Ball being hit starts to slide and roll

[Karol]

Homework Statement

[/B] A billiard ball of mass m and radius r is being hit and starts to slide
on a surface with equivalent of friction μ.
What is the friction force
The distance to rolling only
What is the energy loss in the sliding phase

Homework Equations

Kinetic energy of a solid body: $E_k=\frac{1}{2}I\omega^2$
Torque and angular acceleration: $M=I\alpha$
Angular velocity as a function of angular acceleration and angle: $\omega^2=\omega_0^2+2\alpha\theta$
Angle as a function of angular acceleration and time: $\theta=\frac{1}{2}\alpha t^2$
Shteiner's theorem: $I_c=I_{c.o.m.}+Mr^2$

The Attempt at a Solution

$$M=I_c\alpha~~\rightarrow~~\alpha=\frac{M}{I_c}=...=\frac{g\mu}{kr}$$
$$\left\{ \begin {array} {l} \omega ^2=\omega_0^2-2\alpha \theta \\ \alpha =\frac {f\cdot r}{I_c} \\ \theta=\frac {1}{2}\alpha t^2 \end {array} \right. $$
$$\omega^2=\omega_0^2-\left( \frac{g\mu}{kr} \right)^2t^2$$
Rolling without sliding: $\omega r=v$:
$$\left\{ \begin {array} {l} \omega^2=\omega_0^2-\left( \frac{g\mu}{kr} \right)^2t^2 \\ \omega r=v \end {array} \right. $$
$$\rightarrow \frac{v^2}{r^2}=\omega_0^2-\left( \frac{g\mu}{kr} \right)^2t^2$$
And $v=at=g\mu\cdot t$:
$$\left\{ \begin {array} {l} \frac{v^2}{r^2}=\omega_0^2-\left( \frac{g\mu}{kr} \right)^2t^2 \\ v=g\mu\cdot t \end {array} \right. $$
$$\rightarrow ~...t^2=\frac {(v_0k)^2}{(g\mu) ^2(1+k^2)}$$
Distance:
$$s=\theta r=\frac {v_0^2k}{2g\mu (1+k^2)}$$
Shteiner's:
$$I_A=I_c+mr^2=(1+k)mr^2$$
Wasted energy:
$$\Delta E=E_i-E_f=\frac{1}{2}mv_0^2-\frac{1}{2}I_A\omega^2=...$$

 

[ehild]

What is the initial angular velocity?

 

[Karol]

The initial angular velocity is zero, the ball starts from stand.

 

[ehild]

The initial angular velocity is zero, the ball starts from stand.

No, the ball has an initial linear velocity vo after the hit. So your formula stating that v=μgt is not true.

 

[Karol]

$$v=v_0-at=v_0-g\mu\cdot t,~~v_0=\omega_0 r$$
$$\left\{ \begin {array} {l} \frac{v^2}{r^2}=\omega_0^2-\left( \frac{g\mu}{kr} \right)^2t^2 \\ v=v_0-g\mu\cdot t \end {array} \right.$$
$$\rightarrow t_{12}=kv_0\frac{k\pm \sqrt{3k^2+2}}{g\mu\sqrt{1+k^2}}$$
Not nice

 

[ehild]

$$v=v_0-at=v_0-g\mu\cdot t,~~v_0=\omega_0 r$$

No, ω₀=0. The ball skids initially. The torque of friction makes it rotate later.
The initial angular speed is zero. How does it change with time?

 

[Karol]

$$\left\{ \begin {array} {l} \omega ^2=2\alpha \theta \\ \alpha =\frac {f\cdot r}{I_c} \\ \theta=\frac {1}{2}\alpha t^2 \end {array} \right\}~~\rightarrow~~\omega^2=\left( \frac{g\mu}{kr} \right)^2t^2$$
$$\left\{ \begin {array} {l} \omega^2=\left( \frac{g\mu}{kr} \right)^2t^2 \\ \omega r=v \end {array} \right\}~~\rightarrow~~\frac{v^2}{r^2}=\left( \frac{g\mu}{kr} \right)^2t^2$$
$$\left\{ \begin {array} {l} \frac{v^2}{r^2}=\left( \frac{g\mu}{kr} \right)^2t^2 \\ v=v_0-g\mu\cdot t \end {array} \right\}~~\rightarrow~~t=\frac{k^2(v_0^2-2g\mu)}{g^2\mu^2}$$
$$\theta=\frac{1}{2}\alpha t^2=...=\frac{k^3(v_0^2-2g\mu)}{2rg^3\mu}$$

