Ball being hit starts to slide and roll

[ehild]

$$x=v_0t-\frac{1}{2}at^2=v_0 \frac{v_0 k}{(1+k)g\mu}-\frac{1}{2}g\mu \left( \frac{v_0 k}{(1+k)g\mu} \right)^2=\frac{v_0^2 k}{(1+k)g\mu} \left( 1-\frac{k}{2(1+k)} \right)$$

You have done it, at last!
Now the third question comes. How much did the KE change during sliding?

 

[haruspex]

I am surprised that you discourage attempts to assist as 'unnecessary'.

I believe ehild was only discouraging specific aspects of specific attempts.
When trying to help on a thread, it is very tempting to dismiss the OP's valid but inefficient approach and just show a better one. It is more useful, I believe, to first help the student get the right answer using their original method and then point out better methods. Otherwise, they never get to understand where they went wrong.

Steiners rule is the parallel axis theorem and another post (#11) has suggested that it plays no part in this problem.

No, that post pointed out that if the idea was to take moments about the point of contact then friction would not feature in the equation.

All of the equations could be made easier by replacing 'k' with 2/5

That's a strange claim. Three keystrokes instead of one? Or maybe 5: (2/5).

 

[Karol]

Now the third question comes. How much did the KE change during sliding?

The loss in KE is the work done by friction:
$$\Delta KE=Work=F\cdot x=mg\mu\cdot x=mg\mu \frac{v_0^2 k}{(1+k)g\mu} \left( 1-\frac{k}{2(1+k)} \right)=\frac{mv_0^2 k}{(1+k)} \left( 1-\frac{k}{2(1+k)} \right)$$

 

[ehild]

The loss in KE is the work done by friction:
$$\Delta KE=Work=F\cdot x=mg\mu\cdot x=mg\mu \frac{v_0^2 k}{(1+k)g\mu} \left( 1-\frac{k}{2(1+k)} \right)=\frac{mv_0^2 k}{(1+k)} \left( 1-\frac{k}{2(1+k)} \right)$$

It is very tempting to calculate the energy loss this way, using Work-Energy Theorem. But it is not valid when both force and torque act on a rigid body, changing both the translational and rotational velocity. https://en.wikipedia.org/wiki/Work_(physics)#Work.E2.80.93energy_principle

It is safer to calculate directly the change of KE, including both the translational and rotational one.

 

[ehild]

@lychette:
You addressed me several times and criticised my hints and comments to Karol's work (Posts ##16, 19, 21, 29) You were wrong, and your comments might have confused the OP, and naturally, they annoyed me. I had the impression you did not really follow the thread or misunderstood it . Although your hints for the solution were correct, Karol knew and applied the same method, with mistakes, which were corrected during the thread.
When somebody guides the OP, It is not polite trying to push him/her away. If the helper makes mistakes, it is right to note them, but there were no mistakes in my helping.
As @haruspex pointed out, it is better to help the OP to work according to his/her own way, if it is basically a correct one, and correcting the mistakes, instead of dismissing his/her whole attempt. Helping does not mean that the helper shows how much he/she knows.
At the end when the OP arrived at the solution, you can show your method, or can comment the problem.

 

[Karol]

$$\left\{ \begin {array} {l} \frac{1}{2}mv^2+\frac{1}{2}I_c \omega^2=W \\ \omega r=v \end {array} \right\}~~\rightarrow~~\omega^2=\frac{2W}{mr^2(1+k)}=\frac{2v_0^2k}{(k+1)^2r^2}\left[ 1-\frac{k}{2(k+1)} \right]$$
The reduction in speed:
$$\Delta v=\omega r=\frac{v_0}{k+1}\sqrt{2\left( 1-\frac{k}{2(k+1)} \right)}$$
The initial speed minus Δv should, to my opinion, give v at the end of the sliding phase, which is:
$$v=v_0-at=v_0-g\mu\frac{v_0k}{g\mu(k+1)}=v_0\left[ 1-\frac{k}{k+1} \right]$$
They aren't the same. either i made a mistake in the last line or i don't understand the loss of KE, meaning the reduction in velocity of COM.

