Friction Required for Billiard Ball to Roll without Slipping

[domephilis]

You are missing that when you throw something up and it comes back down where it started the net change in potential energy is zero and the net work done by gravity is also zero because it must be equal to the net change in potential energy.

With you assumed friction potential function the change in potential energy is zero but the work done by friction is not as I indicated in post #20.

Why is the net work done by friction not zero when something goes from x-1 back to x-1?

 

[haruspex]

By Conservation of Angular Momentum, $$|\vec{L}| = |\vec{r}\times\vec{p}|$$ (the norm is there because the direction of the momentum is the same before and after and it simplifies the calculations).
Since $\vec{v} \perp \vec{r}$, $$L_i = Rmv_0$$ $$L_f = Rmv$$

You are leaving out the rotational component that is acquired.

Consider a mass element dm at $\vec r$ displacement from the centre of the ball, C.
If the ball is rotating at rate $\omega$ then its velocity relative to C is $\vec r\times\vec \omega$.
If the ball is at $\vec r_P$ from the axis P and has velocity $\vec v$ then dm is at $\vec r_P+\vec r$ and has velocity $\vec v+\vec \omega\times\vec r$ relative to P.
Its angular momentum about P is therefore $((\vec r_P+\vec r)\times(\vec v-\vec r\times\vec \omega))dm=(\vec r_P\times\vec v+\vec r\times\vec v-\vec r_P\times(\vec r\times\vec \omega)-\vec r\times(\vec r\times\vec \omega))dm$.
By definition of mass centre, $\int \vec r\cdot dm=0$, so the middle two terms vanish:
$\vec L=\int (\vec r_P\times\vec v-\vec r\times(\vec r\times\vec \omega))\cdot dm=M\vec r_P\times\vec v-\int \vec r\times(\vec r\times\vec \omega))dm$.
If we break $\vec r$ into components normal and parallel to the rotation, the triple product reduces to $-r_\perp^2\vec\omega$, hence $\vec L= M\vec r_P\times\vec v+\int r_\perp^2\cdot dm\vec \omega= M\vec r_P\times\vec v+I\vec \omega$

 

[kuruman]

Why is the net work done by friction not zero when something goes from x-1 back to x-1?

Because kinetic friction is always opposite to the displacement. The work done by friction is $$W_{\!f}=\mu_k mgd\cos\!\theta=\mu_k mgd\cos(180^{\circ})=-\mu_k mg d.$$ The force and displacement vectors are antiparallel regardless of the direction of the displacement. So, unlike gravity, the round-trip work done by friction is twice the one-way work if you retrace the path exactly.

 

[domephilis]

I have attached a photo of what my textbook says on the matter of friction and the work-energy theorem. This is from Chapter 5 Section 5.4 of the Fundamentals of Physics I. There are probably more context required. But clearly it says that the work done by friction is … and the energy equation is …

 

[domephilis]

You are leaving out the rotational component that is acquired.

Consider a mass element dm at $\vec r$ displacement from the centre of the ball, C.
If the ball is rotating at rate $\omega$ then its velocity relative to C is $\vec r\times\vec \omega$.
If the ball is at $\vec r_P$ from the axis P and has velocity $\vec v$ then dm is at $\vec r_P+\vec r$ and has velocity $\vec v+\vec \omega\times\vec r$ relative to P.
Its angular momentum about P is therefore $((\vec r_P+\vec r)\times(\vec v-\vec r\times\vec \omega))dm=(\vec r_P\times\vec v+\vec r\times\vec v-\vec r_P\times(\vec r\times\vec \omega)-\vec r\times(\vec r\times\vec \omega))dm$.
By definition of mass centre, $\int \vec r\cdot dm=0$, so the middle two terms vanish:
$\vec L=\int (\vec r_P\times\vec v-\vec r\times(\vec r\times\vec \omega))\cdot dm=M\vec r_P\times\vec v-\int \vec r\times(\vec r\times\vec \omega))dm$.
If we break $\vec r$ into components normal and parallel to the rotation, the triple product reduces to $-r_\perp^2\vec\omega$, hence $\vec L= M\vec r_P\times\vec v+\int r_\perp^2\cdot dm\vec \omega= M\vec r_P\times\vec v+I\vec \omega$

