Friction Required for Billiard Ball to Roll without Slipping

[domephilis]

[Mentor Note: Two threads merged below and the OP has fixed up their LaTeX in later posts]

TL;DR Summary: A billiard ball with zero angular velocity and linear velocity v_0 is released on a horizontal surface with coefficient of friction mu_k. It begins to roll without slipping after travelling a distance d. Show that mu_k = \frac{12v_0^2}{49gd}.

Question: A billiard ball with zero angular velocity and linear velocity v_0 is released on a horizontal surface with coefficient of friction mu_k. It begins to roll without slipping after travelling a distance d. Show that $mu_k = \frac{12v_0^2}{49gd}$.

Using energy principles, I was able to get to this point:

$\mu_k = \frac{1}{gd}(\frac{v_0^2}{2}-\frac{7}{10}v^2)$

What I am currently struggling to do is relating the velocity at distance d and the velocity at the start. I have tried using the torque to find the angular velocity at distance d and use $v = r\omega$ (we can do so since we are rolling without slipping at that point). That didn't lead anywhere because I have no idea how long it takes to reach distance d nor do I know what angle theta has been rotated after travelling d. I am really stuck here and would greatly appreciate any hint or help. Thank you.

 

[domephilis]

TL;DR Summary: A billiard ball with zero angular velocity and linear velocity v_0 is released on a horizontal surface with coefficient of friction mu_k. It begins to roll without slipping after travelling a distance d. Show that $\mu_k = \frac{12v_0^2}{49gd}$.

Question: A billiard ball with zero angular velocity and linear velocity v_0 is released on a horizontal surface with coefficient of friction mu_k. It begins to roll without slipping after travelling a distance d. Show that $$\mu_k = \frac{12v_0^2}{49gd}$$

Using energy principles, I was able to get to this point:

$$\mu_k = \frac{1}{gd}(\frac{v_0^2}{2}-\frac{7}{10}v^2)$$

What I am currently struggling to do is relating the velocity at distance d and the velocity at the start. I have tried using the torque to find the angular velocity at distance d and use $v = r\omega$ (we can do so since we are rolling without slipping at that point). That didn't lead anywhere because I have no idea how long it takes to reach distance d nor do I know what angle theta has been rotated after travelling d. I am really stuck here and would greatly appreciate any hint or help. Thank you.

 

[Ibix]

Change each of the $ signs to a pair of $ signs and refresh the page to get LaTeX working.

 

[Ibix]

What do you mean by "using energy principles" in this case? What exactly did you do?

 

[domephilis]

I guess I was being a bit lazy. Here is what I did:

$$K_1+U_1 = K_2+U_2$$
Since at first, we have zero angular velocity, $$K_1 = \frac{1}{2}v_0^2$$
At distance d, since we started to roll without slipping, $$v = r\omega$$
Hence, $$K_2 = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \\ = \frac{1}{2}mv^2 + \frac{1}{2}\frac{2}{5}mr^2\frac{v^2}{r^2}\\ = \frac{7}{10}mv^2$$
Since the only force doing work is the frictional force, the change in potential energy is $-\mu_k mgd$.
Therefore, $$\mu_k = \frac{1}{gd}(\frac{1}{2}v_0^2-\frac{7}{10}v^2)$$

 

[kuruman]

Is mechanical energy conserved when there is friction? In what form is this potential energy $U$ as a function of distance? Why is it not converted back to kinetic energy when the ball comes to rest as is the case when I throw something up in the air and it comes back down?

You need to do some kinematics. Start by finding the acceleration. You can also use a conservation law, but it's not mechanical energy that is conserved.

 

[domephilis]

Friction is conservative so long as it is a function of displacement. It is not conservative if the normal force acting on it changes.

 

[kuruman]

Friction is conservative so long as it is a function of displacement. It is not conservative if the normal force acting on it changes.

Do you have a reference for that? What is the potential energy function of displacement from which the force of friction can be derived by taking the negative gradient?

 

[A.T.]

Friction is conservative so long as it is a function of displacement. It is not conservative if the normal force acting on it changes.

Friction is conservative if there is no slippage. It is dissipative if there is slippage.

 

[domephilis]

Once there is no slippage, static friction applies. Then, the force of friction depends on what is required to maintain 0 velocity at the bottom of the wheel, which isn’t necessarily conservative. On the other hand, kinetic friction is proportional to the normal force on the object. In our case, this is constant. Hence it is conservative, until we reach d where static friction would take over.