 

[ehild]

It is wrong. Check the dimensions. Why do you work with angular displacement instead of the linear displacement of the CM?
Have you copied the problem text correctly? What does it mean "the distance to rolling only"? It is not a question. But it certainly refers to the linear displacement until the ball starts pure rolling motion.
What do you mean by t? $t=\frac{k^2(v_0^2-2g\mu)}{g^2\mu^2}$ has no sense.

 

[Karol]

What is the distance the ball travels until it only rolls, without sliding.
I find the time till it happens:
$$\left\{ \begin {array} {l} \omega^2=\left( \frac{g\mu}{kr} \right)^2t^2 \\ \omega r=v_0-g\mu t \end {array} \right\}~~\rightarrow~~t=k\left[ \frac{k\pm\sqrt{k^2-v_0^2(k^2-1)}}{(k^2-1)g\mu} \right]$$
Not nice

 

[haruspex]

What is the distance the ball travels until it only rolls, without sliding.
I find the time till it happens:
$$\left\{ \begin {array} {l} \omega^2=\left( \frac{g\mu}{kr} \right)^2t^2 \\ \omega r=v_0-g\mu t \end {array} \right\}~~\rightarrow~~t=k\left[ \frac{k\pm\sqrt{k^2-v_0^2(k^2-1)}}{(k^2-1)g\mu} \right]$$
Not nice

I do not understand how you get the right hand side from the two equations on the left.
Start by simplifying $\omega^2=\left( \frac{g\mu}{kr} \right)^2t^2$.

 

[rcgldr]

Shteiner's theorem: $I_c=I_{c.o.m.}+Mr^2$

I'm not sure how you would use this. If using the point of contact as the axis of rotation, then the only external force, the friction force is at the point of contact so no external torque is generated by the friction force.

A posted by ehild, how does angular a speed change with time. Can you think of a simpler formula that relates $\omega$ and $\alpha$ other than the one using $\omega^2 ...$ formula?

 

[ehild]

What is the distance the ball travels until it only rolls, without sliding.
I find the time till it happens:
$$\left\{ \begin {array} {l} \omega^2=\left( \frac{g\mu}{kr} \right)^2t^2 \\ \omega r=v_0-g\mu t \end {array} \right\}~~\rightarrow~~t=k\left[ \frac{k\pm\sqrt{k^2-v_0^2(k^2-1)}}{(k^2-1)g\mu} \right]$$
Not nice

I is totally wrong. Do you know what is the dimension of k ? You subtract quantities with different dimensions under the square root.
Why do you use the squared equation $\omega^2=\left( \frac{g\mu}{kr} \right)^2t^2$ instead of $\omega=\frac{v}{r}= \frac{g\mu}{kr} t$? And you have the other equation for v : $v=v_0-μgt$. It is easy to eliminate v and get t.

 

[haruspex]

Do you know what is the dimension of k ?

Karol appears to have defined k by I_c = mkr², making it dimensionless.

 

[ehild]

Karol appears to have defined k by I_c = mkr², making it dimensionless.

Of course, I knew, as other things I usually ask from an OP. The question was meant to Karol. It was a hint to check the dimensions. As you see, he subtracted velocity-squared from a dimensionless quantity.

 

[Karol]

$$\left\{ \begin {array} {l} \omega=\alpha t=\left( \frac{g\mu}{kr} \right)t \\ \omega r=v_0-g\mu t \end {array} \right\}~~\rightarrow~~t=\frac{v_0k}{(1+k)g\mu}$$
$$x=\frac{1}{2}at^2=\frac{1}{2}g\mu \left( \frac{v_0k}{(1+k)g\mu} \right)^2=\frac{v_0^2k^2}{2g\mu(1+k)^2}$$

 

[lychette]

No, the ball has an initial linear velocity vo after the hit. So your formula stating that v=μgt is not true.

