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This is the question-

Consider a cylinder of mass $M$ resting on a rough horizontal rug that is pulled out from under it with acceleration $a$ perpendicular to the axis of the cylinder. What is $F$ (friction) at a point P? It is assumed that the cylinder does not slip.

Diagram

Options-

A) $F = M g$

B) $F = M a$

C) $F = \frac{M}{2} a $

D) $F = \frac{M}{3} a $

My attempt-

Since rolling is nothing but rotation about the point of contact (P in this case), according to me P should be at relative rest, therefore $F=Ma$. This should have been a short and crisp question, help me out please answer given is $F=\frac{M}{3} a$, and if you can perhaps please point me out to articles for such rolling friction question, I get stumbled whenever they ask friction in rolling. Thanks in advance..

Edit- This idea also came to my mind- Moment of inertia for a solid cylinder= I = $\frac{MR^2}{2}$

Let the frictional force be F, then,

$F*R=(torque)=I*\alpha$

$F*R=\frac{(M*R^2)*(a)}{2R}$ (as the cylinder is in pure rolling)

therefore- $F=\frac{Ma}{2}$

which is again incorrect any pointers to where I went wrong here again would be appreciated.. thanks

Kaustubh
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  • Welcome to Physics.SE! Please refer to the help section for tips on how to post a good homework-like question. – Jim May 23 '14 at 13:49
  • @Jim saw that, and tried to follow it perhaps you could guide as to what I lacked... – Kaustubh May 23 '14 at 13:50
  • Well, asking for general references on rolling friction is definitely a good addition and will probably make people be more inclined to help. However, the general question needs to show some effort on your part (which it does) AND be about a physics concept. Some idea you need help understanding. Questions asking us to solve a homework problem or do some math question will most likely be closed. Additionally, so of the less enthusiastic members tend to look at a question and if it looks at first glance like you've just copy and pasted a question from a textbook, they'll close it – Jim May 23 '14 at 13:55
  • Keep in mind when you're asking something that the people you are expecting to answer questions are volunteers. We take time out of our day to answer questions and try to help people understand a bit more about the ideas of physics. So if you show a homework question and say "I got this but I don't understand what this aspect of it means", we're happy to help. But we simply cannot devote time to questions that are "I got this but it's wrong, do it the right way and explain that to me". – Jim May 23 '14 at 13:59
  • @Jim I think it is about a physics concept, though it is a homework question (as tagged) I have explained my attempt and since it is faulty in the end I presume I have made an error, so actually I don't mean to ask to solve the question but please atleast point out where the error may be in my attempt,... and I have copied it, couldn't avoid that. thanks... – Kaustubh May 23 '14 at 14:00
  • actually anybody rather than perhaps giving the answer, just giving links which might clear my doubts would be even well appreciated..... – Kaustubh May 23 '14 at 14:05
  • Please consider math formatting for better readability. I have edited the question with $...$ where needed. – John Alexiou May 23 '14 at 15:46
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    There are some useful notes on rolling here: http://www.physics.ohio-state.edu/~gan/teaching/spring99/C12.pdf which has an example of a ball rolling down a slope which is closely related to your problem. – nivag May 23 '14 at 15:52

1 Answers1

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The job of the friction is to enforce the no slip constraint. So let us find the relative acceleration of the two parts and find the force $F$ which makes it zero.

The cylinder EOM are (positive is to the left) $ F = m \dot{v} $ and $r F = I \dot{\omega}$ and the tangential cylinder acceleration at P is $\dot{v}_P = \dot{v} + r \dot{\omega}$

To make the tangential acceleration equal to the rug you have

$$ a = \dot{v}_P = \frac{F}{m} + \frac{r^2 F}{I} \\ F = \left(\frac{1}{m} + \frac{r^2}{I}\right)^{-1} a $$

where the part in the parenthesis is the effective mass of a rolling cylinder on its surface. For a solid cylinder the mass moment of inertia is $I=\frac{m}{2} r^2$ and thus

$$ F = \left(\frac{1}{m} + \frac{2}{m}\right)^{-1} a = \frac{m}{3} a$$

John Alexiou
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