A cylinder of snow rolls down a hill gathering more snow -- calculate its speed

[erobz]

You are attempting to solve with calculus? Why?

You can write down a valid energy balance and read off the velocity for any position on the slope from that.

But the center of mass velocity is not collinear with the coordinate velocity. The mass is accreting, the center of mass has two orthogonal components of velocity.

This is in response to the complications @kuruman brings up in #21. I haven't locked down how to tackle those, but in trying to do so this became apparent along the way. If we are worrying about correcting angular kinetic energies, we should correct translational kinetic energy as well?

 

[erobz]

For the angular velocity at any instant in time if the wheel is not slipping, I can't see how the relationship is invalid?

$$ \hat{v} = \hat{r} \times \hat{\omega} = | \hat{r} | ~ |\hat{\omega}| \sin \theta = r \omega \sin \theta \hat s = v(s) \hat s $$

I think the angular kinetic energy is simply as it was:

$$KE_{rot.} = \frac{1}{4} m(s) v(s)^2 $$

and it was the translational energy that needed to be adjusted?

 

[kuruman]

I will be silent on this for a short period of time while I organize and LaTeX my thoughts along a parallel line. Please stay tuned.

 

[jbriggs444]

But the center of mass velocity is not collinear with the coordinate velocity. The mass is accreting, the center of mass has two orthogonal components of velocity.

This is in response to the complications @kuruman brings up in #21. I haven't locked down how to tackle those, but in trying to do so this became apparent along the way. If we are worrying about correcting angular kinetic energies, we should correct translational kinetic energy as well?

If we make the [factually incorrect] assumption that we are dealing with a rolling cylinder then splitting up the total energy into translation and rotation is trivial, regardless of the curvature of the track of the center of mass. We have an instantaneous center of rotation, a moment of inertia relative to that center and we know the energy that is embodied in that rotation. So we know the rotation rate. So we know the state of motion of every body-fixed point on the rolling shape.

Life is more difficult if we shift from the hypothetical cylinder to a more realistic archimedean spiral. The moment of inertia is trickier. The local curvature of the slope-facing surface of the spiral is more difficult to calculate. One would want that to compute the velocity with which the point of contact is advancing down the slope.

 

[erobz]

If we make the [factually incorrect] assumption that we are dealing with a rolling cylinder then splitting up the total energy into translation and rotation is trivial, regardless of the curvature of the track of the center of mass. We have an instantaneous center of rotation, a moment of inertia relative to that center and we know the energy that is embodied in that rotation. So we know the rotation rate. So we know the state of motion of every body-fixed point on the rolling shape.

So, it seems to me like #32 is OK under the [factually incorrect] simplification of a rolling cylinder.
I see I mixed up some things, #32 is not ok.

Life is more difficult if we shift from the hypothetical cylinder to a more realistic archimedean spiral. The moment of inertia is trickier. The local curvature of the slope-facing surface of the spiral is more difficult to calculate. One would want that to compute the velocity with which the point of contact is advancing down the slope.

We have to stop somewhere or (I think you will agree) there practically is no end to this.

 

[erobz]

For the angular velocity at any instant in time if the wheel is not slipping, I can't see how the relationship is invalid?

$$ \hat{v} = \hat{r} \times \hat{\omega} = | \hat{r} | ~ |\hat{\omega}| \sin \theta = r \omega \sin \theta \hat s = v(s) \hat s $$View attachment 329251I think the angular kinetic energy is simply as it was:

$$KE_{rot.} = \frac{1}{4} m(s) v(s)^2 $$

and it was the translational energy that needed to be adjusted?

Disregard. I think I have some wires crossed in my cross product...

I'm not sure how to fix it formally, but its only the component of $\hat v$ perpendicular to $\hat r$ and $\hat \omega$ which contributes to the angular kinetic energy, correct?

I still think the angular energy term is

$$ \frac{1}{4} m(s) v(s)^2 $$

But I can't seem to show what I want formally.

 

[kuruman]

For the angular velocity at any instant in time if the wheel is not slipping, I can't see how the relationship is invalid?

$$ \hat{v} = \hat{r} \times \hat{\omega} = | \hat{r} | ~ |\hat{\omega}| \sin \theta = r \omega \sin \theta \hat s = v(s) \hat s $$View attachment 329251I think the angular kinetic energy is simply as it was:

$$KE_{rot.} = \frac{1}{4} m(s) v(s)^2 $$

and it was the translational energy that needed to be adjusted?

See retraction in post #21. Sorry about the confusion.

 

[erobz]

See retraction in post #21. Sorry about the confusion.

Ok.

Well my final go would be:

$$ m_o g\left( h_o + r_o \cos (\alpha) \right) = \frac{1}{2} ( m_o + \beta s ) \left( 1 + \left( \frac{\beta}{2 \rho \pi L } \right)^2 \frac{1}{ r_o^2 + \frac{\beta}{\rho \pi L} s} \right) v(s)^2 + \frac{1}{4}( m_o + \beta s )v(s)^2 + ( m_o + \beta s ) g \left( h_o - \sin (\alpha)~ s + \sqrt{ r_o^2+ \frac{\beta}{\pi \rho L} s }~ \cos (\alpha) \right) $$

Its not pretty, but I'll plot $v(s)$ when I get a chance to see if it fixes the issues I found in the last model.

 

[kuruman]

I will be silent on this for a short period of time while I organize and LaTeX my thoughts along a parallel line. Please stay tuned.

Please see https://www.physicsforums.com/threa...-a-horizontal-surface-gathering-snow.1054059/

 

[erobz]

Ok.

Well my final go would be:

View attachment 329260

$$ m_o g\left( h_o + r_o \cos (\alpha) \right) = \frac{1}{2} ( m_o + \beta s ) \left( 1 + \left( \frac{\beta}{2 \rho \pi L } \right)^2 \frac{1}{ r_o^2 + \frac{\beta}{\rho \pi L} s} \right) v(s)^2 + \frac{1}{4}( m_o + \beta s )v(s)^2 + ( m_o + \beta s ) g \left( h_o - \sin (\alpha)~ s + \sqrt{ r_o^2+ \frac{\beta}{\pi \rho L} s }~ \cos (\alpha) \right) $$

Its not pretty, but I'll plot $v(s)$ when I get a chance to see if it fixes the issues I found in the last model.

The model doesn't seem to correct the issue with $\beta$ having an upper limit ( in fact it shrinks it). I can't figure out what it means.

Here they are plotted: