What is the relationship between rotating objects and their surfaces?

[fisico30]

Hello Forum,

Consider a rigid disk that is rotating on a surface. If the surface is elastic but not symmetric, the rotating disk will eventually slow down.
If the surface was perfectly and symmetrically elastic the object will continue to rotate (the front deformation of the surface would hinder the motion and the rear deformation of the surface would help the rotation).

For an object to roll, does the surface need to have a nonzero coefficient of static friction?
I would think so since the contact point is instantaneously at rest...

But what would happen if the rotating disk was rotating on a rigid surface with nonzero coeff of static friction and moved into a region whose surface has zero coeff. of static friction?
I was told that the object would continue to move and rotate at the same rate suffering no torque that would increase or decrease its angular momentum.
But the fact the the new surface has zero coeff. of friction leads me to believe that the rotating disk has no grip.
Like a person that goes from the concrete to ice: it will not continue its motion. Why would the rotating disk continue to rotate and move forward instead?

thanks
fisico30

 

[mfb]

If the surface is elastic but not symmetric, the rotating disk will eventually slow down.

Are you sure you mean "elastic"? And symmetry of what? And how should this brake the disk?

For an object to roll, does the surface need to have a nonzero coefficient of static friction?

No, but with zero static friction the angular velocity and the linear velocity have to be fine-tuned (v=\omega r) to avoid slipping.

But what would happen if the rotating disk was rotating on a rigid surface with nonzero coeff of static friction and moved into a region whose surface has zero coeff. of static friction?

Rotating as in "rolling"? It would continue to roll.

Like a person that goes from the concrete to ice: it will not continue its motion.

It will. Jump from "not ice" on ice to test, as humans have no wheels or similar devices to roll.

 

[tiny-tim]

hello fisico30!

But the fact the the new surface has zero coeff. of friction leads me to believe that the rotating disk has no grip.
Like a person that goes from the concrete to ice: it will not continue its motion. Why would the rotating disk continue to rotate and move forward instead?

because of good ol' Newton's first law (linear and rotational versions) …

any body on which there is no external force will continue to move with constant velocity

any body on which there is no external torque will continue to rotate with constant angular velocity ​

 

[mfb]

any body on which there is no external torque will continue to rotate with constant angular velocity ​

... unless it changes its shape to change its moment of inertia. However, it will keep its angular momentum.

 

[fisico30]

Thanks everyone.
I am convinced that the wheel will continue to rotate at the same rate even when it passes on ice...

How about this: if the wheel rotates on a perfectly rigid surface (so no rolling friction, no air drag, etc...) some say that the wheel will never stop. Others say that the force of static friction, which arises from the wheel pushing backward on the surface, slowly reduces the angular velocity and translational velocity of the wheel by applying a torque...
Is that true?

thanks
fisico30

 

[jbriggs444]

If there is no rolling friction, no air drag, etc then why would you expect the wheel to be "pushing back on the surface".

 

[tiny-tim]

I am convinced that the wheel will continue to rotate at the same rate even when it passes on ice...

yes, if it's not being driven or braked, and if there's no friction, then it will rotate at the same rate

if the wheel rotates on a perfectly rigid surface (so no rolling friction, no air drag, etc...) some say that the wheel will never stop. Others say that the force of static friction, which arises from the wheel pushing backward on the surface, slowly reduces the angular velocity and translational velocity of the wheel by applying a torque...
Is that true?

if the (linear) speed v, the angular speed ω, and the radius are related by the rolling constraint v = rω, and if there is no applied force or torque,* then there will be no friction force whatever the coefficient of friction is

(*in practice, there is always a small amount of air resistance, friction with the axle etc, which will prevent the acceleration being zero unless a small torque is supplied from the engine)

 

[fisico30]

Thanks tiny-tim...

What do you think about this: if an object is rolling on a surface with a finite static coefficient of friction, zero rolling friction and not sliding, will the cylinder eventually slow down (decrease in translational velocity and angular velocity) or will it continue with its constant speed?

Some told me that even if the rolling friction is zero, the static friction at the contact point will cause a torque that will eventually slow down the rotation and speed of the rolling cylinder...

thanks
fisico30

 

[tiny-tim]

hi fisico30!

… if an object is rolling on a surface with a finite static coefficient of friction, zero rolling friction and not sliding, will the cylinder eventually slow down (decrease in translational velocity and angular velocity) or will it continue with its constant speed?