Relating the two definitions of angular momentum was another big question I had in my mind after learning this subject. I really appreciate you clearing that up. The derivation was incredibly enlightening. That said, coming back to the problem, to use this conservation law I would have to know what the angular velocity at x=d is (since, as you say, the rotational component is necessary). Since at x=d the ball rolls without slipping and the slippage is a continuous function, knowing the angular velocity is equivalent knowing the linear velocity of the CM at x=d and that, with energy conservation, would give us the answer. To find the angular velocity or the linear velocity, we would have to know the linear acceleration of the CM, the number of radians traveled from x=0 to x=d, or the time it took to go from 0 to d. None of which is easy to compute (at least to me). That, I believe, is where I was stuck on. I do sincerely apologize for taking up so much of your time. I am a bit slow with these things.

 

[Steve4Physics]

Can I add this to the discussion.

With an energy-based approach here, work done by kinetic (sliding) friction is a combination of the heat generated, the change in linear KE, the change in rotational KE. This ‘3-way split’ makes it messy.

The problem can be solved quite easily using $F=ma,\tau = I \alpha$ and simple kinematics.

During the initial (slipping-and-rolling) phase:
a) the linear velocity, $v$, decreases uniformly from $v_0$;
b) the angular velocity, $\omega$, increases uniformly from 0.
Rolling-without-slipping starts when these match, i.e. when $v = \omega R$.

I’d recommend first finding the time (not the distance) taken till $v = \omega R$. Then take it from there.

 

[domephilis]

Can I add this to the discussion.

With an energy-based approach here, work done by kinetic (sliding) friction is a combination of the heat generated, the change in linear KE, the change in rotational KE. This ‘3-way split’ makes it messy.

The problem can be solved quite easily using $F=ma,\tau = I \alpha$ and simple kinematics.

During the initial (slipping-and-rolling) phase:
a) the linear velocity, $v$, decreases uniformly from $v_0$;
b) the angular velocity, $\omega$, increases uniformly from 0.
Rolling-without-slipping starts when these match, i.e. when $v = \omega R$.

I’d recommend first finding the time (not the distance) taken till $v = \omega R$. Then take it from there.

That was an idea that I tried. But, at least, I couldn't do it. Then again, that isn't saying much. If this was a block, simple kinematics would suffice. We know the forces on it and everything is simple. But some of the frictional force is going to rotating the ball and some is going to slowing things down. That was where I got stuck on in this approach. I have no idea what portion of the force is doing what.

 

[haruspex]

Relating the two definitions of angular momentum was another big question I had in my mind after learning this subject. I really appreciate you clearing that up. The derivation was incredibly enlightening. That said, coming back to the problem, to use this conservation law I would have to know what the angular velocity at x=d is (since, as you say, the rotational component is necessary). Since at x=d the ball rolls without slipping and the slippage is a continuous function, knowing the angular velocity is equivalent knowing the linear velocity of the CM at x=d and that, with energy conservation, would give us the answer. To find the angular velocity or the linear velocity, we would have to know the linear acceleration of the CM, the number of radians traveled from x=0 to x=d, or the time it took to go from 0 to d. None of which is easy to compute (at least to me). That, I believe, is where I was stuck on. I do sincerely apologize for taking up so much of your time. I am a bit slow with these things.

When rolling, what equation relates the linear velocity to the angular velocity?

 

[Steve4Physics]

Use a free body diagram (FBD). Always!

But some of the frictional force is going to rotating the ball

During the rolling-and-slipping phase, the torque is provided by the (full) kinetic frictional force - as you can see on the FBD. So you can easily find the torque and angular acceleration .

and some is going to slowing things down.

During the rolling-and-slipping phase, the tranlational (linear) accelerating force is provided by the (full) kinetic frictional force. So you can easily find the translational acceleration.

 

[domephilis]

Use a free body diagram (FBD). Always!


During the rolling-and-slipping phase, the torque is provided by the (full) kinetic frictional force - as you can see on the FBD. So you can easily find the torque and angular acceleration .


