Speed of cylinder rolling along a horizontal surface gathering snow (Solved)

[kuruman]

Homework Statement

A cylinder with an initial radius $r_0$ rolls without slipping on a horizontal surface with speed $v_0$ when it encounters snow region, still horizontal. As it rolls, snow sticks onto the cylinder which makes its radius slowly increase. The amount of gathered snow is proportional to the distance the snow cylinder has traveled.

Find the linear speed of the cylinder as a function of distance.

Relevant Equations

$\tau=\frac{dL}{dt}$

This is a spin-off of a similar problem posted here in which the cylinder gathers snow as it rolls down an incline. I think one has to understand the snow-gathering process before attempting the more complicated case. A horizontal surface makes that easy but because it is a different problem, I opted to post my thoughts on this on a separate thread.

Here I will adopt a dynamic approach and consider that the angular momentum of the cylinder relative to the point of contact is conserved because the surface can exert no torque. For the reader's convenience, I reproduce derivations of relevant equations taken from the post quoted above.

Let $\beta=$ the rate at which the cylinder gathers snow. It is expedient to define dimensionless parameter $u=\dfrac{\beta s}{m_0}$ and write

Mass of cylinder: $m=m_0+\beta s=(1+u)m_0.$

Radius of cylinder
To find $r(s)$ consider adding to the cylinder mass $dm$ in the form of a cylindrical tube of thickness $dr$ and radius $r$. Then $$\begin{align} & \beta~ds=dm=\rho dV=\rho(2\pi r)(dr)L=\frac{m_0}{\pi r_0^2~L}(2\pi r)( dr)L=\frac{m_0}{ r_0^2}2rdr \nonumber \\
&\implies \frac{m_0}{ r_0^2} \int_{r_0}^{r}2rdr = \beta \int_0^s ds \implies r^2=r_0^2\left(1+\frac{\beta s}{m_0} \right)\implies r=(1+u)^{1/2}r_0.
\nonumber \end{align}$$Moment of inertia
The moment of inertia about the CM of the cylinder is $$I_{cm}=\frac{1}{2}mr^2=\frac{1}{2}(1+u)m_0~(1+u)r_0^2=\frac{1}{2}(1+u)^2m_0r_0^2.$$ Angular momentum
The angular momentum of the cylinder has two parts, spin about the CM and orbital of the CM
(a) About the CM $$L_{spin}=I_{cm}\omega =I_{cm}\frac{v}{r}=\frac{1}{2}(1+u)^2m_0r_0^2\frac{v}{(1+u)^{1/2}r_0}=\frac{1}{2}(1+u)^{3/2}m_0vr_0.$$(b) Of the CM (orbital) $$L_{orb}=mvr=(1+u)^{3/2}m_0vr_0.$$The total angular momentum is $$L=L_{spin}+L_{orb}=\frac{3}{2}(1+u)^{3/2}m_0vr_0.$$Angular momentum is conserved as discussed above. $$0=\frac{dL}{dt}=v\frac{dL}{ds}=v\frac{\beta}{m_0}\frac{dL}{ds}=\frac{v\beta}{m_0} \frac{d}{du}\left(\frac{3}{2}(1+u)^{3/2}m_0vr_0\right).$$Dropping the irrelevant multiplicative parameters gives the ODE $$ \frac{d}{du}\left((1+u)^{3/2}v\right)=0\implies (1+u)^{3/2}v=const.$$ Answer
Since the initial speed is $v_0$, we write $$v=\frac{v_0}{(1+u)^{3/2}}.$$A plot is shown below

Is mechanical energy conserved here?
The answer is no. A look at the speed verifies that. Here we have a deformable system in which one part of the system does work on the other. Any approach that relies on energy conservation must take that into account.

Edited to add the link to the other similar thread.

 

[jbriggs444]

Is mechanical energy conserved here?
The answer is no. A look at the speed verifies that. Here we have a deformable system in which one part of the system does work on the other. Any approach that relies on energy conservation must take that into account.

If we take care to specify the behavior of the layer of snow as it deforms and then conforms to the gathering roll, we may be able to preserve conservation of mechanical energy. I would suggest a few rules to advance that goal.

1. The layer of snow is not vertically compressed when it adheres to the roll. If there was 1 cm of snow on the ground, there will be a spiral layer 1 cm thick on the roll.

2. There is an imaginary mesh plane halfway through the thickness of the snow layer. The deformation of the snow does not cause any compression or elongation of the material along the length (or width) of this imaginary plane.