(assuming no air resistance, and a horizontal surface) constant speed …

if it's already rolling, then the static friction will be zero, and the only external force is vertical (gravity)

Some told me that even if the rolling friction is zero, the static friction at the contact point will cause a torque that will eventually slow down the rotation and speed of the rolling cylinder...

tell them there's no static friction

static friction is less than or equal to µN …

it adjusts itself to fit the starting conditions, and the starting conditions are perfectly happy without it! ​

 

[fisico30]

Maybe I am starting to get it:

When a person walks (I know it is not rolling), the static friction is important because the foot pushes backward and the effect is forward motion, as long as the backward directed push is less that the max static frictional force...

In the case of a cylinder rolling on a surface having nonzero coefficient of static friction, I would think that a static frictional force must exist at the contact point between the cylinder and the surface, since that point is at rest (instantaneously)...
So static friction must exist for rolling to take place, doesn't it? The cylinder, in a sense, is trying to push the surface backward, but the static friction makes the cylinder roll forward instead...

thanks
fisico30

 

[tiny-tim]

In the case of a cylinder rolling on a surface having nonzero coefficient of static friction, I would think that a static frictional force must exist at the contact point between the cylinder and the surface, since that point is at rest (instantaneously)...
So static friction must exist for rolling to take place, doesn't it?

no no no …

rolling will happen so long as the (linear) speed v and the angular speed ω are related by v = ωr

if they start like that, and if there are no external horizontal forces (or torques), then v and ω will stay the same (good ol' Newton's first law) …

so the rolling automatically continues!

(of course, in practice there are losses to rolling resistance and air resistance, eg the net air resistance is almost exactly horizontal, and almost exactly through the centre of the cylinder, so it decreases v very slightly, but leaves ω the same … so there must be a very slight forward friction force to reduce ω, to compensate )

 

[fisico30]

thanks tiny-tim.

I see how v=omega r. It must be like that in the case of rolling.

So the cylinder does not need to speed energy in the form of work against the force of friction at the contact point?
Does that static friction not cause any torque?

fisico30

 

[tiny-tim]

friction at the contact point must cause torque, since it's not through the centre of mass

if the rotation rate is not constant, then there must be net torque (so if all the other forces are through the centre of mass, then there must be friction at the contact point)

 

[fisico30]

Ok , so static friction does cause torque. Does that torque not slow the rotation down?

If rolling exists, like a cylinder on a surface (no rolling friction), that static friction torque must be there too...

 

[tiny-tim]

Ok , so static friction does cause torque. Does that torque not slow the rotation down?

if there is an applied horizontal force C through the centre, and a friction force F, and if the mass is m, and the "rolling mass" (I/r²) is m_r, and if there is no slipping, then:

C + F = (m + m_r)a

energy = 1/2 (m + m_r)v²

F = -(I/r)α = -m_ra​

so the work done is ∫ (C + F).dx …

yes, both the applied force C and the friction force F do work (and F is in the opposite direction to C, so yes it always reduces the good work being done by the applied force )

 

[fisico30]

tiny-tim,
I guess I am implying that there is no force F. The cylinder is given push and set into motion. Will it continue to roll at that translational velocity or will it slow down (rolling friction is zero)?

 

[tiny-tim]

The cylinder is given push and set into motion.

when it has stopped sliding, and started rolling, it will continue to roll forever at the same linear speed and angular speed, and there will be no friction force, if we assume there is no rolling resistance or applied force

 

[rcgldr]

Note that the term is normally called "rolling resistance", not "rolling friction". Wiki article:

http://en.wikipedia.org/wiki/Rolling_resistance

In the idealized case of zero rolling resistance, then the forces involved during deformation and restoration cancel each other, resulting in zero net torque or force.

If there is zero friction (and no other external forces), then the cylinder just continues to rotate at it's current rate and will appear to be "rolling" if the cylinder's forward speed corresponds to it's surface speed = ω x r. The cylinder could be moving in space, free of any external forces, parallel to a plane but not touching the plane. If it's forward speed corresponds to ω x r, then it appears to be "rolling" while moving near the plane, even though it's not touching the plane.

 

[fisico30]

Thanks everyone!
I am finally convinced. My wrong intuition was telling if that if the point on the rim goes from nonzero speed to zero speed at the contact point, a deceleration must have been present. But that acceleration is not caused by external forces. The point slows down because it is part of the wheel.
In pure rolling (no air drag, no rolling friction) the wheel gently rolls and only touches the floor without applying a tangential, shear force on the floor, hence no reaction from the floor in the form of a static friction. My intuition was telling me that if the object is no sliding there must be a static friction keeping it from sliding. That force would have been the static friction. But it is not so...