During the rolling-and-slipping phase, the tranlational (linear) accelerating force is provided by the (full) kinetic frictional force. So you can easily find the translational acceleration.

If I understand correctly, that would be to assume that rolling and slipping are two binary states. If it’s rolling it’s not slipping. If it’s slipping, none of the rotational velocity contribute to linear velocity. Is that true?

 

[domephilis]

When rolling, what equation relates the linear velocity to the angular velocity?

v=r\omega is the equation. But the issue is that I know neither the linear nor the angular velocity at x=d. I need at least one of these to plug in to one of the conservation equations. The linear velocity changes due to friction, but I don’t know how much. Although we do know the torque, we do not know theta or the time elapsed from x=0 to x=d to solve for the omega.

 

[haruspex]

I need at least one of these to plug in to one of the conservation equations.

No, you only need as many independent equations as there are unknowns.
Angular momentum conservation relates initial velocity and angular velocity to final velocity and angular velocity. Rolling contact relates final velocity to final angular velocity.
Two equations, two unknowns.

 

[domephilis]

No, you only need as many independent equations as there are unknowns.
Angular momentum conservation relates initial velocity and angular velocity to final velocity and angular velocity. Rolling contact relates final velocity to final angular velocity.
Two equations, two unknowns.

Let me try again.

By Conservation of Angular Momentum $$mv_0R = Rmv+\frac{2}{5}mR^2\omega$$
Since $v = \omega r$, $$mv_0R = Rmv+\frac{2}{5}mR^2\frac{v}{R}$$
$$\implies v_0R = Rv+\frac{2}{5}Rv$$
$$\implies v_0 = v+\frac{2}{5}v$$
$$\implies v = \frac{5}{7}v_0$$
Plugging this into our conservation of energy equation:
$$\mu_k = \frac{1}{gd}(\frac{1}{2}v_0^2-\frac{7}{10}(\frac{25}{49}v_0^2)) =\frac{v_0^2}{7gd}$$

Did I get the angular momentum conservation right? The right answer should be $\mu_k = \frac{12v_0^2}{49gd}$

 

[haruspex]

Let me try again.

By Conservation of Angular Momentum $$mv_0R = Rmv+\frac{2}{5}mR^2\omega$$
Since $v = \omega r$, $$mv_0R = Rmv+\frac{2}{5}mR^2\frac{v}{R}$$
$$\implies v_0R = Rv+\frac{2}{5}Rv$$
$$\implies v_0 = v+\frac{2}{5}v$$
$$\implies v = \frac{5}{7}v_0$$
Plugging this into our conservation of energy equation:
$$\mu_k = \frac{1}{gd}(\frac{1}{2}v_0^2-\frac{7}{10}(\frac{25}{49}v_0^2)) =\frac{v_0^2}{7gd}$$

Did I get the angular momentum conservation right? The right answer should be $\mu_k = \frac{12v_0^2}{49gd}$

Your problem is the $\mu_kNd$ term. The work done by a force is the force multiplied by the distance it advances, but what is that distance when the two surfaces it acts on are moving differently?
During rolling, there can be static friction, but the two surfaces have the same velocity. So what matters for the work done is the difference in displacements of the points of contact. Your d is the displacement of the point of contact on the ground, but what distance has it travelled around the surface of the ball?

 

[domephilis]

Your problem is the $\mu_kNd$ term. The work done by a force is the force multiplied by the distance it advances, but what is that distance when the two surfaces it acts on are moving differently?
During rolling, there can be static friction, but the two surfaces have the same velocity. So what matters for the work done is the difference in displacements of the points of contact. Your d is the displacement of the point of contact on the ground, but what distance has it travelled around the surface of the ball?

That is really brilliant. That really was a nuance I would have never thought of.