 

[ehild]

$$\left\{ \begin {array} {l} \frac{1}{2}mv^2+\frac{1}{2}I_c \omega^2=W \\ \omega r=v \end {array} \right\}~~\rightarrow~~\omega^2=\frac{2W}{mr^2(1+k)}=\frac{2v_0^2k}{(k+1)^2r^2}\left[ 1-\frac{k}{2(k+1)} \right]$$
The reduction in speed:
$$\Delta v=\omega r=\frac{v_0}{k+1}\sqrt{2\left( 1-\frac{k}{2(k+1)} \right)}$$
The initial speed minus Δv should, to my opinion, give v at the end of the sliding phase, which is:
$$v=v_0-at=v_0-g\mu\frac{v_0k}{g\mu(k+1)}=v_0\left[ 1-\frac{k}{k+1} \right]$$
They aren't the same. either i made a mistake in the last line or i don't understand the loss of KE, meaning the reduction in velocity of COM.

$\frac{1}{2}mv^2+\frac{1}{2}I_c \omega^2$ is not the work, but the kinetic energy of the rolling ball. You can express it in terms of v, m, and k. And you know the final velocity if you substitute t into the equation v=v₀-μgt.
I do not follow what you tried to do.
You need the change of the KE, KE(final) - KE(initial). Initially, the ball had only translational kinetic energy, sliding with velocity v₀ and having zero angular velocity.
You do not get this change of energy equal to the work of friction. Read the part
Work of forces acting on a rigid body
in the Wikipedia article https://en.wikipedia.org/wiki/Work_(physics)#Work.E2.80.93energy_principle
Part of the work done by friction decreased the translational KE, but increased the rotational one.

 

[Karol]

$$\left\{ \begin {array} {l} \frac{1}{2}m\Delta v^2+\frac{1}{2}I_c \Delta \omega^2=Work \\ \Delta\omega\cdot r=\Delta v \end {array} \right\}~~\rightarrow~~\frac{m\Delta v^2}{2}(k+1)=\frac{mv_0^2 k}{(1+k)} \left( 1-\frac{k}{2(1+k)} \right)$$
$$\Delta v^2=\frak{v_0 k(k+2)}{(k+1)^3}$$

 

[ehild]

$$\left\{ \begin {array} {l} \frac{1}{2}m\Delta v^2+\frac{1}{2}I_c \Delta \omega^2=Work \\ \Delta\omega\cdot r=\Delta v \end {array} \right\}$$

v=ωr is valid only for pure rolling. Δv=v(rolling) -v₀=Δωr-v₀.

$$\Delta v^2=\frak{v_0 k(k+2)}{(k+1)^3}$$

What is the meaning of this line?

Remember, you have to calculate the change of energy which is entirely kinetic. What is the KE of the ball initially? What is its KE when it starts pure rolling?

 

[Karol]

The angular velocity at the moment of pure rolling:
$$\omega=\alpha t=\frac{g\mu}{kr}\frac{v_0k}{g\mu(k+1)}=\frac{v_0}{(k+1)r}$$
The linear velocity (COM) there:
$$v=\omega r=\frac{v_0}{k+1}$$
$$\Delta EK=\frac{1}{2}mv_0^2-\frac{1}{2}mv^2=\frac{mv_0^2}{2}\left( 1-\frac{1}{(k+1)^2} \right)$$
But i made in another way. again, ω is at the point of end of sliding (and pure rolling):
$$\left\{ \begin {array} {l} \frac{1}{2}m(\Delta v)^2+\frac{1}{2}I_c \omega^2=Work \\ \Delta v=\omega\cdot r-v_0 \end {array} \right.$$
$$\rightarrow~~\Delta v=v_0 \left( \frac{-k+\sqrt{k^2-\frac{1}{k+1}}}{2(1+k)} \right)$$
I don't know why those 2 answers aren't equal.

 

[ehild]

The angular velocity at the moment of pure rolling:
$$\omega=\alpha t=\frac{g\mu}{kr}\frac{v_0k}{g\mu(k+1)}=\frac{v_0}{(k+1)r}$$
The linear velocity (COM) there:
$$v=\omega r=\frac{v_0}{k+1}$$
$$\Delta EK=\frac{1}{2}mv_0^2-\frac{1}{2}mv^2=\frac{mv_0^2}{2}\left( 1-\frac{1}{(k+1)^2} \right)$$

You forgot the rotational KE.

But i made in another way. again, ω is at the point of end of sliding (and pure rolling):
$$\left\{ \begin {array} {l} \frac{1}{2}m(\Delta v)^2+\frac{1}{2}I_c \omega^2=Work

It is wrong.
Read the Wikipedia article.

 

[Karol]

$$\Delta EK=\frac{1}{2}mv_0^2-\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2=\frac{kmv_0^2}{2(k+1)}$$
I think $\frac{1}{2}m(\Delta v)^2+\frac{1}{2}I_c \omega^2=Work$ is wrong because the friction work also produces heat, so not all of it enters KE (translational and rotational).