In the case of pure rolling, the tangential speed of a point on the rim is equal to the translation speed of the center of mass: v_t =v_cm

In the case of v_t > v_cm we have both sliding and translation: car moving in the snow, the tires spin fast but the translation is not that much.

How about the case v_t < v_cm? In which physical situation would that happen?

Thanks
fisico30

 

[tiny-tim]

How about the case v_t < v_cm? In which physical situation would that happen?

car braking in the snow?

 

[ForceBoard]

Sounds like a perfect test case for a multibody simulation program, a couple of years ago while working in the automotive industry I used a program called ADAMS which you could simulate this scenario in pretty easily.

 

[fisico30]

Tiny Tim,

I have an elementary question: why does an object (any object) rotate about its CM if an initial torque is applied to it and then thrown in the air? What theorem or math demonstration of that?

thanks
fisico30

 

[tiny-tim]

why does an object (any object) rotate about its CM if an initial torque is applied to it and then thrown in the air?

because the centre of mass follows the same parabola that the same non-rotating object would follow, so we naturally say that its motion is a combination of its non-rotating motion together with a rotation about the centre of mass

 

[fisico30]

Good reply.

So, we start by remembering that the center of mass of the object is that theoretical point where we can imagine all the mass to be concentrated. That is the idealization of making an extended object a point-like object.
The center of gravity is the same thing but if the acceleration of gravity changes (we are very high) the CM and CG are slightly displaced.

A point like the CM, being a point, cannot rotate about itself, so when launched in the air it will follow the parabolic trajectory. I know that it must be that way. But why does the CM be exactly that point that follows the parabolic trajectory. Why can it not be another point?

An extended object, rotating while flying in the air, will have its CM follow the parabolic trajectory. If that was not the case the CM would also rotate about that pivotal point...

But there are cases (when the object has its rotation constrained about a specific axis) where the CM is rotating too.

thanks,
fisico30

 

[tiny-tim]

But why does the CM be exactly that point that follows the parabolic trajectory. Why can it not be another point?

if the body is rigid, then ∑ F_i = ∫∫∫ ρ(x,y,z)v(x,y,z) dxdydz

= ∫∫∫ ρ(x,y,z)v_c.o.m dxdydz + ∫∫∫ ρ(x,y,z)(v(x,y,z) - v_c.o.m) dxdydz

= Mv_c.o.m + d/dt ∫∫∫ ρ(x,y,z)((x,y,z) - (x,y,z)_c.o.m) dxdydz

= Mv_c.o.m + d/dt 0 (by definition) ​

 

[fisico30]

Thank you again.

I guess you missed a d/dt... did you.

Force is the time derivative of momentum (change in momentum). The first integral in your explanation should have a d/dt, correct?

thanks
fisico30

 

[tiny-tim]

oops!

oops! yes, an extra d/dt after every "equals"!

 

[fisico30]

Great, thanks.

So, v(x,y,z) is the velocity of each point, v_com is the velocity of the CM. Both velocity are with respect to the lab frame of reference.

The sum of the forces F_i include all the forces regardless of where these forces are applied on the body (at which points (x,y,z))...


why would the second integral cancel? I am missing that part. Why would (x,y,z) cancel (x,y,z) of the CM?

thanks again for the patience,
fisico30

 

[tiny-tim]

= Mv_c.o.m + d/dt ∫∫∫ ρ(x,y,z)((x,y,z) - (x,y,z)_c.o.m) dxdydz

= Mv_c.o.m + d/dt 0 (by definition) ​

why would the second integral cancel? I am missing that part. Why would (x,y,z) cancel (x,y,z) of the CM?

because the position (x,y,z)_c.o.m is defined by (x,y,z)_c.o.m = (1/∫∫∫ ρ(x,y,z) dxdydz) ∫∫∫ ρ(x,y,z)*(x,y,z) dxdydz

 

[fisico30]

Hi again tiny-tim.

I found this old post (of mine):https://www.physicsforums.com/showthread.php?t=252436

Doc A was saying that:

In general, the instantaneous axis of rotation can be anywhere, not necessarily passing through the center of mass. What is true, and maybe this is what you are thinking of, is that any arbitrary motion of a rigid body can be viewed as a rotation about the center of mass plus the translation of the center of mass.

Is he right? what do you think?
i thought that the instantaneous axis of rotation can only and always pass through the CM..