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What is the relation between the accelerations of the ring and the disc (see image)?

Both the ring and the disc have mass $M$. The ring has a radius $R$ and disc has radius $2R$. They are connected by a light inextensible string. A force $F$ acts on topmost point of the disc.

The question actually asks the minimum value of the coefficient of friction for rolling without slipping to be possible. I have formulated the other equations, using torque or forces.

But, I need the relation between the acclerations of the ring and disc to solve them (I have 7 variables in my current set of equations, but only 6 equations). I have no idea how to relate the acclerations.

Also, I looked at the hint in my book, and it says $a_{disc} = 2 a_{ring}$. There is no explanation about how they arrived at this result.

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    $\begingroup$ Hi Pratyush and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Apr 29 '16 at 7:48
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    $\begingroup$ @John I believe I have adhered to the guidelines in the meta post. If not, point me in the right direction. $\endgroup$ – Pratyush Yadav Apr 29 '16 at 7:53
  • $\begingroup$ Whether homework or not, Pratyush is asking for help with thinking through the motion of the discs. He is not asking for the homework answer, which is the minimum coefficient of friction. I agree with Pratyush that this is within the guidelines. $\endgroup$ – sammy gerbil Apr 29 '16 at 9:04
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At the instantaneous moment shown in the diagram, we can write: $$2R\alpha_{ring}=a_{disc}$$ as both are in pure rolling. This also tells us that the point on the ring where the thread is attached has an acceleration $=2R\alpha_{ring}=2a_{ring}$ so we find that: $$a_{disc}=2a_{ring}$$ Note that when the string moves to another position this will not be true, but it might be enough for you to solve the question for now!

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    $\begingroup$ Thanks for your answer. I was already halfway through writing my own answer, as I figured the problem on my own, when you answered. I'll let mine stay anyway. Also, I understand what you are trying to say, but being a little more clearer might help someone else. Thanks. $\endgroup$ – Pratyush Yadav Apr 29 '16 at 9:39
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    $\begingroup$ But do note the note in my answer! :) $\endgroup$ – FreezingFire Apr 29 '16 at 9:45
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Okay, so I figured it out myself. Here's what I think:

Take $P$ to be the point on top of the ring, where the string is attached. Now, two things contribute to $P$'s acceleration : the acceleration of the centre of mass of the ring, and the acceleration due to the angular motion.

So, $a_P = a_{ring} + \alpha \times R$, where $\alpha$ is the angular acceleration of the ring.

But, the acceleration of $P$ must be equal to the acceleration of the string, since they are connected. The acceleration of the string is the acceleration of the centre of mass of the disc.

So, $$\begin{align} a_{disc} &= a_P \\ a_{disc} &= a_{ring} + \alpha \times R \\ &= a_{ring} + a_{ring} && \text{(Because the ring is in pure rolling, $a = R \alpha$)} \\ a_{disc} &= 2 a_{ring} \\ \end{align}$$

This conforms with the hint in my book. Correct me if I'm wrong.

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  • $\begingroup$ With this sort of problem you will have to find a relationship between the accelerations (or velocities or displacements) of the bodies based on the geometry of the problem. This is what you have found out above and that is the basis of the hint given in the question. $\endgroup$ – Farcher Apr 29 '16 at 14:25
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As the ring moves forward, the string unwinds from it. When the ring has completed one revolution, every point on it has moved forward by the distance of its circumference. The string has unwound by an amount equal to the ring's circumference. So while the centre of the ring has moved forward by one ring circumference, the end of the string (where the disc is attached at its centre) has moved forward by the same distance plus the amount of string unwound, ie a total of 2 ring circumferences. So distances - and therefore also velocities and accelerations - of ring and disc are in the ratio of 1:2.

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