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infinite_sodium
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- Homework Statement
- A coin of radius $b$ and mass $M$ rolls on a horizontal surface at speed $V$. If the plane of the coin is vertical the coin rolls in a straight line. If the plane is tilted, the path of the coin is a circle of radius $R$ . Find and expression for the tilt angle of the coin $\alpha$ in terms of the given quantities. (Because of the tilt of the coin the circle traced by its center of mass is slightly smaller than $R$ but you can ignore the difference)
- Relevant Equations
- $\mathbf{R} = R\mathbf{r} + b\sin \alpha \mathbf{z}$
$R\dot \theta = b\omega_s$
$\mathbf{\omega} = \omega_r \mathbf{r} + \omega_z \mathbf{z}$
$\mathbf{\omega} = (\omega_s \cos \alpha)\mathbf{\hat r} + (\omega_s \sin \alpha + \dot \theta)\mathbf{\hat z}$,
Because the object precesses, it has a vertical contribution to its angular momentum, whom their contributions to the angular momentum isn't vertical but rather rotates (thus needing the torque to do so), because it doesn't align with the principal axes. Thus why can I assume these are constant? The moment of inertia about the axis of 'spin' and the perpendicular axis are clearly not equal, and it seems unlikely that the contribution from precession somehow manages to be vertical. Image: (left is the rolling coin) https://i.sstatic.net/Y094CRx7.png
Use a cylindrical coordinate system with the origin at the bottom.
By using coordinate transformations and the fact that the moments of inertia about the main symmetry axis and the 'side' principal axis is $$\frac{1}{2} Mb^2$$ and $$\frac{1}{4} Mb^2$$ respectively, we have, denoting ,
$$\mathbf{L} = \mathbf {R} \times M\mathbf {\dot R} + \frac{Mb^2}{2}(\omega_r \cos \alpha + \omega_z \sin \alpha)(\cos \alpha \mathbf{r} + \sin \alpha \mathbf {z}) + \frac{Mb^2}{4}(-\omega_r \sin \alpha + \omega_z \cos \alpha)(-\sin \alpha \mathbf{r} + \cos \alpha \mathbf{z}) $$
which simplifies to
$$\mathbf{L} = -MbR\dot \theta \cos \alpha \mathbf {r} + \frac{1}{4} Mb^2 (2\omega_r \cos^2 \alpha + \omega_r \sin^2 \alpha + \omega_z \sin \alpha \cos \alpha)\mathbf{r} + k\mathbf{z}$$
(where $k\mathbf{z}$ contains the constant z-components who play no dynamical role). Thus $$\mathbf{\dot L} = -MbR \dot \theta ^ 2 \cos \alpha \mathbf{\hat \theta} + \frac{1}{4}Mb^2 \dot \theta (2\omega_r \cos^2 \alpha + \omega_r \sin^2 \alpha + \omega_z \sin \alpha \cos \alpha) \mathbf {\hat \theta}$$
Substituting $$\mathbf{\omega} = (\omega_s \cos \alpha)\mathbf{\hat r} + (\omega_s \sin \alpha + \dot \theta)\mathbf{\hat z}$$, we have,
$$\mathbf{\dot L} = [\frac{-3}{2} MRb \dot \theta^2 \cos \alpha - \frac{1}{4} \dot \theta^2 \sin \alpha \cos \alpha] \mathbf {\hat \theta}$$
The torque about the origin is $$\mathbf {\tau} = mgR\mathbf{\hat \theta} - mg(R + b\sin \alpha)\mathbf{\hat \theta} = -mgb\sin \alpha \mathbf{\hat \theta}$$. Thus
$$\frac{3}{2} R\dot \theta ^ 2 \cos \alpha + \frac{1}{4} b \dot \theta^2 \sin \alpha \cos \alpha = g\sin \alpha$$
Now, the solution from the back of the book states $$\tan{\alpha}=\frac{3V^2}{2Rg}=\frac{3R\dot\theta^2}{2g}$$. What went wrong? [Note: this can be obtained by simply neglecting the term ##\dot \theta## in my angular velocity equation.]
Use a cylindrical coordinate system with the origin at the bottom.
By using coordinate transformations and the fact that the moments of inertia about the main symmetry axis and the 'side' principal axis is $$\frac{1}{2} Mb^2$$ and $$\frac{1}{4} Mb^2$$ respectively, we have, denoting ,
$$\mathbf{L} = \mathbf {R} \times M\mathbf {\dot R} + \frac{Mb^2}{2}(\omega_r \cos \alpha + \omega_z \sin \alpha)(\cos \alpha \mathbf{r} + \sin \alpha \mathbf {z}) + \frac{Mb^2}{4}(-\omega_r \sin \alpha + \omega_z \cos \alpha)(-\sin \alpha \mathbf{r} + \cos \alpha \mathbf{z}) $$
which simplifies to
$$\mathbf{L} = -MbR\dot \theta \cos \alpha \mathbf {r} + \frac{1}{4} Mb^2 (2\omega_r \cos^2 \alpha + \omega_r \sin^2 \alpha + \omega_z \sin \alpha \cos \alpha)\mathbf{r} + k\mathbf{z}$$
(where $k\mathbf{z}$ contains the constant z-components who play no dynamical role). Thus $$\mathbf{\dot L} = -MbR \dot \theta ^ 2 \cos \alpha \mathbf{\hat \theta} + \frac{1}{4}Mb^2 \dot \theta (2\omega_r \cos^2 \alpha + \omega_r \sin^2 \alpha + \omega_z \sin \alpha \cos \alpha) \mathbf {\hat \theta}$$
Substituting $$\mathbf{\omega} = (\omega_s \cos \alpha)\mathbf{\hat r} + (\omega_s \sin \alpha + \dot \theta)\mathbf{\hat z}$$, we have,
$$\mathbf{\dot L} = [\frac{-3}{2} MRb \dot \theta^2 \cos \alpha - \frac{1}{4} \dot \theta^2 \sin \alpha \cos \alpha] \mathbf {\hat \theta}$$
The torque about the origin is $$\mathbf {\tau} = mgR\mathbf{\hat \theta} - mg(R + b\sin \alpha)\mathbf{\hat \theta} = -mgb\sin \alpha \mathbf{\hat \theta}$$. Thus
$$\frac{3}{2} R\dot \theta ^ 2 \cos \alpha + \frac{1}{4} b \dot \theta^2 \sin \alpha \cos \alpha = g\sin \alpha$$
Now, the solution from the back of the book states $$\tan{\alpha}=\frac{3V^2}{2Rg}=\frac{3R\dot\theta^2}{2g}$$. What went wrong? [Note: this can be obtained by simply neglecting the term ##\dot \theta## in my angular velocity equation.]
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