Find the distance the ball travels before slipping ceases to occur

[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

 

[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.

 

[koa]

i don`t 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 don`t know if it`s right or wrong

 

[ghlzce]

i

i think you are in the right way, good luck

 

[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.

 

[ghlzce]

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

 

[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 traveled then you might be able to do it.

 

[koa]

hey tide ,
first of all i`d like to thank u for all ur help

The horizontal motion will be

v = v_0 - \mu g t
how u got the acceleration?
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.

I can`t really see why

 

[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.

 

[koa]

check this

N=MG
Fr=uN=uMG
TORQUE=Fr.a(the radius)=I(ALPHA)
BUT IN THE PROBLEM THEY DIDN`T GIVE US THE COIFICCIENT BETWEEN THE BALL AND THE TABLE??

 

[Tide]

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

 

[koa]

i think i got it

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

 

[koa]

cauld somebody please check my answer

hi,
cauld somebody please check my answer i`m 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

D=(25A^2Wo^2)/(48mG)... THE ANSWER

IS IT THE ANSWER?