# Wheel Rolling Motion

Homework Statement:
If you...arrange for Mgsin(θ) to exceed fs-max, then you eliminate the smooth rolling and the body slides down the ramp.
Relevant Equations:
Can't understand this sentence.
Halliday's book says the following about a wheel rolling down a ramp:
"Note that the pull by the gravitational force causes the body to come down the ramp, but it is the frictional force that causes the body to rotate and thus roll. If you eliminate the friction (by, say, making the ramp slick with ice or grease) or arrange for Mgsin(θ) to exceed fs-max, then you eliminate the smooth rolling and the body slides down the ramp." [Chapter 11, Page 300]

I understand how eliminating friction (ice) would cause sliding (pure translational motion) instead of rolling.

But, I can't understand this part:
"If you...arrange for Mgsin(θ) to exceed fs-max, then you eliminate the smooth rolling and the body slides down the ramp.".

Shouldn't Mgsin(θ) be greater than fs-max at all times?

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Homework Statement: If you...arrange for Mgsin(θ) to exceed fs-max, then you eliminate the smooth rolling and the body slides down the ramp.
Homework Equations: Can't understand this sentence.

...

Shouldn't Mgsin(θ) be greater than fs-max at all times?
The rolling (of perfectly smooth and rigid body) may start at arbitrarily small angle θ. The criteria above is for start of sliding motion.

TSny
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But, I can't understand this part:
"If you...arrange for Mgsin(θ) to exceed fs-max, then you eliminate the smooth rolling and the body slides down the ramp.".
I agree with you. This statement doesn't look correct to me.

Shouldn't Mgsin(θ) be greater than fs-max at all times?
I'm not really clear on what you're asking here. "Shouldn't Mgsin(θ) be greater than fs-max at all times?"... in order for what to happen?

The rolling (of perfectly smooth and rigid body) may start at arbitrarily small angle θ. The criteria above is for start of sliding motion.

Finally!
Thank You.
So, if the fs-max is broken instantaneously in the beginning somehow by Mgsin(θ), you'll step into fk territory and thereby begin pure translational motion.
On the other hand, during smooth rolling motion, fs should be lower than Mgsin(θ)?

• trurle
The wording in this Halliday book could have been better. But, the way I understand this now is that some nudge has to be given by Gravity or another force, or else a wheel will slide rather than roll down.

I agree with you. This statement doesn't look correct to me.

I'm not really clear on what you're asking here. "Shouldn't Mgsin(θ) be greater than fs-max at all times?"... in order for what to happen?

In order for rolling motion instead of pure translational (slipping) motion to occur for a wheel going down a ramp.

kuruman
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Gold Member
Homework Statement: If you...arrange for Mgsin(θ) to exceed fs-max, then you eliminate the smooth rolling and the body slides down the ramp.
Homework Equations: Can't understand this sentence.

Shouldn't Mgsin(θ) be greater than fs-max at all times?
Simply put, no. Mgsin(θ) is not what one should be looking at. For the wheel to roll without slipping, a certain amount of static friction ##f_s## is required. Rolling without slipping will occur if ##f_s <f_s^{max}##, where ##f_s^{max}=\mu_s N=\mu_s mg\cos\theta##. If you draw a free body diagram and figure out the required static friction, you will find ##f_s=\frac{1}{3}mg\sin\theta## under the assumption that the rolling body is a cylinder. With a little bit of algebra you will find that the condition for rolling without slipping is ##\mu >\frac{1}{3}\tan\theta##. The statement in the textbook is poorly worded if not nonsensical.

• PeroK and lightlightsup
Simply put, no. Mgsin(θ) is not what one should be looking at. For the wheel to roll without slipping, a certain amount of static friction ##f_s## is required. Rolling without slipping will occur if ##f_s <f_s^{max}##, where ##f_s^{max}=\mu_s N=\mu_s mg\cos\theta##. If you draw a free body diagram and figure out the required static friction, you will find ##f_s=\frac{1}{3}mg\sin\theta## under the assumption that the rolling body is a cylinder. With a little bit of algebra you will find that the condition for rolling without slipping is ##\mu >\frac{1}{3}\tan\theta##. The statement in the textbook is poorly worded if not nonsensical.