We can start with the formula: $$\omega^2 - \omega_0^2 = 2\alpha\Delta\theta$$

Since $\omega_0 = 0$, the distance travelled around the surface of the ball in terms of radians is
$$\Delta\theta = \frac{\omega^2}{2\frac{\mu_kmgR}{I}}$$
(where the $\alpha$ was found by using $\tau = I\alpha$)
After some algebra we have $$R\Delta\theta = \frac{(R\omega)^2}{5\mu_kg}$$ (This is the "linear" distance travelled around the surface of the ball)
Going back to the conservation of energy equations, we make the corresponding alterations and get:
$$\frac{7}{10}mv^2-\frac{1}{2}mv_0^2 = -\mu_kmg(d-\frac{v^2}{5\mu_kg})$$
Hence,$$gd\mu_k = \frac{1}{2}v_0^2-\frac{5}{10}v^2$$
Using the results derived earlier using conservation of angular momentum, we get
$$\mu_k = \frac{12v_0^2}{49gd}$$

Q.E.D.

 

[haruspex]

Going back to the conservation of energy equations, we make the corresponding alterations and get:
$$(d+\frac{v^2}{5\mu_kg})$$

but I wrote

the difference in displacements of the points of contact.

 

[domephilis]

I am such an idiot. Thank you so much. I have made the proper edits up top. Frankly, this question is way out of my depth. I couldn't have done it at all without your help. I really appreciate your time and patience in solving this problem. AP physics is really tough sometimes. But, I have learnt a lot from doing this here. And I really love the Eureka moments. Adios.

 

[haruspex]

I have made the proper edits up top.

You did well.
Yes, it is quite an educational question.

 

[haruspex]

If it’s slipping, none of the rotational velocity contribute to linear velocity. Is that true?

I assume you mean contributes to linear acceleration.
For the purpose of my reply, I'll take velocity and angular velocity as both positive when rolling (though in a common convention they would have opposite signs).
If the rotation is too slow or backwards ($v>\omega r$) then the frictional force ($\mu_kN$) causes positive angular acceleration and negative linear acceleration; if too fast ($v<\omega r$) then the frictional force acts the other way and causes negative angular acceleration and positive linear acceleration.

 

[Steve4Physics]

Since a solution is now posted (and since I'd already done one for my own entertainment) here’s an alternative, if of interest.

Spoiler

$\mu_k$ is abbreviated to $\mu$ to reduce clutter.

Initially there is both rolling and slipping.

The linear acceleration is $a = -\frac {\mu mg}m = -\mu g$. The linear velocity at time $t$ is $v= v_0 -\mu g t$.

The angular acceleration is $\alpha = \frac {\tau}I = \frac {\mu mgR} {(2/5)mR^2}= \frac {5\mu g} {2R}$. The angular velocity at time $t$ is $\omega = \frac {5\mu g} {2R} t$.

Rolling without slipping starts when $v = \omega R$. That's the critical condition to understand. Say this happens at time $t=T$.

$v_ 0 - \mu g T = \frac {5\mu g} {2R} T R$.

A bit of algebra gives:
$T = \frac {2v_0}{7\mu g}$

Distance covered in this time is:
$d = v_0 T + ½ aT^2$
$= v_0 \frac {2v_0}{7\mu g}- ½ (\mu g) (\frac {2v_0}{7\mu g})^2$

A bit of algebra gives:
$d = \frac {12 v_0^2}{49 \mu g}$

$\mu = \frac {12 v_0^2}{49 g d}$

Edited - typo's corrected: $v_02$ changed to $v_0^2$.

 

[domephilis]

Since a solution is now posted (and since I'd already done one for my own entertainment) here’s an alternative, if of interest.

Spoiler

$\mu_k$ is abbreviated to $\mu$ to reduce clutter.

Initially there is both rolling and slipping.

The linear acceleration is $a = -\frac {\mu mg}m = -\mu g$. The linear velocity at time $t$ is $v= v_0 -\mu g t$.

The angular acceleration is $\alpha = \frac {\tau}I = \frac {\mu mgR} {(2/5)mR^2}= \frac {5\mu g} {2R}$. The angular velocity at time $t$ is $\omega = \frac {5\mu g} {2R} t$.

Rolling without slipping starts when $v = \omega R$. That's the critical condition to understand. Say this happens at time $t=T$.

$v_ 0 - \mu g T = \frac {5\mu g} {2R} T R$.