3. The local density of the snow does not change as a result of the deformation. We may imagine that snow from the more-compressed inner side of a layer on the roll is squeezed through the imaginary plane into the less-compressed outer side of the layer.

4. The flow associated with the migration of snow through the midle layer is laminar and there is zero viscosity.

If the equations and the graph show an energy loss, we should be able to trace that energy loss to an inelastic interaction. I see none in evidence.

Here I will adopt a dynamic approach and consider that the angular momentum of the cylinder relative to the point of contact is conserved because the surface can exert no torque.

Note that there is an unbalanced retarding torque from gravity. The trailing side of the snow cylinder is thicker than the leading side.

 

[kuruman]

If we agree that angular momentum is conserved because there is no torque, then for the horizontally rolling cylinder we also have to agree that mechanical energy is not conserved. That's because

• All the energy is in the form of kinetic energy.
• The kinetic energy can be written as $KE=\dfrac{L^2}{2I_p}$ where $I_p$ is the moment of inertia about the point of contact.
• $I_p$ increases as the cylinder gathers more snow.

 

[erobz]

• All the energy is in the form of kinetic energy.

The roll is gaining potential energy as a function of $s$?

 

[jbriggs444]

If we agree that angular momentum is conserved because there is no torque,

We do not agree that angular momentum is conserved. Because there is a torque.

 

[kuruman]

We do not agree that angular momentum is conserved. Because there is a torque.

Are you referring to this torque?

Note that there is an unbalanced retarding torque from gravity. The trailing side of the snow cylinder is thicker than the leading side.

Yes, that does not conserve angular momentum. The work done by gravity on the cylinder in that sense accounts for some of the loss of kinetic energy. There is also the work done by gravity when the CM is raised by $dr$ as has been mentioned earlier. However, I think that the leading contribution to the reduction of kinetic energy is the rotational impulse in the simple-minded model of bringing a ring of thickness $dr$ to speed after one revolution.

 

[jbriggs444]

Are you referring to this torque?

Yes, that is the one. Regardless of the thickness of the snow layer, that torque integrates to an identical angular momentum contribution by the time the cylinder has attained a given radius.

Cut the thickness in half and you double the number of rotations.

However, I think that the leading contribution to the reduction of kinetic energy is the rotational impulse in the simple-minded model of bringing a ring of thickness $dr$ to speed after one revolution.

If there were a dissipative force involved in that interaction, I could agree that a reduction in mechanical energy would take place. But I see no such dissipative force.

 

[haruspex]

Homework Statement: A cylinder with an initial radius $r_0$ rolls without slipping on a horizontal surface with speed $v_0$ when it encounters snow region, still horizontal. As it rolls, snow sticks onto the cylinder which makes its radius slowly increase. The amount of gathered snow is proportional to the distance the snow cylinder has traveled.

Find the linear speed of the cylinder as a function of distance.

Answer
Since the initial speed is $v_0$, we write $$v=\frac{v_0}{(1+u)^{3/2}}.$$

So it can't start rolling from rest on the snow?

If it starts as a perfect cylinder then there is an awkward step in the radius each revolution. If the snow layer picked up (undeformed) has thickness $d$ then I would take the initial "cylinder"'s section as an arithmetic spiral $r=r_0+\frac d{2\pi}\theta$ for $r_0<r<r_c$, where $r_c$ is the distance from the spiral's origin, 0, to the point Q where it meets the top of the plane of traversed snow, P.
Note that Q is the instantaneous centre of rotation, but O does not lie on the normal to P at Q. Thus the velocity of O is not parallel to the plane.

Next steps would be to find:
- the KE associated with that rotation
- the fall in height of the original "cylinder"'s mass centre
- the fall in height of the collected snow.
To simplify these, I suggest taking the original cylinder as having the same density as the snow but with an added point mass at its origin.

 

[kuruman]

If there were a dissipative force involved in that interaction, I could agree that a reduction in mechanical energy would take place. But I see no such dissipative force.

Here is why I think there is a dissipative force. I will use a physical situation that is analogous to the rolling cylinder to model it without the nitty-gritty details of asymmetrically positioned snow, gravity, etc.

