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I try to calculate the radius of the rolling coin. The task was found in physics book https://www.amazon.co.uk/General-Course-Physics-1-Mechanics/dp/5922102257 (only in Russian). But unfortunately my solution leads to the answer that differ from the answer provided in the book (by coefficient 3). I am trying to figure out if I made a mistake or there is a typo in the tutorial.

Disk with radius $r$ rolls skidless at an angular velocity $w$ in an inclined position towards the ground. Thus, it makes a circle on the ground of radius $R$. The angle between the horizontal plane of the disc and the ground is $\alpha$. In addition $R>>r$. Find $R$.

My solution: enter image description here variation of angular momentum: $$\vec M=\dot {\vec L}$$ where $M$ - moment of force and $L$ - angular momentum of the disk.

There are three forсes affect the disk: (1) the force of gravity = mg downward, (2) earth pressure force that = - mg upwards and (3) force of static friction towards the center of the circle R. The first two forces create angular momentum $\vec M=[\vec {O_1O_2},\vec gravity]+[\vec {O_1O_3},\vec{earth pressure}]=[\vec{O_1O_2},\vec{mg}]−[\vec{O_1O_3},\vec{mg}]=-[\vec r,\vec{mg}]=mgr\,cos(\alpha)$

$$mgr\,cos(\alpha)=[\vec \Omega \vec L]$$ where $m$ is a weight of the disc and $\Omega$ - angular velocity related to the movement of the disk around the circle. $$\Omega R=wr $$ $$\Omega =wr/R $$ $$\vec L=I_\parallel \vec{w_\parallel}+I_\bot \vec{w_\bot}$$ where $I_\parallel$ is the moment of inertia passing through the center of the disk along the axis of symmetry of the disk, $I_\bot$ is the moment of inertia passing perpendicular to the axis of symmetry of the disk. $\vec w_\parallel$ and $\vec w_\bot$ are components of the vector $\vec w$ passing along the axis of symmetry of the disk and perpendicular.

$w_\parallel\approx w\sqrt{1-r^2/R^2}$

The fact that $R>>r$ leads to the $w_\parallel\approx w$: $$\vec L=I_\parallel \vec{w_\parallel}+I_\bot \vec{w_\bot} \approx I_\parallel \vec{w}$$

$$I_\parallel=mr^2/2$$ $$[\vec \Omega \vec L]=\Omega L \, sin(\alpha)$$ Thus: $$R=\frac{w^2r^2}{2g}tg(\alpha)$$

But in the book where i found this task the answer is $R=\frac{3w^2r^2}{2g}tg(\alpha)$ I think there is a mistake in the book. It is not clear where the coefficient three came from. Can anyone check?

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    $\begingroup$ Shouldn't the $R>>r$ fact be used somewhere? @Alexey $\endgroup$
    – Lelouch
    Jul 23, 2020 at 13:24
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    $\begingroup$ @Lelouch I think R >> r is used. I think the derviation is already using the approximation that is valid only when $\Omega$ is much smaller than $\omega$. In order for $\Omega$ to be be much smaller than $\omega$: R must be much larger than r. $\endgroup$
    – Cleonis
    Jul 23, 2020 at 14:01
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    $\begingroup$ @Lelouch I add additional point where R>>r is used $\endgroup$
    – Alex
    Jul 23, 2020 at 14:13
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    $\begingroup$ @Cleonis The rolling coin is initially in an inclined position towards the ground. There are three forсes affect coin: (1) the force of gravity = mg downward, (2) earth pressure force that = - mg upwards and (3) force of static friction towards the center of the circle R. The first two forces create angular momentum $M=[O_1 O_2, mg] - [O_1 O_3, mg]=[r, mg] $ $\endgroup$
    – Alex
    Jul 23, 2020 at 21:44
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    $\begingroup$ @Cleonis The setup described in the task is absolutely physical. The question is an interesting problem of dynamics of a rigid body, especially if solved (hard work!) without the restriction R >> r. In the simple case R >> r the solution of the book is not wrong, but unfortunately the question was unexpectedly closed. $\endgroup$
    – Pangloss
    Jul 24, 2020 at 9:58

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