 

[domephilis]

Note that the bottom of the wheel has zero velocity when there is no slippage, because the center of mass is moving forward at v and the rotation is bringing the wheel back at a velocity v. I think the last piece of the puzzle here is to find out what v is when the ball is at d.

 

[domephilis]

Do you have a reference for that? What is the potential energy function of displacement from which the force of friction can be derived by taking the negative gradient?

Friction is not a vector here. In this case, friction is a constant force that can be easily integrated to give us the potential energy function U. Here friction acts on a horizontal surface that is flat.

 

[kuruman]

Friction is not a vector here. In this case, friction is a constant force . . .

What is it then, a scalar? You seem to think that a constant force cannot be a vector. Forces always have magnitudes and directions regardless of whether they are constant or not.

. . . that can be easily integrated to give us the potential energy function U. Here friction acts on a horizontal surface that is flat.

If it's that easy can you show me how to do it? Remember that the work done by friction must depend on the end points and be the negative of the change in potential energy between the points.

 

[domephilis]

What is it then, a scalar? You seem to think that a constant force cannot be a vector. Forces always have magnitudes and directions regardless of whether they are constant or not.

If it's that easy can you show me how to do it? Remember that the work done by friction must depend on the end points and be the negative of the change in potential energy between the points.

In the strictest sense, it is a vector. But it is a one-dimensional vector, and therefore it requires less vector calculus to deal with. Since the force is constant, we have $$U = - \int -\mu_k mg dx$$, so that when you differentiate it U is equal to negative the derivative of the force function (by the fundamental theorem of calculus). The key thing here is that we are using kinetic friction not static friction. Since the distance between the points is d, the integral evaluates to what I put up there in the beginning (I think).

 

[haruspex]

TL;DR Summary: A billiard ball with zero angular velocity and linear velocity v_0 is released on a horizontal surface with coefficient of friction mu_k. It begins to roll without slipping after travelling a distance d. Show that $\mu_k = \frac{12v_0^2}{49gd}$.

Question: A billiard ball with zero angular velocity and linear velocity v_0 is released on a horizontal surface with coefficient of friction mu_k. It begins to roll without slipping after travelling a distance d. Show that $$\mu_k = \frac{12v_0^2}{49gd}$$

Using energy principles, I was able to get to this point:

$$\mu_k = \frac{1}{gd}(\frac{v_0^2}{2}-\frac{7}{10}v^2)$$

What I am currently struggling to do is relating the velocity at distance d and the velocity at the start. I have tried using the torque to find the angular velocity at distance d and use $v = r\omega$ (we can do so since we are rolling without slipping at that point). That didn't lead anywhere because I have no idea how long it takes to reach distance d nor do I know what angle theta has been rotated after travelling d. I am really stuck here and would greatly appreciate any hint or help. Thank you.

Any other conservation law you can think of that you might be able to use?

 

[domephilis]

Any other conservation law you can think of that you might be able to use?

Since there is an external torque (i.e. the friction), I don't think conservation of momentum would help that much (I could be wrong). Can you give me some more hints?

 

[domephilis]

I have used one conservation law (conservation of energy) to get at the equation I gave a few posts ago. I'm not sure if you have seen that. I don't know what else I can use at this point.

 

[haruspex]

Since there is an external torque (i.e. the friction), I don't think conservation of momentum would help that much (I could be wrong). Can you give me some more hints?

You mention torque and momentum, so I assume you mean angular momentum, not linear momentum.
The thing about torques and angular momentum is that they are relative to an axis. Choice of axis matters.

 

[domephilis]

Yes, I meant angular momentum. I'm assuming you mean that if I choose the axis at a convenient location the torque would disappear. But I would have to choose an axis that is always at the edge of the billiard ball (which is in contact with the surface). Can I choose a moving axis?

 

[kuruman]

In the strictest sense, it is a vector. But it is a one-dimensional vector, and therefore it requires less vector calculus to deal with. Since the force is constant, we have $$U = - \int -\mu_k mg dx$$, so that when you differentiate it U is equal to negative the derivative of the force function (by the fundamental theorem of calculus). The key thing here is that we are using kinetic friction not static friction. Since the distance between the points is d, the integral evaluates to what I put up there in the beginning (I think).