A bit of algebra gives:
$T = \frac {2v_0}{7\mu g}$

Distance covered in this time is:
$d = v_0 T + ½ aT^2$
$= v_0 \frac {2v_0}{7\mu g}- ½ (\mu g) (\frac {2v_0}{7\mu g})^2$

A bit of algebra gives:
$d = \frac {12 v_02}{49 \mu g}$

$\mu = \frac {12 v_02}{49 g d}$

I think there is an extra two in the final answer. I think that's just a typo. Regardless, there is one difficult thing for me to understand in this approach. The same force of friction causes both linear and angular acceleration/deceleration as if there are two copies of the same force. It is doing both things at the same time with just one force. Is this what actually happens? Or is this just mathematically convenient to derive and there is some underlying simplification? It seems like the reason it doesn't violate energy principles is that if one considers the whole track rather than the instantaneous force applied a faster linear deceleration would "shorten the work" for the angular acceleration. But, at any instant, the force applied can only do one of two things in full or both things with some apportionment between them. How does it do both things? If these underlying assumptions are answered, I agree this is a much more elegant way of doing this problem. Thank you for posting this alternate solution.

 

[domephilis]

I assume you mean contributes to linear acceleration.
For the purpose of my reply, I'll take velocity and angular velocity as both positive when rolling (though in a common convention they would have opposite signs).
If the rotation is too slow or backwards ($v>\omega r$) then the frictional force ($\mu_kN$) causes positive angular acceleration and negative linear acceleration; if too fast ($v<\omega r$) then the frictional force acts the other way and causes negative angular acceleration and positive linear acceleration.

Thank you for the explanation!

 

[Steve4Physics]

I think there is an extra two in the final answer. I think that's just a typo.

Not sure where you mean - can't see anything wrong. But am happy to make a correction if there's a mistake.

Regardless, there is one difficult thing for me to understand in this approach. The same force of friction causes both linear and angular acceleration/deceleration as if there are two copies of the same force. It is doing both things at the same time with just one force. Is this what actually happens?

The friction is simultaneously doing two jobs - changing the linear momentum and (because it provides a torque) changing the angular momentum. This is what actually happens! The same (net) force does both jobs at the same time. And each job can be analysed independently of the other.

Or is this just mathematically convenient to derive and there is some underlying simplification?

Well, all the laws of physics are arguably just mathematical conveniences to describe how reality works. But it's an accurate reflection of reality - not a trick.

Of course this happens without violating conservation of energy (or of anything else).

I'm no expert in rigid body dynamics, but maybe someone who is can chip in an explain further. Or consider posting a new thread about this. Once you are comfortable with the idea, it can simplify many problems.

 

[domephilis]

Not sure where you mean - can't see anything wrong. But am happy to make a correction if there's a mistake.


The friction is simultaneously doing two jobs - changing the linear momentum and (because it provides a torque) changing the angular momentum. This is what actually happens! The same (net) force does both jobs at the same time. And each job can be analysed independently of the other.


Well, all the laws of physics are arguably just mathematical conveniences to describe how reality works. But it's an accurate reflection of reality - not a trick.

Of course this happens without violating conservation of energy (or of anything else).

I'm no expert in rigid body dynamics, but maybe someone who is can chip in an explain further. Or consider posting a new thread about this. Once you are comfortable with the idea, it can simplify many problems.

I think you meant to write v_0 squared but forgot the ^ in the latex. Also, the point I'm trying to make is that, I agree that it's doing two jobs, but I don't get how it gives each job its full force. Let's say f is the frictional force. In calculating the rotation we used fr to compute the torque. If that's true, that should have used up all of the force, and we won't have any leftover to give to the linear acceleration. In your calculation, it seems like at each instant we have fr torque and f linear and thus the effect of 2f. My question is where the other f came from.

 

[Steve4Physics]

I think you meant to write v_0 squared but forgot the ^ in the latex.

Many thanks. Eyes (and brain) going downhill. Have edited.

Also, the point I'm trying to make is that, I agree that it's doing two jobs, but I don't get how it gives each job its full force. Let's say f is the frictional force. In calculating the rotation we used fr to compute the torque. If that's true, that should have used up all of the force, and we won't have any leftover to give to the linear acceleration. In your calculation, it seems like at each instant we have fr torque and f linear and thus the effect of 2f. My question is where the other f came from.