Karol's statement is true...the initial angular velocity is zero

 

[ehild]

$$\left\{ \begin {array} {l} \omega=\alpha t=\left( \frac{g\mu}{kr} \right)t \\ \omega r=v_0-g\mu t \end {array} \right\}~~\rightarrow~~t=\frac{v_0k}{(1+k)g\mu}$$
$$x=\frac{1}{2}at^2=\frac{1}{2}g\mu \left( \frac{v_0k}{(1+k)g\mu} \right)^2=\frac{v_0^2k^2}{2g\mu(1+k)^2}$$

The time is correct at last.
How do you define x? what is it?
You wrote v=vo-μgt, for the velocity of the center of mass. Does the CM accelerate or decelerate? What is the acceleration? Positive or negative? What is the initial velocity of the CM? How do you calculate displacement in case of uniformly accelerating motion?
You should ask these questions from yourself before you substitute expressions into random equations.

 

[lychette]

when the ball is first struck centrally it has only linear velocity and, just like a block of wood, it will slide and experience a friction force. Friction will cause a linear decceleration so that the velocity of the ball across the surface will decrease. straight forward calculation a = F/m
The friction force has another effect..the moment of the friction force about the centre of the ball will cause an angular acceleration (clockwise if the ball is moving to the right)...again straight forward ang acc α = T/I , T = torque = Fr and I = moment of inertia about the centre of the ball = 2Mr²/5
After a time,t, the linear velocity will have decreased and the angular velocity will have increased.
The ball will slide along the surface until the linear velocity has decreased and the angular velocity has increased so that rolling occurs.
The ball will roll when v = ωr
This should help you to find the time,t, taken for rolling to occur.
To find the distance use s = ut + ½at²
I think it is not allowed to give any help in homework other than to point you in the right direction.
You seem to be tied up in equations without recognising the physics of the situation...hope this helps

 

[Karol]

How do you define x? what is it?

X is the distance the ball travels till it only rotates and doesn't skid.
$$x=v_0t-\frac{1}{2}at^2=v_0\sqrt {\frac{v_0k}{(1+k)g\mu}}-\frac{1}{2}g\mu \left( \frac{v_0k}{(1+k)g\mu} \right)^2=v_0\sqrt {\frac{v_0k}{(1+k)g\mu}}-\frac{v_0^2k^2}{2g\mu(1+k)^2}$$

 

[ehild]

X is the distance the ball travels till it only rotates and doesn't skid.
$$x=v_0t-\frac{1}{2}at^2=v_0\sqrt {\frac{v_0k}{(1+k)g\mu}}-\frac{1}{2}g\mu \left( \frac{v_0k}{(1+k)g\mu} \right)^2=v_0\sqrt {\frac{v_0k}{(1+k)g\mu}}-\frac{v_0^2k^2}{2g\mu(1+k)^2}$$

Why the square root?

 

[lychette]

Why the square root?

this is going nowhere fast ..

 

[Karol]

$$x=v_0t-\frac{1}{2}at^2=v_0 \frac{v_0k}{(1+k)g\mu}-\frac{1}{2}g\mu \left( \frac{v_0k}{(1+k)g\mu} \right)^2=\frac{v_0^2k}{(1+k)g\mu} \left( 1-\frac{v_0k^2}{2g\mu(1+k)^2}\right)$$

 

[haruspex]

$$x=v_0t-\frac{1}{2}at^2=v_0 \frac{v_0k}{(1+k)g\mu}-\frac{1}{2}g\mu \left( \frac{v_0k}{(1+k)g\mu} \right)^2=\frac{v_0^2k}{(1+k)g\mu} \left( 1-\frac{v_0k^2}{2g\mu(1+k)^2}\right)$$

Some errors in the final factorisation. The last term should be smpler.

 

[ehild]

$$x=v_0t-\frac{1}{2}at^2=v_0 \frac{v_0k}{(1+k)g\mu}-\frac{1}{2}g\mu \left( \frac{v_0k}{(1+k)g\mu} \right)^2=\frac{v_0^2k}{(1+k)g\mu} \left( 1-\frac{v_0k^2}{2g\mu(1+k)^2}\right)$$

Wrong again, and you could have seen at once if you checked the dimensions inside the parentheses. You subtract ([L]/[T])/([L]/[T]²) from 1. Does not bother you showing your silly mistakes to the whole world?
Write out the squared term first and simplify with gμ. Pull out the identical terms. Apply the distribution law a-ab= a(1-b). Check the factorization by looking at the dimensions.

 

[lychette]

The linear decelleration = F/M = μMg/M = μg

Angular acceleration α = T/I = Fr/I = μMgr/I I = 2/5Mr² about centre of ball so α =5/2μg/r

after t secs linear velocity has decreased to v = V - μgt
after t secs angular velocity has increased to ω = αt = 5μgt/2r
eventually, when v has decreased and ω has increased so that only rolling occurs V - μgt = (5μgt/2r)r
this gives t = 2V/7μg
once this time is known it should be straight forward to calculate distance travelled.
I do not know what 'k' is in the analysis but I hope that this time will be revealed once you have sorted out your equations

 

[ehild]

@lychette
All your hints are unnecessary, as the OP knew how to solve such problems before you arrived. He only made numerous mistakes. Just try to read the thread. He wrote the distance traveled in his last post, with some silly mistakes in factorizing. No need to press him to start again from the beginning. And you must not solve the problem instead of him.
After he arrived to the correct solution, you can summarize the process in a nice and logical way.
By the way, k is the factor relating mr² to the moment of inertia I, I=kmr². It is a ball, but it is not sure, a homogeneous one.

 

[Karol]

$$x=v_0t-\frac{1}{2}at^2=v_0 \frac{v_0 k}{(1+k)g\mu}-\frac{1}{2}g\mu \left( \frac{v_0 k}{(1+k)g\mu} \right)^2=\frac{v_0^2 k}{(1+k)g\mu} \left( 1-\frac{k}{2g\mu(1+k)^2} \right)$$
But the dimensions in the brackets now are [T²]/[L], whilst they should have no dimensions at all since i subtract them from one and since the term $\frac{v_0^2k}{(1+k)g\mu}$ that precedes the brackets has dimension of length, and there should be no interference.

 

[ehild]

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

Wrong again.

 

[lychette]

@lychette
All your hints are unnecessary, as the OP knew how to solve such problems before you arrived. He only made numerous mistakes. Just try to read the thread. He wrote the distance traveled in his last post, with some silly mistakes in factorizing. No need to press him to start again from the beginning. And you must not solve the problem instead of him.
After he arrived to the correct solution, you can summarize the process in a nice and logical way.
By the way, k is the factor relating mr² to the moment of inertia I, I=kmr². It is a ball, but it is not sure, a homogeneous one.

I am surprised that you discourage attempts to assist as 'unnecessary'. Perhaps if more people had offered advice the problem would be solved by now.(it is not difficult!) If the OP decided to ignore my contribution I am OK with that.
Steiners rule is the parallel axis theorem and another post (#11) has suggested that it plays no part in this problem.
The original post indicates that the ball is a billiard ball and this is a homogeneous ball. All of the equations could be made easier by replacing 'k' with 2/5

In post #6 you write :
No, ω0=0. The ball skids initially. The torque of friction makes it rotate later.
The initial angular speed is zero.
At t = 0 the ball has no angular velocity but immediatley friction applies a torque and the ball's angular velocity increases. The ball is skidding i.e rotating against the surface as well as sliding over the surface. The OP seems to want to use the angular distance moved by the ball and this will not give the linear distance because of the skidding. This is visible in videos of snooker balls being struck.
The same effect can be seen when steam locomotives start to move. The wheels sometimes spin and slip as the engine accelerates. In this case a large torque is provided by the engine rather than friction with the rails. Wheel spin in cars is the same sort of thing.
I will make no further posts to this thread.

 

[Karol]

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

 

[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.