Excellent explanation. I put in some work to derive what you did and it does check out.
So:
My only question remains: How did you arrive at this main "law", exactly: ##f_s <f_s^{max}##

Can they be equal?

haruspex
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Excellent explanation. I put in some work to derive what you did and it does check out.
So:
My only question remains: How did you arrive at this main "law", exactly: ##f_s <f_s^{max}##

Can they be equal?
Since these are real physical quantities of a continuous nature (not discrete), exact equality is meaningless. It doesn't matter whether you take it as < or ≤.

PeroK
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2020 Award
Excellent explanation. I put in some work to derive what you did and it does check out.
So:
My only question remains: How did you arrive at this main "law", exactly: ##f_s <f_s^{max}##

Can they be equal?

Static friction acts to the extent it is required - up to its maximum value. If you have an object on a flat surface, then there is the potential for a static friction force up to ##f_s^{max}##. If you apply no horizontal force to the box, then there is no force of static friction. As you apply an increasing horizontal force, the static friction opposes this force equally, up to its maximum. As long as the object does not move, you have:

##F = f_s \le f_s^{max}##,

where ##F## is the applied horizontal force. Which simply says that the static friction is equal in magnitude to the force trying to move the box.

If ##F## exceeds ##f_s^{max}##, then the object will start to move and the resisting force will become kinetic friction, ##f_k##, which is typically less than ##f_s^{max}##. That means that once you get an object moving, the friction decreases.

This ties in with what you might expect that once you get an object moving it is easier to keep it moving; but if you let it stop, it's harder to get it moving again.

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• lightlightsup
kuruman
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Gold Member
Excellent explanation. I put in some work to derive what you did and it does check out.
So:
My only question remains: How did you arrive at this main "law", exactly: ##f_s <f_s^{max}##

Can they be equal?
To augment what others have said, the equality is used when the object is neither slipping nor at rest, i.e. right at the threshold of slipping. It's like asking a question to which a "simple" yes/no answer is expected when the actual answer is "maybe".

• lightlightsup
Static friction acts to the extent it is required - up to its maximum value. If you have an object on a flat surface, then there is the potential for a static friction force up to ##f_s^{max}##. If you apply no horizontal force to the box, then there is no force of static friction. As you apply an increasing horizontal force, the static friction opposes this force equally, up to its maximum. As long as the object does not move, you have:

##F = f_s \le f_s^{max}##,

where ##F## is the applied horizontal force. Which simply says that the static friction is equal in magnitude to the force trying to move the box.

If ##F## exceeds ##f_s^{max}##, then the object will start to move and the resisting force will become kinetic friction, ##f_k##, which is typically less than ##f_s^{max}##. That means that once you get an object moving, the friction decreases.

This ties in with what you might expect that once you get an object moving it is easier to keep it moving; but if you let it stop, it's harder to get it moving again.

Thank you everyone for your help.
I know I keep beating a dead horse here, but, to summarize:

This case deals with a wheel starting at the top of an incline with friction.

If there is no friction, then, obviously, the wheel will slide down instead of rolling down.
No confusion there.

BUT: For a wheel to roll down an incline that has friction:
My conclusion (perhaps erroneous) is that:
mgsin(θ) and/or a force applied parallel to it must be FAR FAR GREATER than the ##f_s^{max}## in order for the wheel to slip and for ##f_k## to kick in.
That's what the Halliday book got wrong?
It should have said "FAR FAR GREATER" or used the ">>" sign.
This is why a wheel would slip on an extremely low-friction surface.
The instantaneous ##f_s^{max}## would be minuscule compared to the instantaneous force applied to get the wheel rolling. Thus, at the small point of contact, ##f_k## kicks in and slipping (translation instead of rotation) occurs.

What is the threshold for this difference between the instantaneous force applied and ##f_s^{max}## that causes ##f_k## (slipping) to kick in? I don't know. I suppose that is a bit more experimental.

All of these rolling physics problems in my introductory mechanics class always have us assume that the wheel will be in rolling motion. No problem asks me to calculate whether or not slipping will occur. I'm guessing this is done for the reasons I mentioned above.

Please let me know what you think.

TSny
Homework Helper
Gold Member
For reference, the page from Halliday is here .

The relevant quote is from the bottom of the page: "If you...arrange for Mg sin θ to exceed fs,max, then you eliminate the smooth rolling and the body slides down the ramp."

But, this isn't true. ##Mg\sin\theta## can exceed ##f_{\rm s,max}## and still have rolling without slipping.

