# Ball being hit starts to slide and roll

1. Aug 7, 2016

### Karol

1. The problem statement, all variables and given/known data
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

2. Relevant 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: $I_c=I_{c.o.m.}+Mr^2$

3. 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:
$$\Delta E=E_i-E_f=\frac{1}{2}mv_0^2-\frac{1}{2}I_A\omega^2=...$$

2. Aug 7, 2016

### ehild

What is the initial angular velocity?

3. Aug 7, 2016

### Karol

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

4. Aug 7, 2016

### ehild

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

5. Aug 7, 2016

### 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

6. Aug 7, 2016

### ehild

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?

7. Aug 7, 2016

### Karol

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

8. Aug 7, 2016

### 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.

Last edited: Aug 7, 2016
9. Aug 7, 2016

### 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

10. Aug 7, 2016

### haruspex

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

11. Aug 7, 2016

### rcgldr

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?

12. Aug 7, 2016

### ehild

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.

13. Aug 7, 2016

### haruspex

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

14. Aug 8, 2016

### ehild

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.

15. Aug 8, 2016

### Karol

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

16. Aug 8, 2016

### lychette

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

17. Aug 8, 2016

### ehild

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.

18. Aug 8, 2016

### 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

19. Aug 8, 2016

### Karol

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

20. Aug 8, 2016

### ehild

Why the square root?