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

[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: