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A bowling ball of mass $M$ and radius $r_0$ is thrown along a level surface so that initially ($t = 0$) it slides with a linear speed $v_0$ but does not rotate. As it slides, it begins to spin, and eventually rolls without slipping. How long does it take to begin rolling without slipping?

I am confused with the textbook solution. I understand that kinetic friction first acts on the ball. The linear velocity is given by $$ V_{CM} = v_0 - \mu g t$$

The angular acceleration is given by $$I \alpha = \Sigma \tau \implies \alpha = 5 \mu_k g/2r_0 $$

The book states this angular acceleration is constant (presumably from $t=0$). How can one arrive at this conclusion?

I'm confused how this can this be true if the force causing the torque changes from kinetic friction to static friction (when the ball starts to roll without slipping)?

See the full book solution here. Source: Giancoli's Physics for Scientists and Engineers.

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I think that your skepticism comes about because you intuitively think that the force of kinetic friction should change gradually to static friction as the ball speeds up, since the relative motion between the ball's spinning surface and the ground decreases to zero. Your textbook assumes that this transition is actually instantaneous, and that the kinetic friction force is exactly the same until there is no relative motion at all, at which point the friction is entirely static. It seems counter-intuitive, but that's actually how it is.

If you've ever seen someone play a cello or any stringed instrument with a bow, you can see a little better how this works. A cellist steadily moves his bow across the string, which vibrates as it slips and catches on the sticky bow, switching between kinetic and static friction. If friction had a gradual transition, I imagine that the bow would just push the string to a certain displacement and reach equilibrium, and we'd be robbed of a great instrument, and a lot of excellent music.

Does your problem make more sense now? The angular acceleration and the torque are constant, because the kinetic friction force is constant until it disappears. Once the ball spins with the floor, there's no torque on the ball because it's just rolling—to the floor it seems to be still. But as the ball is spinning up (and the slipping is slowing down), the force on it remains the same. Friction is just weird like that.

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at which point the friction is entirely static. Are you saying that after the kinetic friction disappears, there IS static friction? After talking with jkej, I came to agree that after the ball starts to roll there are no frictional forces acting on it. I now understand static friction to be a reactive force, and there is nothing "fighting against" the ball's motion for static friction to react to. –  jp24 Mar 27 '13 at 4:46
    
Right, you could say that. Static friction is there in that it will react to any opposing forces because it's "locked in." Like you say, though, the friction force on the ball is zero because it causes no acceleration. I like to think of the static friction as there, but it's not doing anything, so you've got the right idea. –  krs013 Mar 27 '13 at 5:23

The angular acceleration is constant as long as the ball slips. When the ball stops slipping the angular acceleration drops to zero. I don´t see where the solution states anything else.

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How do you know that the angular acceleration = 0 when it stops slipping? Wouldn't the no slip condition imply $\alpha = a_{CM} / R $ for that portion of the ball's trip? –  jp24 Mar 26 '13 at 23:24
    
@user1850672 $\alpha=a_{CM}/r$ holds all the time, but when the slipping stops $a_{CM}$ becomes zero beacuse $F_{fr}$ becomes zero. –  jkej Mar 26 '13 at 23:45
    
I'm sorry I'm a bit skeptical. $v_{CM} = R \omega$ iff the ball rolls without slipping. Why would $\alpha = a_{CM}/R$ hold all the time? I thought that it only followed from differentiating the first condition w.r.t. to time. Furthermore, after the slipping stops, $F_{fr}$ is still present I believe. It's just that it is static friction, no longer kinetic friction. –  jp24 Mar 27 '13 at 0:11
    
@user1850672 Sorry, the identity that holds all the time that I was thinking of is actually $Mar=I\alpha=2Mr^2\alpha/5$ which gives $\alpha=5a/2r$. But never mind that. Static friction only occurs as a reaction to another force. Since there is no such force here there is no static friction. Similarly there is no static friction on the ball when it lies still on the ground. –  jkej Mar 27 '13 at 0:44
    
Both the book and my instructor have gone to great extent to emphasize: Static friction must be present to make round objects roll. mit.edu/~8.01/gen98/zachar.html –  jp24 Mar 27 '13 at 0:55

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