Place a puck of mass $M$ and radius $R$ on a horizontal frictionless table. Give a purely angular impulse to the puck to set it spinning with angular speed $\omega.$ Draw a line on the table that is tangent to the circumference of the puck. Give a purely linear impulse to the puck that is parallel to the line and results in a speed $v=\omega ~R.$ Note that the puck is moving at constant speed while the tangent point to the line instantaneously at rest with respect to the table and the line.

Now imagine mass $m$ at rest at some point on the line. As the puck goes by, mass $m$ simply sticks to the puck at the tangent point that is instantaneously at rest. Is mechanical energy conserved? Well, if I were to solve this problem, I would conserve linear momentum and angular momentum about the location of mass $m$. Specifically, the linear momentum equation would be the usual $$Mv_0=(M+m)v_{\!f} \implies v_{\!f}=\frac{Mv_0}{M+m}.$$That's what one gets for a perfectly inelastic collision in which linear momentum is conserved but not kinetic energy. Similar considerations apply for the rotational collision.

The rolling and picking up snow action looks mechanical energy conserving to me. Nothing inelastic going on. We ditch one energy conserving constraint and pick up a different energy conserving constraint. No sudden jerks or vibrations to dissipate mechanical energy.

There may be no sudden jerks, but I think that focusing on them perhaps keeps you from seeing the bigger picture. We have mass of snow $dm$ at rest in an inertial frame O. We want it to be at rest in a non-inertial rotating frame O' that is also translating relative to O. I don't see how that can be accomplished without the expenditure of mechanical energy.

 

[kuruman]

So it can't start rolling from rest on the snow?

Should it? It's on a horizontal surface.

 

[jbriggs444]

Here is why I think there is a dissipative force. I will use a physical situation that is analogous to the rolling cylinder to model it without the nitty-gritty details of asymmetrically positioned snow, gravity, etc.

Place a puck of mass $M$ and radius $R$ on a horizontal frictionless table. Give a purely angular impulse to the puck to set it spinning with angular speed $\omega.$ Draw a line on the table that is tangent to the circumference of the puck. Give a purely linear impulse to the puck that is parallel to the line and results in a speed $v=\omega ~R.$ Note that the puck is moving at constant speed while the tangent point to the line instantaneously at rest with respect to the table and the line.

I think I am with you so far. We have a spinning puck with an instantaneous center of rotation on its rim.

Now imagine mass $m$ at rest at some point on the line. As the puck goes by, mass $m$ simply sticks to the puck at the tangent point that is instantaneously at rest.

RIght. So at the relevant instant, we expand our system boundary to encompass a new bit of mass that is at rest.

Is mechanical energy conserved?

Yes, absolutely. We imported a bit of mass that has zero mechanical energy. We had one rigid object with a certain amount of kinetic energy. Now we have a slightly different rigid object with the same amount of kinetic energy. We can integrate $\frac{1}{2}v^2 dm$ across both objects and get identical results. Mechanical energy is conserved prior to the interaction. Mechanical energy is conserved across the interaction. Mechanical energy is conserved subsequent to the interaction. Mechanical energy is conserved.

One assumes that the imported mass has zero gravitational potential energy. As indeed it will in the spinning puck scenario that you propose.

Any argument to the contrary must be flawed.

Edit to add:

There is one caveat to the argument that I am making. The added mass must be pointlike. Otherwise there is a rotational impulse associated with bringing a non-rotating mass $m$ into the system where it becomes a rotating mass element. If we conserve angular momentum, that impulse will be inelastic -- we are zeroing out the rotational kinetic energy associated with the dissimilar rotation rates.

In the context of the original problem, one can mitigate this caveat by making the snow layer thin. The lost mechanical energy from a thick layer scales (I think) quadratically with the snow thickness if we contemplate a fixed mass of added snow.

 

[kuruman]

Mechanical energy is conserved.

I totally agree and disavow my previous statements that mechanical energy is not conserved. I did my own mathematical calculations on the side and the scales of preconception fell off my eyes.

In the case of the rotating and translating puck we have two initially separate masses, one moving and one at rest, that come together and move as one thereafter. I falsely generalized that whenever this happens, mechanical energy cannot be conserved. I've had blind spots before, but this takes the cake. Thank you for helping me prove myself wrong by trying to prove you wrong.

 

[haruspex]

Should it? It's on a horizontal surface.

Ah - I didn't notice you'd switched it to horizontal, whereas the original thread was for a slope.

 

[kuruman]

Ah - I didn't notice you'd switched it to horizontal, whereas the original thread was for a slope.

Yes, I branched out to a separate thread because I changed the question.