Angular momentum question

radagast_

1. The problem statement, all variables and given/known data


a cylinder with a given radius [tex]R[/tex] rolls to the right without slipping ([tex]\omega= \frac{v}{R}[/tex]). It hits a nail fixed to the pillar. find [tex]h[/tex], the height of the nail above the floor, in which the cylinder will roll back without slipping.
[tex]I=\frac{MR^2}{2}[/tex] for the cylinder.
- The nail applies lateral force only.
- the collision with the nail is ellastic, there's no loss of kinetic E whatsoever.
- express [tex]h[/tex] with [tex]R[/tex] only.

2. Relevant equations
conservation of linear momentum, conservation of angular momentum.
[tex]J=r\times P[/tex]
3. The attempt at a solution
OK.
there is no loss of kinetic energy. so if v is the speed before the collision and u is the speed of the ball after, [tex]v=-u[/tex].
Thus, there is impact as follows: [tex]2mv=\Delta P[/tex]
Also, I want the cylinder to roll backwards still with the term of not-slipping - [tex]\omega=\frac{v}{R}[/tex] so it's the same with angular momentum: [tex]2F\mu*R=\Delta J[/tex]
Now, I was thinking about solving this with [tex]J=(h-R)\times P[/tex], but then I don't know what to do with [tex]F\mu[/tex].

I'd appreciate help in this :)

Thanx.

Oerg

What happens when a rolling object slips?

It slips when the amount of tangenial force is greater than the maximum friction.

Also, how high h is affects the amount of tangenial force generated by the impulse during the collision.

radagast_

Do you mean I need to break the F generated from the nail to perpendicular and tangential, and relate to the tangential one? and then I need to do (Ft is the tangential F) [tex]F_t*R=2F\mu*R[/tex]?
About your second line - do you mean the linear impulse or angular?

tiny-tim

Hi !

You are specifically told: "The nail applies lateral force only."

So you can assume there is only a horizontal force (very unrealistic, I would think )

And forget friction.

The question is asking for the height needed for the cylinder not to want to slip!

As you say, conservation of energy means that the speed out will be the same as the speed in.

For the cylinder not to want to slip, the angular momentum, A, must therefore also be the same - but of course in the opposite direction.

So what height must the nail be to produce a torque which changes the angular momentum by 2A?

radagast_

That's my problem: I get from that this equation: the nail's torque needs to equal both directions' friction torque: [tex](h-R)F=2F\mu*R[/tex], by the angular momentum conservation.
from this I get [tex]\frac{F}{F\mu}=\frac{2R}{h-R}[/tex]
and I remain with a ratio of the forces I don't know how to get.

tiny-tim

Ah! No …

There is no "friction torque", because there is no friction.

The cylinder is rolling with steady speed - it's not pushing at the floor - the friction is zero.

The question gives you I, the moment of inertia.

Use it!

radagast_

I don't understand - the cylinder is spinning - so there's a torque which generates [tex]\omega[/tex], right? how can you say there isn't (im not talking about force, but torque).

So you're saying I should do something like this: [tex]2I\omega=F(h-r)[/tex]. I need also to use [tex]\omega=\frac{v}{R}[/tex]. but then what can I use for [tex]F[/tex]?

Oerg

tiny-tim!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

you need friction to roll!!!!!!!!!!!!!!!!!!!!!!!!!!!!! thats how wheels work!!!!!!

Oerg

radagast, how would you separate the force exerted by the nail into its tangenial and centripetal components?

Also, bear in mind conservation of angular momentum. This will give you a clue as to how much force was exerted on the cylinder during collision.

radagast_

Oreg, I don't understand why I need to separate the nail force, instead of separating the r vector to the center of mass, in which I'll get N=(r-h)*F.
Conservation of angular momentum is what I started with. I still remain with a missing ratio - F/Fmu ... I don't know where to get more information.

tiny-tim

No … that's only the way wheels work if they don't work …

Imagine a log rolling on ice - it will keep going at the same speed without any friction, and it will keep rolling without slipping (so long as it started off rolling without slipping, of course).

Wheels only need friction with the ground during acceleration or braking (or turning).
I'm saying there's no friction.

Yes, there is a torque when the cylinder hits the nail (but not before or after), and that's what you have to calculate.

What can you use for F? You tell us … what equation do you have that involves F, but doesn't involve angular momentum?

radagast_

well, I have F=ma, and 2mv=integral{f*dt}. from both I can't get anything... because the acceleration is 0. what am I missing?

tiny-tim

You are missing … the full version of Newton's second law.

It's often referred to as force = mass x acceleration.

But that's not actually what good ol' Netwon said.

He said total force = change of momentum.

If the change of momentum is gradual (not sudden), then the rate of force equals the rate of change of momentum, which is mass x acceleration!

(That's why force = mass x acceleration is so useful - most forces are graudal.)

But if the change of momentum is sudden, then we use the total force (usually called the impulse), instead of the rate of force.

In this case, the cylinder reverses instantaneously, so we use the impulse (total force), F say, and we put it equal to momentum after minus momentum before.

So F = … ?

radagast_

ahh. so [tex]F=2mv[/tex].
For some reason I got confused thinking the momentum difference is integral of F and not F itself... ~(_8^[I] D'oh!
Thus
[tex]2I\omega=F(h-r) => 2*\frac{mR^2}{2}*\frac{v}{R}=2mv(h-r)[/tex]
Thus
[tex]h=\frac{3R}{2}[/tex]

correct?

tiny-tim

radagast_

Thanks tiny-tim and Oerg!

Oerg

it still doesnt make sense to me. How does a wheel slip when its angular momentum has changed by 2a.

Doesnt a wheel slip when the torque is greater than the torque caused by the friction???