Why can I neglect angular momentum due to precession here?

infinite_sodium
Messages
2
Reaction score
0
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.]
 
Last edited by a moderator:
Physics news on Phys.org
Welcome!
I believe that you can neglect angular momentum due to precession here because the coin is rolling in balance of forces and moments.
 
Can you elaborate? The principal axes are tilted to even if the angular velocity due to precession is constant, the angular momentum is not.
 
I've stayed out of this thread because I have trouble deciphering the meaning of some parts…
infinite_sodium said:
Because the object precesses, it has a vertical contribution to its angular momentum, whom their contributions
whom? their? I have no idea what is meant by that.
infinite_sodium said:
it seems unlikely that the contribution from precession somehow manages to be vertical
I do not see how the precession axis can be other than vertical. If you compare the state of the disc at one time with that at another, there is a horizontal displacement and a rotation about a vertical axis. If the rotation were about a non-vertical axis then the angle of the plane of disc to the vertical will change.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top