I understand your difficulty. Have you ever played snooker/billiards/pool? The cue ball moves in the direction of the cue’s motion – even if you hit the ball off-centre, giving it spin.

The force from the cue (for an off-centre impact) delivers both linear momentum and angular momentum simultaneously.

Next paragraph struck-through as I'm not convinced it's correct.
To conserve energy, the distance moved by the cue (hence the work done by it) while in contact with the ball will be slightly larger when making the off-centre impact compared to a centre impact.

I found an old thread which is closely related and you may find it a useful read,
https://www.physicsforums.com/threa...r-acceleration-of-free-body-not-fixed.935617/

 

[domephilis]

Many thanks. Eyes (and brain) going downhill. Have edited.


I understand your difficulty. Have you ever played snooker/billiards/pool? The cue ball moves in the direction of the cue’s motion – even if you hit the ball off-centre, giving it spin.

The force from the cue (for an off-centre impact) delivers both linear momentum and angular momentum simultaneously.

Next paragraph struck-through as I'm not convinced it's correct.
To conserve energy, the distance moved by the cue (hence the work done by it) while in contact with the ball will be slightly larger when making the off-centre impact compared to a centre impact.

I found an old thread which is closely related and you may find it a useful read,
https://www.physicsforums.com/threa...r-acceleration-of-free-body-not-fixed.935617/

I haven't played pool in a while. Last tried it at some clubhouse almost a decade ago. Regardless, I do get what you mean. I guess the question is this: "Are $F=ma$ and $FR=I\alpha$ independently true for each force considered? If so, is there a corresponding proof for energy conservation?". I think "Factchecker" in the thread that you gave agrees with this claim. But I was wondering about a proof.

We know that kinetic energy can be divided as follows: $$K = K_{rot} + K_{CM}$$ To simplify things, let's say that $K_1 = 0$. Hence by energy conservation, we have $$K_{rot} + K_{CM} = \int_{x_1}^{x_2}f(x)dx$$ Since $F=ma$ (suppose that the force is constant), $$K_{rot} + K_{CM} = (ma)d$$. Hence, changes in rotational and linear kinetic energy should add up to the change in only the "linear" potential (ma)d. So then, the only way our claim can be true is if there's something funny with d (or if rotational KE is 0).

That said I think we did find the solution for the problem. Do you know how to close a thread? Thank you.

 

[Steve4Physics]

Hence, changes in rotational and linear kinetic energy should add up to the change in only the "linear" potential (ma)d.

Probably best to avoid the use of the word 'potential' as it has specific meanings in physics. In fact '(ma)d' equals the work done by the net force (if I understand you correctly).

So then, the only way our claim can be true is if there's something funny with d (or if rotational KE is 0).

If a fixed force is applied for a fixed time (i.e. a given impulse) then I think 'd ' must increase - so that there is additional work done by the force to provide the additional rotational kinetic energy. This would happen 'naturally' because of the motion of the point of application and the rotation.

That said I think we did find the solution for the problem. Do you know how to close a thread? Thank you.

I think the correct way is to ask a Mentor to do this. If you click 'Report' on the bottom left of a message you can make this request.

 

[domephilis]

Probably best to avoid the use of the word 'potential' as it has specific meanings in physics. In fact '(ma)d' equals the work done by the net force (if I understand you correctly).


If a fixed force is applied for a fixed time (i.e. a given impulse) then I think 'd ' must increase - so that there is additional work done by the force to provide the additional rotational kinetic energy. This would happen 'naturally' because of the motion of the point of application and the rotation.


I think the correct way is to ask a Mentor to do this. If you click 'Report' on the bottom left of a message you can make this request.

Makes sense. Thank you. I will try the 'Report' button.

 

[berkeman]

That said I think we did find the solution for the problem. Do you know how to close a thread? Thank you.

No need to close the thread. It can remain open for the time being.

 

[domephilis]

No need to close the thread. It can remain open for the time being.

OK. Thanks.