 

[ehild]

$$\Delta EK=\frac{1}{2}mv_0^2-\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2=\frac{kmv_0^2}{2(k+1)}$$

The result is correct, but you forgot a pair of parentheses.
$$\Delta EK=\frac{1}{2}mv_0^2-\left(\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2\right)=\frac{kmv_0^2}{2(k+1)}$$

I am reading the Wikipedia article now, but i would like to explain my thought when i wrote:
$$\left\{ \begin {array} {l} \frac{1}{2}m(\Delta v)^2+\frac{1}{2}I_c \omega^2=Work \\ \Delta v=\omega\cdot r-v_0 \end {array} \right.$$
The $\Delta v$ is the reduction in linear velocity. if there were no friction v₀ would continue. the second term $\frac{1}{2}I_c \omega^2$ is the rotational energy gained by friction. by kinetic energy KE you mean both those energies, right?

Yes, the KE is the sum of the translational and rotational one, but 1/2 m (Δv)² is not the difference of the translational KE-s as a²-b²≠(a-b)².
What do you mean on "Work"?

 

[Karol]

How can the result be correct, indeed i added $\frac{1}{2}I_c\omega^2$, in accordance with my $\Delta EK=\frac{1}{2}mv_0^2-\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2$. it also makes sense since friction makes rotational energy, whilst in your:
$$\Delta EK=\frac{1}{2}mv_0^2-\left(\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2\right)$$
I have to subtract it. but it also makes sense, so which sense is correct?...
I changed the last post:

I think $\frac{1}{2}m(\Delta v)^2+\frac{1}{2}I_c \omega^2=Work$ is wrong because the friction work also produces heat, so not all of it enters KE (translational and rotational).

By Work i meant friction work $F\cdot x=mg\mu\cdot x$

 

[ehild]

How can the result be correct, indeed i added $\frac{1}{2}I_c\omega^2$, in accordance with my $\Delta EK=\frac{1}{2}mv_0^2-\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2$. it also makes sense since friction makes rotational energy, whilst in your:
$$\Delta EK=\frac{1}{2}mv_0^2-\left(\frac{1}{2}mv^2+\frac{1}{2}I_c\omega^2\right)$$
I have to subtract it. but it also makes sense, so which sense is correct?...

When you write Δ(something) it means the change of that something. ΔKE means the difference between the final and initial kinetic energies. ΔKE=KE(final)-KE(initial). KE(initial)=1/2 mv₀². KE(final)=1/2 mv²+1/2 I ω², and v=ωr in the final stage, therefore KE(final) = 1/2 m v²+ 1/2 k m v² =1/2 (k+1) v²
The difference is ΔKE=1/2 (k+1) v²-1/2 mv₀². It is negative, so the ball losses energy. That lost amount of energy is the question.

By Work i meant friction work $F\cdot x=mg\mu\cdot x$

When you speak about work you have to say what does what kind of work on what. The work of the force of friction on a sliding object is negative. mgμx is not the work done by the friction on the ball. The force of friction does translational work on the sliding ball and the torque of friction does rotational work.

 

[rcgldr]

By Work i meant friction work $F\cdot x=mg\mu\cdot x$

Some of the "work" done by the (sliding) friction force is lost into heat while the ball is sliding. The rest is a conversion of linear energy into angular energy with no gain or loss in the total energy (no work done).

 

[Karol]

So the work of friction is (i denote torque as M): $Work=F\cdot x+M\theta$, and only part of the member $F\cdot x$ is turned into KE, since a part is "wasted". but the member $M\theta$ is entirely transformed, right?

 

[ehild]

One

So the work of friction is (i denote torque as M): $Work=F\cdot x+M\theta$, and only part of the member $F\cdot x$ is turned into KE, since a part is "wasted". but the member $M\theta$ is entirely transformed, right?

I do not understand what you try to say.
The total work done on the ball is equal to the change of its KE. One part of the work decreases the KE, the other part increases it.
According to the Wikipedia article, the elementary work is dW=(Fv+Tω)dt, where F is the resultant force and T is its torque. You know F and T, and the time dependence of v and ω. Integrate between t=0 and t_r, the time when pure rolling starts.

 

[J Hann]

You might try using conservation of momentum about the point where slipping stops.
That would quickly give you the final angular velocity.
Knowing that it seems that the problem would be greatly simplified (energy loss, etc.)

