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

• guv
In summary, the given answer stating that graph III best represents the speed of the snow cylinder is based on faulty reasoning. The final speed of a cylinder rolling down an incline does not depend on its radius, only on the rotational inertia, which is the same for both the original cylinder and the snow-gathering cylinder. Therefore, the snow-gathering cylinder will have a lower acceleration due to the added mass of snow that needs to be accelerated. This is best represented by graph II, which shows a lower initial speed and a slower increase in speed compared to graph III.
erobz said:
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.

Last edited:
erobz said:
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.

kuruman said:
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.

MatinSAR
erobz said:
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:

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