So then you say that one can write a so-called potential energy function for friction $$U_f(x)=\mu_k mgx.$$ The problem with this is that if you start at point $x_1$, move to $x_2$ and then back to $x_1$ the change in potential energy is $$\Delta U_f=\mu_k mgx_1-\mu_k mgx_1=0$$whereas the work done by friction is $$W_f=-2\mu_k mg|x_2-x_1|.$$Do you see the problem with assuming that friction can be derived from a scalar potential?

 

[haruspex]

Yes, I meant angular momentum. I'm assuming you mean that if I choose the axis at a convenient location the torque would disappear. But I would have to choose an axis that is always at the edge of the billiard ball (which is in contact with the surface). Can I choose a moving axis?

It doesn’t have to be moving. It just has to be always on the line of action of the frictional force.

 

[domephilis]

So then you say that one can write a so-called potential energy function for friction $$U_f(x)=\mu_k mgx.$$ The problem with this is that if you at point $x_1$, move to $x_2$ and then back to $x_1$ the change in potential energy is $$\Delta U_f=\mu_k mgx_1-\mu_k mgx_1=0$$whereas the work done by friction is $$W=-2\mu_k mg|x_2-x_1|.$$Do you see the problem with assuming that friction can be derived from a scalar potential?

The same thing happens with gravity. You throw a rock up, it comes down, the potential energy mgh has not changed if you compare the endpoints. That doesn't mean that nothing happened in between. But gravity is widely held as a conservative force. Is there something I am missing here?

 

[domephilis]

It doesn’t have to be moving. It just has to be always on the line of action of the frictional force.

Let me try that to see if it works. Thanks for the hint.

 

[domephilis]

It doesn’t have to be moving. It just has to be always on the line of action of the frictional force.

I tried doing the question again. It got a bit confusing. If we use the $$L = I\omega$$ definition of angular momentum, then what is the angular momentum at the beginning where there is no angular velocity. My professor noted on the blackboard that even objects moving linearly has an angular momentum, but he used the $$\vec{L} = \vec{r} \times \vec{p}$$ definition which (without extensive use of vector calculus) can only be used with point masses rather than the sphere of a billiard ball we have here. The only other thing I can think of is to consider the billiard ball as a point mass at the CM and use the vector definition, but I am not sure if I can do that. I tried to do it this way, but got another wrong answer.

 

[haruspex]

consider the billiard ball as a point mass at the CM and use the vector definition

That should work. Please post your working.

 

[domephilis]

I have attached a photo of the whiteboard. I can type it in latex in just a moment.

 

[domephilis]

That should work. Please post your working.

In case the photo isn't clear, here is the latex version.

We have from above: $$\mu_k = \frac{\frac{1}{2}v_0^2-\frac{7}{10}v^2}{gd}$$
Hence, $$v^2 = \frac{10}{7}(\frac{1}{2}v_0^2-gd\mu_k)$$
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 $\vec{v} \perp \vec{r}$, $$L_i = Rmv_0$$ $$L_f = Rmv$$
(Note that the angular momentum is consider relative the point of contact between the wheel and the surface.)
Hence $v^2 = v_0^2$
Hence, by the second equation in the post $$v_0^2 = \frac{10}{7}(\frac{1}{2}v_0^2-gd\mu_k)$$
Hence $$\mu_k = -4\frac{v_0^2}{14gd}$$

 

[kuruman]

Hence $v^2 = v_0^2$

Nonsense. This says that the ball does not slow down. You need to rethink your expression for the angular momentum. Remember that $L=I_p\omega$ where $I_p$ is the moment of inertia about the point you use to calculate the angular momentum.

 

[kuruman]

Is there something I am missing here?

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.

 

[domephilis]

Nonsense. This says that the ball does not slow down. You need to rethink your expression for the angular momentum. Remember that $L=I_p\omega$ where $I_p$ is the moment of inertia about the point you use to calculate the angular momentum.

If I use that definition, I need to know what omega is at the beginning. I was explicitly told that there was no angular velocity at the beginning. I have to somehow convert “linear momentum” into angular momentum. The only way I know how to do that is by using the vector definition of angular momentum. But that does not involve moment of inertia, which is why haruspex suggested that I can consider the ball as a point mass at the CM to avoid trying to reinvent the wheel of integrating over the momentum of each infinitesimal constituent of the ball.

 

[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 $v_0^2$.