 

[rcgldr]

So the work of friction is (i denote torque as M): $Work=F\cdot x+M\theta$, and only part of the member $F\cdot x$ is turned into KE, since a part is "wasted". but the member $M\theta$ is entirely transformed, right?

The equation you have in post #44 is the proper approach, the energy lost is the initial energy minus the final energy. I'm not sure that $F\cdot x$ is very useful in this case.

 

[ehild]

You might try using conservation of momentum about the point where slipping stops.
That would quickly give you the final angular velocity.
Knowing that it seems that the problem would be greatly simplified (energy loss, etc.)

Yes, it is a much simpler method to get the final speed. You can show it, as the problem is already solved (Post #40)

 

[Karol]

$$dW=(Fv+T\omega)dt,~~v=at=g\mu t,~~\omega=\alpha t=\frac{g\mu}{kr}t$$
$$Work=\int dW=...=\frac{mv_0^2k}{2(k+1)}$$
But that's the KE loss i found in post #42, so, no heat generated?

You might try using conservation of momentum about the point where slipping stops.
That would quickly give you the final angular velocity.

Which momentum, angular? it is mvr. now, what does it help me? which condition to apply at that point and what initial information exists? if i don't know the velocity, since i guess i have to find it with the conservation of momentum, then what else do i know?

 

[ehild]

$$dW=(Fv+T\omega)dt,~~v=at=g\mu t,~~\omega=\alpha t=\frac{g\mu}{kr}t$$
$$Work=\int dW=...=\frac{mv_0^2k}{2(k+1)}$$
But that's the KE loss i found in post #42, so, no heat generated?

Your formula for v is wrong. v= v₀-μgt. You miss a negative sign in the formula for W. The work is negative, and equal to the change of KE.
Energy is not lost. The mechanical energy lost is converted to other kind of energy, heat now.

 

[J Hann]

$$dW=(Fv+T\omega)dt,~~v=at=g\mu t,~~\omega=\alpha t=\frac{g\mu}{kr}t$$
$$Work=\int dW=...=\frac{mv_0^2k}{2(k+1)}$$
But that's the KE loss i found in post #42, so, no heat generated?

Which momentum, angular? it is mvr. now, what does it help me? which condition to apply at that point and what initial information exists? if i don't know the velocity, since i guess i have to find it with the conservation of momentum, then what else do i know?

M V R is the initial angular momentum.
Now what is the angular momentum about the point of contact when there is no more slipping?
(Hint: Use the parallel axis theorem)

 

[Karol]

Your formula for v is wrong. v= v0-μgt

$$dW=(Fv+T\omega)dt,~~v=v_0-at=v_0-g\mu t,~~\omega=\alpha t=\frac{g\mu}{kr}t$$
$$Work=\int dW=...=\frac{mv_0^2k(k+3)}{2(k+1)}>\frac{mv_0^2k}{2(k+1)}=\Delta EK$$
This time it makes sense to me since the friction work reduces total KE and also produces heat.

 

[ehild]

$$dW=(Fv+T\omega)dt,~~v=v_0-at=v_0-g\mu t,~~\omega=\alpha t=\frac{g\mu}{kr}t$$
$$Work=\int dW=...=\frac{mv_0^2k(k+3)}{2(k+1)}>\frac{mv_0^2k}{2(k+1)}=\Delta EK$$
This time it makes sense to me since the friction work reduces total KE and also produces heat.

Wrong. What did you take for F?
If the work is positive it increases the kinetic energy. But it is decreased as part of it was converted to heat.

 

[Karol]

$$dW=(Fv+T\omega)dt,~~v=v_0-at=v_0-g\mu t,~~\omega=\alpha t=\frac{g\mu}{kr}t$$
$$F=-mg\mu,~~t=\frac{v_0k}{(k+1)g\mu}$$
$$Work=\int dW=...=-\frac{mv_0^2k}{2(k+1)}=-\Delta EK$$
The work is entirely converted to decrease in KE? and which part is the heat?
I expected $Work>\vert \Delta KE \vert$

 

[ehild]

Work is not energy. The work done by all forces acting on an object is equal to the change of the KE of the object.
Some of the work is that of conservative forces. That work changes the potential energy. Some of the work is due to non-conservative forces.
If the potential energy does not change, but the KE decreases, the lost KE is transformed to internal energy, the KE of the random motion of particles.

 

[Karol]

I thank you very much ehild for your patience, Karol

 

[rcgldr]

Link to one of the prior threads for this problem:

https://www.physicsforums.com/threads/rolling-sliding-with-friction.159337