For example, take

##I_{\rm com} = \frac{1}{2}MR^2, \,\,\,\,\,\, \mu_s = \frac{1}{2}, \,\,\,\,\,\, \theta = \sin^{-1}\frac{3}{5}##

If I didn't make any mistakes, the body will roll down without slipping with

##Mg\sin\theta = \frac{3}{5}Mg, \,\,\,\,\,\, f_{\rm s, max} = \frac{2}{5}Mg , \,\,\,\,\,\, f_s = \frac{1}{5}Mg##

So, here we have ##Mg\sin\theta > f_{\rm s, max}##, yet the body rolls without slipping with friction force ##f_s##.

What is the threshold for this difference between the instantaneous force applied and ##f_s^{max}## that causes ##f_k## (slipping) to kick in? I don't know.
@kuruman outlined how to do this in post #7.

For the example above, the body would be on the verge of slipping if you raised the angle to ##\theta = \tan^{-1} \frac{3}{2}## . Then,

##f_s = f_{\rm s, max}## and ## Mg\sin\theta = 3 f_{\rm s, max}## (for this example).

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PeroK
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2020 Award
Thank you everyone for your help.
I know I keep beating a dead horse here, but, to summarize:

This case deals with a wheel starting at the top of an incline with friction.

If there is no friction, then, obviously, the wheel will slide down instead of rolling down.
No confusion there.

BUT: For a wheel to roll down an incline that has friction:
My conclusion (perhaps erroneous) is that:
mgsin(θ) and/or a force applied parallel to it must be FAR FAR GREATER than the ##f_s^{max}## in order for the wheel to slip and for ##f_k## to kick in.
That's what the Halliday book got wrong?
It should have said "FAR FAR GREATER" or used the ">>" sign.
This is why a wheel would slip on an extremely low-friction surface.
The instantaneous ##f_s^{max}## would be minuscule compared to the instantaneous force applied to get the wheel rolling. Thus, at the small point of contact, ##f_k## kicks in and slipping (translation instead of rotation) occurs.

What is the threshold for this difference between the instantaneous force applied and ##f_s^{max}## that causes ##f_k## (slipping) to kick in? I don't know. I suppose that is a bit more experimental.

All of these rolling physics problems in my introductory mechanics class always have us assume that the wheel will be in rolling motion. No problem asks me to calculate whether or not slipping will occur. I'm guessing this is done for the reasons I mentioned above.

Please let me know what you think.

In general, the motion of an object rolling down a slope depends only on its geometry. In particular, the ratio of its moment of inertia to its mass. You could try to solve the general problem to show this.

If the ball is accelerated by another force, perhspa assume through its centre of mass, then it is logical to consider the case where the object is on a flat surface: the incline adds nothing conceptually interesting.

Another task would be to calculate the maximum force that could be applied here. Again this will depend on the ratio of moment of inertia to mass.

Can you attempt either of these problems?

You've been given quite a lot of ideas from the posts above for some specific cases.

vela
Staff Emeritus
Homework Helper
BUT: For a wheel to roll down an incline that has friction:
My conclusion (perhaps erroneous) is that:
mgsin(θ) and/or a force applied parallel to it must be FAR FAR GREATER than the ##f_s^{max}## in order for the wheel to slip and for ##f_k## to kick in.
This isn't the right criterion.

That's what the Halliday book got wrong?
I think you should forget the quote. It's just plain wrong, and it's leading you astray.

Consider rolling a cylinder down an incline with friction. If it doesn't slip, then that means the frictional force was big enough to produce the angular acceleration needed for no slipping. Now increase the angle a bit and let it roll down again. What would be the difference? It would accelerate more quickly, which means the angular acceleration to prevent slipping must be larger, which means the frictional force must be larger. If you keep increasing the angle, eventually the linear acceleration of the cylinder will be so large that the force of friction necessary to prevent slipping will exceed the maximum possible. That's when the cylinder will slip.

• • lightlightsup and TSny
This isn't the right criterion.

I think you should forget the quote. It's just plain wrong, and it's leading you astray.

Consider rolling a cylinder down an incline with friction. If it doesn't slip, then that means the frictional force was big enough to produce the angular acceleration needed for no slipping. Now increase the angle a bit and let it roll down again. What would be the difference? It would accelerate more quickly, which means the angular acceleration to prevent slipping must be larger, which means the frictional force must be larger. If you keep increasing the angle, eventually the linear acceleration of the cylinder will be so large that the force of friction necessary to prevent slipping will exceed the maximum possible. That's when the cylinder will slip.

Wow. That paragraph is golden.
That really clears things up.
I am going to attempt to work on the math surrounding this, over the next few days.

kuruman