard both masses as a single body)

… so tr is part of the torque

and you need two equations for T and t …

a standard linear F = ma for the linear motion of the c.o.m. of the cylinder,

and a τ = Iα for the rotation of the cylinder about its c.o.m.

3. Nov 6, 2010

### mmmboh

Well I get (T+t)=ma (I have a problem with this equation though, because if there were no mass m attached then the cylinder wouldn't rotate or move, but this equation predicts it would I think) and (T-t)r=Iα=I*a/r.
Solving I get t=T/2 and for acceleration I get a=T/m, which is the acceleration of the cylinder, but not the mass. So then since t=T/2=mamass...so amass=T/2m... I don't know, that means that if the mass were 0 it would have an infinite acceleration, doesn't make sense..but if the mass were zero wouldn't the tension be zero...I must be missing something.

Last edited: Nov 6, 2010
4. Nov 6, 2010

### tiny-tim

it would rotate, because the rope doesn't slip
I'm not convinced … what I are you using?

5. Nov 6, 2010

### mmmboh

I=mr2/2

I said T-t=0.5ma=(T+t)/2, and solved for t like that.

6. Nov 6, 2010

anyone?

7. Nov 7, 2010

### tiny-tim

(just got up :zzz: …)
I get t = T/3, not T/2

8. Nov 7, 2010

### mmmboh

Ya, you're right, my mistake. So then a=T/3m...hmm...one thing though, since the rope is massless, the net tension everywhere should be zero so that it doesn't undergo infinite accleration...but if the tension on one side of the rope is different than on the other side, how is this resolved?

9. Nov 7, 2010

### tiny-tim

the cylinder is rough, and the friction prevents the rope from slipping

the friction supplies the difference in tension, to make the net force zero

10. Nov 7, 2010

### mmmboh

Ah thanks!..Sorry one more thing, why in this case does α=acylinder/r...shouldn't α=(astring-acylinder)/r to make it not slip? because the relative acceleration of the string to the cylinder is what matters...For instance if the acceleration of the cylinder was equal to the acceleration of the string, then the cylinder wouldn't need to rotate at all and the string still wouldn't slip.

11. Nov 7, 2010

### tiny-tim

no, if the string is not slipping, then the speed of the string equals the speed of the rim of the cylinder (and so also the acceleration of the string equals the acceleration of the rim of the cylinder)

12. Nov 7, 2010

### mmmboh

Right but, in my equation T-t=0.5ma, that acceleration is the radial acceleration of the rim of the cylinder, so the acceleration of the string, but in my F=ma equation, I had T+t=ma, where this a is the translational acceleration of the cylinder....but to get t=T/3 I equated these equations, but that must mean that the radial acceleration of the rim is equal to the translational acceleration, I don't understand why this must be in this case. Because if the translational acceleration is equal to the acceleration of the string, then the rim will still be moving at the speed of the string, but it will have no radial acceleration.

13. Nov 7, 2010

### tiny-tim

oh sorry, I'd completely forgotten that the cylinder was moving (I was thinking of a similar question in another thread)

yes, you're right, it is α=(amass-acylinder)/r

14. Nov 7, 2010

### mmmboh

Doesn't α=(astring-acylinder)/r? (by astring I mean the part that is being pulled) not (amass-acylinder?)/a I mean if anything it would be (amass+acylinder)/r since they are moving in opposite directions, but then again the cylinder can move without the mass moving can't it? This is why I was confused on how to solve it.

Last edited: Nov 7, 2010
15. Nov 8, 2010

### tiny-tim

hi mmmboh!

(just got up :zzz: …)
it's best to use amass, because the mass comes into another equation, while the pulling end of the string doesn't.

let's see … is it + or - ? …

let's concentrate on B, the bit of the cylinder nearest us at the start (and we'll call the mass A and the cylinder C) …

then when the cylinder has rotated an angle θ, B has moved θ clockwise, and the string has moved with it, so the string is now rθ nearer the mass A,

so Ax - Cx = constant minus rθ,

so amass - acylinder = -rα

16. Nov 8, 2010

### mmmboh

Hm ok so now I have 3 unknowns (t,acylinder,amass), but I only have two equations, I solved T=0.25m(amass+acylinder), but I need another equation to get rid of acylinder

One other thing, I don't quite get why Ax - Cx = k-rθ (k would be the original distance between them is what I assume you mean the constant to be), because the cylinder moves also...like if the cylinder just rotated I would get the equation, because the change in position of the mass and cylinder relative to each other would just be how much string gets wrapped around the cylinder, but in this case the cylinder is moving, so don't you have to take into account how much it moves translationally?

17. Nov 9, 2010

### tiny-tim

hi mmmboh!
no, you should have three equations …

the linear equation, the torque equation, and the constraint equation
that's what Cx is …

x-component of the (Centre of the) Cylinder!

18. Nov 9, 2010

### mmmboh

I'm sorry, I feel stupid but I don't really understand. I know that Cx is the x-component of the center of mass, my issue is why Ax-Cx=k-rθ...Because lets say the center of mass of the cylinder doesn't move, then k-rθ is how much closer the mass is to the cylinder, in other words then Ax-Cx=k-rθ..but lets say the cylinder rotates that same amount, but this time Cx also moves, then the equation would predict the same result, but obviously it's not the same because now the cylinder is closer to the mass...

Sorry, but what is my constraint equation?

I have never had this much trouble understanding a problem

Thanks for your help by the way.

19. Nov 9, 2010

### tiny-tim

no, because the mass has also moved, in this case the same amount as the cylinder (if the rotation is the same)
Ax - Cx = k - rθ …

a constraint equation is a geometrical equation rather than a force equation … like a body being constrained to move along a particular surface, or in this case the free length of string being constrained to be connected to the position and rotation of the bodies

20. Nov 9, 2010

### mmmboh

Ok but, I used that constraint equation already to find α to plug it into my torque equation...

The three equations are T+t=ma
(T-t)r=Iα=0.5mr2α
Ax - Cx = k - rθ, so differentiating this twice I can solve for α.

So I get T+t=macylinder, and T-t=0.5m(acylinder-amass), and then I'm stuck.