# Find the distance the ball travels before slipping ceases to occur

1. Dec 2, 2004

### koa

A billiard ball of radius "a" is initially spinning about a horizontal axis with angular speed "w" and with zero forward speed. If the coefficient of sliding friction between the ball and the biliard table is "m",
(A)find the distance the ball travels before slipping ceases to occur.
(B) the work lost to friction

2. Dec 2, 2004

### Tide

What exactly you tried so far?

HINT: Slipping will cease when the speed of the ball is the same as the angular velocity (about the contact point) times its radius.

3. Dec 4, 2004

### koa

i dont know how to actually start
i think
first i have to get the energy equation
which is K1+u1+others(the energy due to the friction)= k2+u2
we know that u1=u2=0
k1= 1/2 Iwo^2
k2= 1/2 Iw^2 +1/2 mv2
others= Fr. d(the distance the ball moved)

second i think we can get the torque due to the friction
t=I.alpha=Fr.a(the radius of the ball)

that how i think
i dont know if its right or wrong

4. Dec 4, 2004

### ghlzce

i

i think you are in the right way, good luck

5. Dec 4, 2004

### Tide

The horizontal motion will be

$$v = v_0 - \mu g t$$

and the rotation rate will be

$$\omega = \omega_0 - \frac {5}{2} \frac {\mu g}{r} t$$

Can you see why? Also, slipping ceases when $\omega r = v$ and you should be able to take it from there.

6. Dec 4, 2004

### ghlzce

Tide, do you think if i use the energy equation, i will get the distance?
U1+K1+w=k2=U2 ENERGY

7. Dec 4, 2004

### Tide

It's not obvious to me how you would do that but I suppose if you're really careful about it and are able to determine speed in terms of distance travelled then you might be able to do it.

8. Dec 4, 2004

### koa

hey tide ,
first of all id like to thank u for all ur help

I cant really see why

9. Dec 4, 2004

### Tide

Koa,

The only horizontal force on the ball is the force of friction which is proportional to the normal force between the ball and the table. Therefore, the frictional force is just the weight of the ball times the coefficient of friction.

10. Dec 4, 2004

### koa

check this

N=MG
Fr=uN=uMG
BUT IN THE PROBLEM THEY DIDNT GIVE US THE COIFICCIENT BETWEEN THE BALL AND THE TABLE??

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11. Dec 4, 2004

### Tide

That would be the "m" that you specified in your original post. I called it $\mu$.

12. Dec 5, 2004

### koa

i think i got it

i think i got it
thanx
ill post the solution after i finish it completely
thank u

13. Dec 6, 2004

### koa

hi,
cauld somebody please check my answer im not sure if it`s right or wrong.

Energy Equation: K1+U1+Wo=K2+U2,U1=U2=0
1/2IWo^2-FfD=1/2IWf^2+1/2MVf^2 , M=MASS, m= THE FRICTION COIFFICIENT

I= 5/2MA^2 , V=AW, Ff=mGM , ACCELARATION= A*ALPHA

D= THE DISTANCE THE BALL MOVED, Vf= FINAL VELOCITY, Wf= Vf/A

ACC(ACCELARATION)=A.ALPHA

SO 1/2(5/2) 5/2MA^2Wo^2- mGD= 1/2(5/2)MA^2 Wf^2+1/2MVf^2

5/2A^2Wo^2-2MGD=5/2Vf^2+Vf^2=7/2Vf^2

Vf^2=5/7A^2Wo^2-4/7MGD.....................EQU (1)

TORQUE=Ff.A=I.ALPHA=5/2MA^2(ACC/A)

Ff=5/2M.ACC=mMG

ACC=2/5mG......................EQU2

Vf^2=Vo^2+2 ACC (X-Xo)= 2(2/5mG)D=4/5mGD.......EQU 3

BY SUBTITUTING EQU 3 IN EQU 1

WE HAVE

4/5 mGD= 5/7 A^2Wo^2- 4/7 mGD

4/5mGD+4/7mGD=5/7A^2 Wo^2

48/35mGD=5/7 A^2Wo^2