Finding the Height of a Nail for a Rolling Cylinder without Slipping

[radagast_]

Homework Statement

http://img503.imageshack.us/img503/3535/11uv8.gif

a cylinder with a given radius R rolls to the right without slipping (\omega= \frac{v}{R}). It hits a nail fixed to the pillar. find h, the height of the nail above the floor, in which the cylinder will roll back without slipping.
I=\frac{MR^2}{2} 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 h with R only.

Homework Equations

conservation of linear momentum, conservation of angular momentum.
J=r\times P

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, v=-u.
Thus, there is impact as follows: 2mv=\Delta P
Also, I want the cylinder to roll backwards still with the term of not-slipping - \omega=\frac{v}{R} so it's the same with angular momentum: 2F\mu*R=\Delta J
Now, I was thinking about solving this with J=(h-R)\times P, but then I don't know what to do with F\mu.

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) F_t*R=2F\mu*R?
About your second line - do you mean the linear impulse or angular?

 

[tiny-tim]

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) F_t*R=2F\mu*R?
About your second line - do you mean the linear impulse or angular?

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: (h-R)F=2F\mu*R, by the angular momentum conservation.
from this I get \frac{F}{F\mu}=\frac{2R}{h-R}
and I remain with a ratio of the forces I don't know how to get.

 

[tiny-tim]

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

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 \omega, 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: 2I\omega=F(h-r). I need also to use \omega=\frac{v}{R}. but then what can I use for F?

 

[Oerg]

tiny-tim!

you need friction to roll! that's 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]

tiny-tim!

you need friction to roll! that's how wheels work!

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 don't understand - the cylinder is spinning - so there's a torque which generates \omega, 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: 2I\omega=F(h-r). I need also to use \omega=\frac{v}{R}. but then what can I use for F?

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]

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?

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

correct?

 

[tiny-tim]

Looks good! ​

 

[radagast_]

Thanks tiny-tim and Oerg!

 

[Oerg]

it still doesn't 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?

 

[tiny-tim]

just geometry … point of contact is stationary

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

No, it slips when the velocity of its point of contact is non-zero.

In perfect rolling, the bit of the wheel that's on the ground is always (instantaneously) stationary.

If it isn't stationary, that means the wheel is slipping along the ground.

If there's no acceleration and no slipping, there should be no friction.

This isn't physics … it's geometry!

 

[Oerg]

wait wait wait!

rolling means it is rotating as it is moving along! So doesn't this mean that the exact same point on the object is not in constant contact with the ground?!

 

[tiny-tim]

wait wait wait!

I'm waiting! ​

rolling means it is rotating as it is moving along! So doesn't this mean that the exact same point on the object is not in constant contact with the ground?!

Yes, you're right, the point of contact changes all the time.

But whichever bit of the rim of the wheel is on the ground at any particular moment is stationary!

Confusing, isn't it?

 

[Oerg]

I'm waiting! ​

But whichever bit of the rim of the wheel is on the ground at any particular moment is stationary!

i don't know what you mean by that, but all I know for rolling motion is that there is rotational kinetic energy that is dependent on the moment of inertia and there is also a translational kinetic energy that is dependent on the velocity of the outer rim of the cylinder.

The rim is rotating as with the rest of the cylinder at the same angular speed. I just cnanot see how the rim remains stationary or any part of the rim.

 

[tiny-tim]

top centre and bottom have different velocities

i don't know what you mean by that, but all I know for rolling motion is that there is rotational kinetic energy that is dependent on the moment of inertia and there is also a translational kinetic energy that is dependent on the velocity of the outer rim of the cylinder.

Forget energy - this is a geometry issue.

The rim is rotating as with the rest of the cylinder at the same angular speed. I just cnanot see how the rim remains stationary or any part of the rim.

Exactly - at the same angular speed - relative to the centre, of course.

So, if the angular velocity is w, and the radius is R, then the bottom of the wheel has a velocity relative to the centre of wR one way, and the top of the wheel has a velocity relative to the centre of wR the other way.

To put it another way - if the wheel is rotating, then the top centre and bottom must have different velocities.

