# What will be the radius of the circle of the rolling coin? Gyroscope task [closed]

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: 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?

• Shouldn't the $R>>r$ fact be used somewhere? @Alexey Jul 23, 2020 at 13:24
• @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. Jul 23, 2020 at 14:01
• @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 $M=[O_1 O_2, mg] - [O_1 O_3, mg]=[r, mg]$