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

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