The top always moves twice as fast as the centre, and the bottom is always stationary.

 

[physixguru]

Forget energy - this is a geometry issue.


Exactly - at the same angular speed - relative to the centre, of course.

So, if the angular velocity is w, and the radius is R, then the bottom of the wheel has a velocity relative to the centre of wR one way, and the top of the wheel has a velocity relative to the centre of wR the other way.

To put it another way - if the wheel is rotating, then the top centre and bottom must have different velocities.

The top always moves twice as fast as the centre, and the bottom is always stationary.

i object mylord.

are you thinking that the bottom is a fixed point?

 

[Oerg]

yes you are right. and that is from a stationary observer's poinbt of view. But in physisc, we do not do that because it gets confusing!

We separate these two velocities into a rotational motion and a linear motion.

Ok, now back to the question of friction. A wheel needs friction to roll doesn't it??/ If there is no friction, the wheel would simply rotate at the same spot rigfht?! isn't this called slipping?

 

[tiny-tim]

A wheel needs friction to roll doesn't it??/ If there is no friction, the wheel would simply rotate at the same spot rigfht?! isn't this called slipping?

It needs friction to start rolling, or to accelerate the vehicle, or to stop rolling.

But it doesn't need friction to roll at a constant rate.

If there is no friction, the wheel's forward velocity will stay the same, and its angular velocity will stay the same.

Consider a log on (frictionless) ice - if you manage to start it rolling without slipping, it will carry on doing so for ever.

 

[Oerg]

It needs friction to start rolling, or to accelerate the vehicle, or to stop rolling.

But it doesn't need friction to roll at a constant rate.

If there is no friction, the wheel's forward velocity will stay the same, and its angular velocity will stay the same.

Consider a log on (frictionless) ice - if you manage to start it rolling without slipping, it will carry on doing so for ever.

a log is different from a wheel!

Why do cars skid on rainy days? Because there is lower friction with the road. Thus when they turn, the torque is too much for the wheels to handle that's why they slip

 

[physixguru]

My take is that friction always assists rolling.

As far as slipping is concerned , slipping does not necessarily imitate rolling. it is rolling along with a component of frictional force acting along and back the wheel.

rolling objects to frictional force at any point of contact.

 

[physixguru]

a log is different from a wheel!

Why do cars skid on rainy days? Because there is lower friction with the road. Thus when they turn, the torque is too much for the wheels to handle that's why they slip

I object.

It is not the torque that causes it to skid. The centripetal force is minimised and thus the tangential component of velocity takes advantage and the car skids.

Friction always necessarily keeps the body in circular motion.

 

[Oerg]

if we were to apply a foce to a wheel on a frictionless surface, the wheel will simply accelerate without turning.

If a wheel is rolling and friction is taken out, then the wheel would contrinue rolling.

when a wheel slips, it means that the torque is too much for the friction to handle.

 

[tiny-tim]

if we were to apply a foce to a wheel on a frictionless surface, the wheel will simply accelerate without turning.

(I take it you mean a stationary wheel)

It depends where you apply the force.

If the point of application is the centre of the wheel (or if the line of application passes through the centre), then I agree. But if the point of application is off-centre, then the wheel will accelerate and will start to turn.

You agree?

Then, if you judge the point of application right, the velocity and the turn may match, and there will be no slipping - even on ice!

 

[Oerg]

(I take it you mean a stationary wheel)

It depends where you apply the force.

If the point of application is the centre of the wheel (or if the line of application passes through the centre), then I agree. But if the point of application is off-centre, then the wheel will accelerate and will start to turn.

:

yes that was what i meant. The difference being that if there was friction, then the wheel would turn.

again, i thought that slipping means that something doesn't have enough grip. And with grip you have friction. So in order to see when the wheel would slip, you got to have friction for that./

 

[tiny-tim]

yes that was what i meant. The difference being that if there was friction, then the wheel would turn.

You're missing my point, that even that if there is no friction, then the wheel can still turn:

But if the point of application is off-centre, then the wheel will accelerate and will start to turn.

You agree?

Then, if you judge the point of application right, the velocity and the turn may match, and there will be no slipping - even on ice!

Do you agree?