Friction Required for Billiard Ball to Roll without Slipping

[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.

 

[domephilis]

One last thing I'd like to post on this thread. I have found the following explanation from Doc AI in response to this thread (https://www.physicsforums.com/threads/linear-and-angular-acceleration.184624/):

"The article is correct. The linear acceleration of the center of mass just depends on the net force on the object, not on where the force is applied. The angular acceleration about the center of mass depends on where the force is applied. (Both statements are just consequences of Newton's 2nd law.)

Realize that the work you do on an object is force times the distance that the contact point moves. When you push the object with an off-center force the contact point moves more (compared to an equal on-center force), thus it takes more work to maintain the force--that extra work goes into the rotational energy."

I think this really cleared it up for me intuitively. In my understanding, torque explains the "how" of the motion of an object, but F=ma is true regardless of how the motion occurs internally. The important thing for energy conservation is that the movement of the contact point is not the same as the movement of the CM. Hope this helps.