When does the ball start rolling without slipping?

[brethman86]

3. A solid rubber ball is thrown so that it has an initial horizontal velocity of 12 m/s on a flat horizontal concrete surface, but is not initially rotating. Eventually friction will induce the ball to roll without slipping. Assuming you can neglect rolling friction, a) determine the amount of time after the ball is thrown that it begins to roll without slipping. b) determine the final center-of mass velocity without slipping.


2. Initially the ball sliding, so -Kinetic friction=ma this equals -mg(uk)=ma. According to my teacher "note that it doesn't initially start rolling without slipping, so you don't have the acceleration constraint. You do have,however, a constraint between their linear and angular speeds when it starts rolling without slipping." I took this as Vx=-w(omega) assuming the ball is thrown to the right, it will be rotating clockwise so angular velocity will be negative. When the ball is rolling w/o slipping Vcm=Iw(omega)R

I=2/5Mr^2

mu k=.0.80

I am really struggling and getting frustrated with this question, but this is what I have done so far. Obviously this can't depend on the radius or mass of this ball so every thing I'm trying to to cancel them out.

Initially the ball is undergoing translational motion only so I used N's 2nd law -kinetic frict=ma, so mg(uk)=ma. mass cancels a= -7.84 m/s^2

Now I'm lost and don't know what to do next, but here is my idea (Vx)f or -w= (Vx)i-a(delta t), I have two unknowns so guess I'm wrong. I need some major help!

 

[Doc Al]

Initially the ball is undergoing translational motion only so I used N's 2nd law -kinetic frict=ma, so mg(uk)=ma. mass cancels a= -7.84 m/s^2

The same friction force that produces the translational acceleration also exerts a torque and thus a rotational acceleration. Hint: As the translational speed slows, the rotational speed increases increases.

 

[brethman86]

Ok so here is what I did next:

-kinetic friction equals Tnet.

alpha=Tnet/I, so -mg(uk)R/(2/5MR^2) mass cancels 1 r cancels. -g(0.80)5/(2R) with the accel constraint ax=-alphaR, -g(0.80)5/(2) so alpha=at which equals -19.6 m/s^2.

I used kinematics after this to solve delta t.

I think this is the constraint Vxf=-wf, so (Vxf)=(Vx)i + a(delta t) and wf=wi + alpha(delta t) wi=0 so since Vxf=-wf then -wf=Vxi + a(delta t) .Substitution -(Vxf)- a(delta t)=-alpha(delta t). So solving for delta t I got 0.437s. Is this right?

 

[asleight]

The same friction force that produces the translational acceleration also exerts a torque and thus a rotational acceleration. Hint: As the translational speed slows, the rotational speed increases increases.

[tex]\vec{v}=\vec{r}\vec{\omega}[/tex], are you sure what you wrote is true?

 

[Doc Al]

alpha=Tnet/I, so -mg(uk)R/(2/5MR^2) mass cancels 1 r cancels.

This is good.

-g(0.80)5/(2R) with the accel constraint ax=-alphaR, -g(0.80)5/(2) so alpha=at which equals -19.6 m/s^2.

You lost me here. After a time "t", the rotational speed will satisfy the condition for rolling without slipping, thus: v = αRt (since its initial rotational speed is zero).

Combine this with the kinematics for the translational motion, which starts at speed 12 m/s and ends up with speed v.

 

[Doc Al]

[tex]\vec{v}=\vec{r}\vec{\omega}[/tex], are you sure what you wrote is true?

v = ωr is not generally true as a relationship between translational speed and rotational speed when something both rolls and slips, which is what we have here. Once the condition for rolling without slipping is met, then it will be true.

Consider that the translational speed starts out at its maximum value, 12 m/s, when the ball isn't rotating at all, ω = 0.

 

[brethman86]

For the second part, would Vcm=I(omega)R?

 

[Doc Al]

For the second part, would Vcm=I(omega)R?

Not sure what you mean by "second part" or what I(omega) stands for. The condition for rolling without slipping is Vcm = ωR, if that's what you mean.

(Find Vcm and ω as functions of time. Then set Vcm = ωR and solve for the time, and then Vcm.)